Why does this VBA-generated QR-Code stutter? (barcode-vba-macro-only) - vba

Context
I am using barcode-vba-macro-only (mentioned in this SO post) in MS Excel 2010 to generate a QR code.
(The bar code will be used to facilitate paying a bill using Girocode, but that is not important here, except to say I need to structure the input exactly the way shown below.)
The problem
The VBA macro creates great QR-Codes, but somehow, when given certain input, the output (encoded in the QR code) "stutters", i.e. repeats part of the text.
E.g., when given this input:
BCD
001
1
SCT
SOLADES1HDB
Recipient First and Last Name
DE86672500200000123456
EUR123.45
it produces this output:
which oddly repeats part of the content:
BCD
001
1
SCT
SOLADES1HDB
Recipient First and Last Name
DE
Recipient First and Last Name
DE86672500200000123456
EUR123.45
(Note the DE and the line Recipient First and Last Name which appear twice.)
What I want
A working, free/GPL solution in Excel to generate such codes ;-) ... for example by understanding why this happens, and fixing the VBA code.
What I have tried (Update 1)
I have played around with different inputs and found that just adding some extra "AAA" to the end of the long number solves the stuttering... so I am intrigued what causes this.
I forked to code on GitHub, added some code comments and translated a few of the existing (Czech) comments
Through some debugging, I found that the implementation messes up the starting position of different encodings (which it stores in array eb): after encoding the "Recipient First and Last Name" including newline and "DE" as "Byte", it probably tries to switch to "Decimal" or "Alphanum" encoding (only 3.33 or 5.5 bit per character instead of 8)... but then falls back to encoding in "Byte" format and thereby gets the starting position wrong.
The code
You can download my test XLSM file here, and access my improved code file on GitHub.
I think the issue is probably in the core function shown below, in the section where the array eb() is filled.
Function qr_gen(ptext As String, poptions As String) As String
Dim encoded1() As Byte ' byte mode (ASCII) all max 3200 bytes
Dim encix1%
Dim ecx_cnt(3) As Integer
Dim ecx_pos(3) As Integer
Dim ecx_poc(3) As Integer
Dim eb(1 To 20, 1 To 4) As Integer 'store how many characters should be in which ECI mode. This is a list of rows, each row corresponding a the next batch of characters with a different ECI mode.
' eb(i, 1) - ECI mode (1 = numeric, 2 = alphanumeric, 3 = byte)
' eb(i, 2) - last character in previous row
' eb(i, 3) - number of characters in THIS row
' eb(i, 4) - number of bits for THIS row
Dim ascimatrix$, mode$, err$
Dim ecl%, r%, c%, mask%, utf8%, ebcnt%
Dim i&, j&, k&, m&
Dim ch%, s%, siz%
Dim x As Boolean
Dim qrarr() As Byte ' final matrix
Dim qrpos As Integer
Dim qrp(15) As Integer ' 1:version,2:size,3:ccs,4:ccb,5:totby,6-12:syncs(7),13-15:versinfo(3)
Dim qrsync1(1 To 8) As Byte
Dim qrsync2(1 To 5) As Byte
ascimatrix = ""
err = ""
mode = "M"
i = InStr(poptions, "mode=")
If i > 0 Then mode = Mid(poptions, i + 5, 1)
' M=0,L=1,H=2,Q=3
ecl = InStr("MLHQ", mode) - 1
If ecl < 0 Then mode = "M": ecl = 0
If ptext = "" Then
err = "Not data"
Exit Function
End If
For i = 1 To 3
ecx_pos(i) = 0
ecx_cnt(i) = 0
ecx_poc(i) = 0
Next i
ebcnt = 1
utf8 = 0
For i = 1 To Len(ptext) + 1
' Decide how many bytes this character has
If i > Len(ptext) Then
k = -5 ' End of text --> skip several code sections
Else ' need to parse character i of ptext and decide how many bytes it has
k = AscL(Mid(ptext, i, 1))
If k >= &H1FFFFF Then ' FFFF - 1FFFFFFF
m = 4
k = -1
ElseIf k >= &H7FF Then ' 7FF-FFFF 3 bytes
m = 3
k = -1
ElseIf k >= 128 Then
m = 2
k = -1
Else ' normal 7bit ASCII character, so it is worth it to check if it belong to the Numeric or Alphanumeric subsets defined in ECI (array qralnum)
m = 1
k = InStr(qralnum, Mid(ptext, i, 1)) - 1
End If
End If
' Depending on k and a lot of other things, increase ebcnt
If (k < 0) Then ' Treat mult-byte case or exit? (bude byte nebo konec)
If ecx_cnt(1) >= 9 Or (k = -5 And ecx_cnt(1) = ecx_cnt(3)) Then ' Until now it was possible numeric??? (Az dosud bylo mozno pouzitelne numeric)
If (ecx_cnt(2) - ecx_cnt(1)) >= 8 Or (ecx_cnt(3) = ecx_cnt(2)) Then ' pred num je i pouzitelny alnum
If (ecx_cnt(3) > ecx_cnt(2)) Then ' Jeste pred alnum bylo byte
eb(ebcnt, 1) = 3 ' Typ byte
eb(ebcnt, 2) = ecx_pos(3) ' Position pozice
eb(ebcnt, 3) = ecx_cnt(3) - ecx_cnt(2) ' delka
ebcnt = ebcnt + 1
ecx_poc(3) = ecx_poc(3) + 1
End If
eb(ebcnt, 1) = 2 ' Typ alnum
eb(ebcnt, 2) = ecx_pos(2)
eb(ebcnt, 3) = ecx_cnt(2) - ecx_cnt(1) ' delka
ebcnt = ebcnt + 1
ecx_poc(2) = ecx_poc(2) + 1
ecx_cnt(2) = 0
ElseIf ecx_cnt(3) > ecx_cnt(1) Then ' byly bytes pred numeric
eb(ebcnt, 1) = 3 ' Typ byte
eb(ebcnt, 2) = ecx_pos(3) ' Position pozice
eb(ebcnt, 3) = ecx_cnt(3) - ecx_cnt(1) ' delka
ebcnt = ebcnt + 1
ecx_poc(3) = ecx_poc(3) + 1
End If
ElseIf (ecx_cnt(2) >= 8) Or (k = -5 And ecx_cnt(2) = ecx_cnt(3)) Then ' Az dosud bylo mozno pouzitelne alnum
If (ecx_cnt(3) > ecx_cnt(2)) Then ' Jeste pred alnum bylo byte
eb(ebcnt, 1) = 3 ' Typ byte
eb(ebcnt, 2) = ecx_pos(3) ' Position pozice
eb(ebcnt, 3) = ecx_cnt(3) - ecx_cnt(2) ' delka
ebcnt = ebcnt + 1
ecx_poc(3) = ecx_poc(3) + 1
End If
eb(ebcnt, 1) = 2 ' Typ alnum
eb(ebcnt, 2) = ecx_pos(2)
eb(ebcnt, 3) = ecx_cnt(2) ' delka
ebcnt = ebcnt + 1
ecx_poc(2) = ecx_poc(2) + 1
ecx_cnt(3) = 0
ecx_cnt(2) = 0 ' vse zpracovano
ElseIf (k = -5 And ecx_cnt(3) > 0) Then ' konec ale mam co ulozit
eb(ebcnt, 1) = 3 ' Typ byte
eb(ebcnt, 2) = ecx_pos(3) ' Position pozice
eb(ebcnt, 3) = ecx_cnt(3) ' delka
ebcnt = ebcnt + 1
ecx_poc(3) = ecx_poc(3) + 1
End If
End If
If k = -5 Then Exit For
If (k >= 0) Then ' We can alphanumeric? (Muzeme alnum)
If (k >= 10 And ecx_cnt(1) >= 12) Then ' Until now it was perhaps numeric (Az dosud bylo mozno num)
If (ecx_cnt(2) - ecx_cnt(1)) >= 8 Or (ecx_cnt(3) = ecx_cnt(2)) Then ' There is also an alphanumeric which is worth it(Je tam i alnum ktery stoji za to)
If (ecx_cnt(3) > ecx_cnt(2)) Then ' Even before it was alnum byte (Jeste pred alnum bylo byte)
eb(ebcnt, 1) = 3 ' Typ byte
eb(ebcnt, 2) = ecx_pos(3) ' Position (pozice)
eb(ebcnt, 3) = ecx_cnt(3) - ecx_cnt(2) ' length (delka)
ebcnt = ebcnt + 1
ecx_poc(3) = ecx_poc(3) + 1
End If
eb(ebcnt, 1) = 2 ' Typ alnum
eb(ebcnt, 2) = ecx_pos(2)
eb(ebcnt, 3) = ecx_cnt(2) - ecx_cnt(1) ' length (delka)
ebcnt = ebcnt + 1
ecx_poc(2) = ecx_poc(2) + 1
ecx_cnt(2) = 0 ' processed everything (vse zpracovano)
ElseIf (ecx_cnt(3) > ecx_cnt(1)) Then ' Previous Num is byte (Pred Num je byte)
eb(ebcnt, 1) = 3 ' Typ byte
eb(ebcnt, 2) = ecx_pos(3) ' Position (pozice)
eb(ebcnt, 3) = ecx_cnt(3) - ecx_cnt(1) ' length (delka)
ebcnt = ebcnt + 1
ecx_poc(3) = ecx_poc(3) + 1
End If
eb(ebcnt, 1) = 1 ' Typ numerix
eb(ebcnt, 2) = ecx_pos(1)
eb(ebcnt, 3) = ecx_cnt(1) ' length (delka)
ebcnt = ebcnt + 1
ecx_poc(1) = ecx_poc(1) + 1
ecx_cnt(1) = 0
ecx_cnt(2) = 0
ecx_cnt(3) = 0 ' processed everything (vse zpracovano)
End If
If ecx_cnt(2) = 0 Then ecx_pos(2) = i
ecx_cnt(2) = ecx_cnt(2) + 1
Else ' possible alnum (mozno alnum)
ecx_cnt(2) = 0
End If
If k >= 0 And k < 10 Then ' Can be numeric (muze byt numeric)
If ecx_cnt(1) = 0 Then ecx_pos(1) = i
ecx_cnt(1) = ecx_cnt(1) + 1
Else
ecx_cnt(1) = 0
End If
If ecx_cnt(3) = 0 Then ecx_pos(3) = i
ecx_cnt(3) = ecx_cnt(3) + m
utf8 = utf8 + m
If ebcnt >= 16 Then ' We have already taken 3 other blocks of bits (Uz by se mi tri dalsi bloky stejne nevesli)
ecx_cnt(1) = 0
ecx_cnt(2) = 0
End If
Debug.Print "Character:'" & Mid(ptext, i, 1) & "'(" & k & _
") ebn=" & ecx_pos(1) & "." & ecx_cnt(1) & _
" eba=" & ecx_pos(2) & "." & ecx_cnt(2) & _
" ebb=" & ecx_pos(3) & "." & ecx_cnt(3)
Next
ebcnt = ebcnt - 1 ' ebcnt now has its final value
Debug.Print ("ebcnt=" & ebcnt)
c = 0
For i = 1 To ebcnt
Select Case eb(i, 1)
Case 1: eb(i, 4) = Int(eb(i, 3) / 3) * 10 + (eb(i, 3) Mod 3) * 3 + IIf((eb(i, 3) Mod 3) > 0, 1, 0)
Case 2: eb(i, 4) = Int(eb(i, 3) / 2) * 11 + (eb(i, 3) Mod 2) * 6
Case 3: eb(i, 4) = eb(i, 3) * 8
End Select
c = c + eb(i, 4)
Next i
Debug.Print ("c=" & c)
' UTF-8 is default not need ECI value - zxing cannot recognize
' Call qr_params(i * 8 + utf8,mode,qrp)
Call qr_params(c, ecl, qrp, ecx_poc)
If qrp(1) <= 0 Then
err = "Too long"
Exit Function
End If
siz = qrp(2)
Debug.Print "ver:" & qrp(1) & mode & " size " & siz & " ecc:" & qrp(3) & "x" & qrp(4) & " d:" & (qrp(5) - qrp(3) * qrp(4))
'MsgBox "ver:" & qrp(1) & mode & " size " & siz & " ecc:" & qrp(3) & "x" & qrp(4) & " d:" & (qrp(5) - qrp(3) * qrp(4))
ReDim encoded1(qrp(5) + 2)
' Table 3 — Number of bits in character count indicator for QR Code 2005:
' mode indicator (1=num,2=AlNum,4=Byte,8=kanji,ECI=7)
' mode: Byte Alphanum Numeric Kanji
' ver 1..9 : 8 9 10 8
' 10..26 : 16 11 12 10
' 27..40 : 16 13 14 12
' UTF-8 is default not need ECI value - zxing cannot recognize
' if utf8 > 0 Then
' k = &H700 + 26 ' UTF-8=26 ; Win1250 = 21; 8859-2 = 4 viz http://strokescribe.com/en/ECI.html
' bb_putbits(encoded1,encix1,k,12)
' End If
encix1 = 0
For i = 1 To ebcnt
Select Case eb(i, 1)
Case 1: c = IIf(qrp(1) < 10, 10, IIf(qrp(1) < 27, 12, 14)): k = 2 ^ c + eb(i, 3) ' encoding mode "Numeric"
Case 2: c = IIf(qrp(1) < 10, 9, IIf(qrp(1) < 27, 11, 13)): k = 2 * (2 ^ c) + eb(i, 3) ' encoding mode "alphanum
Case 3: c = IIf(qrp(1) < 10, 8, 16): k = 4 * (2 ^ c) + eb(i, 3) ' encoding mode "Byte"
End Select
Call bb_putbits(encoded1, encix1, k, c + 4)
Debug.Print "ver:" & qrp(1) & mode & " size " & siz & " ecc:" & qrp(3) & "x" & qrp(4) & " d:" & (qrp(5) - qrp(3) * qrp(4))
j = 0 ' count characters that have been output in THIS row eb(i,...)
m = eb(i, 2) 'Start (after) last character of input from previous row
r = 0
While j < eb(i, 3)
k = AscL(Mid(ptext, m, 1))
m = m + 1
If eb(i, 1) = 1 Then
' parse numeric input - output 3 decimal digits into 10 bit
r = (r * 10) + ((k - &H30) Mod 10)
If (j Mod 3) = 2 Then
Call bb_putbits(encoded1, encix1, r, 10)
r = 0
End If
j = j + 1
ElseIf eb(i, 1) = 2 Then
' parse alphanumeric input - output 2 alphanumeric characters into 11 bit
r = (r * 45) + ((InStr(qralnum, Chr(k)) - 1) Mod 45)
If (j Mod 2) = 1 Then
Call bb_putbits(encoded1, encix1, r, 11)
r = 0
End If
j = j + 1
Else
' Okay, byte mode: coding according to Chapter "6.4.2 Extended Channel Interpretation (ECI) mode" of ISOIEC 18004_2006Cor 1_2009.pdf
If k > &H1FFFFF Then ' FFFF - 1FFFFFFF
ch = &HF0 + Int(k / &H40000) Mod 8
Call bb_putbits(encoded1, encix1, ch, 8)
ch = 128 + Int(k / &H1000) Mod 64
Call bb_putbits(encoded1, encix1, ch, 8)
ch = 128 + Int(k / 64) Mod 64
Call bb_putbits(encoded1, encix1, ch, 8)
ch = 128 + k Mod 64
Call bb_putbits(encoded1, encix1, ch, 8)
j = j + 4
ElseIf k > &H7FF Then ' 7FF-FFFF 3 bytes
ch = &HE0 + Int(k / &H1000) Mod 16
Call bb_putbits(encoded1, encix1, ch, 8)
ch = 128 + Int(k / 64) Mod 64
Call bb_putbits(encoded1, encix1, ch, 8)
ch = 128 + k Mod 64
Call bb_putbits(encoded1, encix1, ch, 8)
j = j + 3
ElseIf k > &H7F Then ' 2 bytes
ch = &HC0 + Int(k / 64) Mod 32
Call bb_putbits(encoded1, encix1, ch, 8)
ch = 128 + k Mod 64
Call bb_putbits(encoded1, encix1, ch, 8)
j = j + 2
Else
ch = k Mod 256
Call bb_putbits(encoded1, encix1, ch, 8)
j = j + 1
End If
End If
Wend
Select Case eb(i, 1)
Case 1:
If (j Mod 3) = 1 Then
Call bb_putbits(encoded1, encix1, r, 4)
ElseIf (j Mod 3) = 2 Then
Call bb_putbits(encoded1, encix1, r, 7)
End If
Case 2:
If (j Mod 2) = 1 Then Call bb_putbits(encoded1, encix1, r, 6)
End Select
'MsgBox "blk[" & i & "] t:" & eb(i,1) & "from " & eb(i,2) & " to " & eb(i,3) + eb(i,2) & " bits=" & encix1
Next i
Call bb_putbits(encoded1, encix1, 0, 4) ' end of chain
If (encix1 Mod 8) <> 0 Then ' round to byte
Call bb_putbits(encoded1, encix1, 0, 8 - (encix1 Mod 8))
End If
' padding
i = (qrp(5) - qrp(3) * qrp(4)) * 8
If encix1 > i Then
err = "Encode length error"
Exit Function
End If
' padding 0xEC,0x11,0xEC,0x11...
Do While encix1 < i
Call bb_putbits(encoded1, encix1, &HEC11, 16)
Loop
' doplnime ECC
i = qrp(3) * qrp(4) 'ppoly, pmemptr , psize , plen , pblocks
Call qr_rs(&H11D, encoded1, qrp(5) - i, i, qrp(4))
'Call arr2hexstr(encoded1)
encix1 = qrp(5)
' Pole pro vystup
ReDim qrarr(0)
ReDim qrarr(1, qrp(2) * 24& + 24&) ' 24 bytes per row
qrarr(0, 0) = 0
ch = 0
Call bb_putbits(qrsync1, ch, Array(&HFE, &H82, &HBA, &HBA, &HBA, &H82, &HFE, 0), 64)
Call qr_mask(qrarr, qrsync1, 8, 0, 0) ' sync UL
Call qr_mask(qrarr, 0, 8, 8, 0) ' fmtinfo UL under - bity 14..9 SYNC 8
Call qr_mask(qrarr, qrsync1, 8, 0, siz - 7) ' sync UR ( o bit vlevo )
Call qr_mask(qrarr, 0, 8, 8, siz - 8) ' fmtinfo UR - bity 7..0
Call qr_mask(qrarr, qrsync1, 8, siz - 7, 0) ' sync DL (zasahuje i do quiet zony)
Call qr_mask(qrarr, 0, 8, siz - 8, 0) ' blank nad DL
For i = 0 To 6
x = qr_bit(qrarr, -1, i, 8, 0) ' svisle fmtinfo UL - bity 0..5 SYNC 6,7
x = qr_bit(qrarr, -1, i, siz - 8, 0) ' svisly blank pred UR
x = qr_bit(qrarr, -1, siz - 1 - i, 8, 0) ' svisle fmtinfo DL - bity 14..8
Next
x = qr_bit(qrarr, -1, 7, 8, 0) ' svisle fmtinfo UL - bity 0..5 SYNC 6,7
x = qr_bit(qrarr, -1, 7, siz - 8, 0) ' svisly blank pred UR
x = qr_bit(qrarr, -1, 8, 8, 0) ' svisle fmtinfo UL - bity 0..5 SYNC 6,7
x = qr_bit(qrarr, -1, siz - 8, 8, 1) ' black dot DL
If qrp(13) <> 0 Or qrp(14) <> 0 Then ' versioninfo
' UR ver 0 1 2;3 4 5;...;15 16 17
' LL ver 0 3 6 9 12 15;1 4 7 10 13 16; 2 5 8 11 14 17
k = 65536 * qrp(13) + 256& * qrp(14) + 1& * qrp(15)
c = 0: r = 0
For i = 0 To 17
ch = k Mod 2
x = qr_bit(qrarr, -1, r, siz - 11 + c, ch) ' UR ver
x = qr_bit(qrarr, -1, siz - 11 + c, r, ch) ' DL ver
c = c + 1
If c > 2 Then c = 0: r = r + 1
k = Int(k / 2&)
Next
End If
c = 1
For i = 8 To siz - 9 ' sync lines
x = qr_bit(qrarr, -1, i, 6, c) ' vertical on column 6
x = qr_bit(qrarr, -1, 6, i, c) ' horizontal on row 6
c = (c + 1) Mod 2
Next
' other syncs
ch = 0
Call bb_putbits(qrsync2, ch, Array(&H1F, &H11, &H15, &H11, &H1F), 40)
ch = 6
Do While ch > 0 And qrp(6 + ch) = 0
ch = ch - 1
Loop
If ch > 0 Then
For c = 0 To ch
For r = 0 To ch
' corners
If (c <> 0 Or r <> 0) And _
(c <> ch Or r <> 0) And _
(c <> 0 Or r <> ch) Then
Call qr_mask(qrarr, qrsync2, 5, qrp(r + 6) - 2, qrp(c + 6) - 2)
End If
Next r
Next c
End If
' qr_fill(parr as Variant, psiz%, pb as Variant, pblocks%, pdlen%, ptlen%)
' vyplni pole parr (psiz x 24 bytes) z pole pb pdlen = pocet dbytes, pblocks = bloku, ptlen celkem
Call qr_fill(qrarr, siz, encoded1, qrp(4), qrp(5) - qrp(3) * qrp(4), qrp(5))
mask = 8 ' auto
i = InStr(poptions, "mask=")
If i > 0 Then mask = val(Mid(poptions, i + 5, 1))
If mask < 0 Or mask > 7 Then
j = -1
For mask = 0 To 7
GoSub addmm
i = qr_xormask(qrarr, siz, mask, False)
' MsgBox "score mask " & mask & " is " & i
If i < j Or j = -1 Then j = i: s = mask
Next mask
mask = s
' MsgBox "best is " & mask & " with score " & j
End If
GoSub addmm
i = qr_xormask(qrarr, siz, mask, True)
ascimatrix = ""
For r = 0 To siz Step 2
s = 0
For c = 0 To siz Step 2
If (c Mod 8) = 0 Then
ch = qrarr(1, s + 24 * r)
If r < siz Then i = qrarr(1, s + 24 * (r + 1)) Else i = 0
s = s + 1
End If
ascimatrix = ascimatrix _
& Chr(97 + (ch Mod 4) + 4 * (i Mod 4))
ch = Int(ch / 4)
i = Int(i / 4)
Next
ascimatrix = ascimatrix & vbNewLine
Next r
ReDim qrarr(0)
qr_gen = ascimatrix
Exit Function
addmm:
k = ecl * 8 + mask
' poly: 101 0011 0111
Call qr_bch_calc(k, &H537)
'MsgBox "mask :" & hex(k,3) & " " & hex(k xor &H5412,3)
k = k Xor &H5412 ' micro xor &H4445
r = 0
c = siz - 1
For i = 0 To 14
ch = k Mod 2
k = Int(k / 2)
x = qr_bit(qrarr, -1, r, 8, ch) ' svisle fmtinfo UL - bity 0..5 SYNC 6,7 .... 8..14 dole
x = qr_bit(qrarr, -1, 8, c, ch) ' vodorovne odzadu 0..7 ............ 8,SYNC,9..14
c = c - 1
r = r + 1
If i = 7 Then c = 7: r = siz - 7
If i = 5 Then r = r + 1 ' preskoc sync vodorvny
If i = 8 Then c = c - 1 ' preskoc sync svisly
Next
Return
End Function ' qr_gen

Why this happens
Through some debugging, I found that the original implementation messes up the starting position of different encodings (which it stores in array eb): after encoding the "Recipient First and Last Name" including newline and "DE" as "Byte", it probably tries to switch to "Decimal" or "Alphanum" encoding (only 3.33 or 5.5 bit per character instead of 8)... but then falls back to encoding in "Byte" format and thereby gets the starting position wrong.
The solution
I have now added some error checking to the code which manually removes the stuttering.
You can find my improved code on Github, see in particular barcody.bas.
The key addition is this part:
i = 1
While i < (ebcnt - 1)
If eb(i, 2) + eb(i, 3) <> eb(i + 1, 2) Then
' oops, this should not happen. First document it:
Debug.Print ("eb() rows " & i & " and " & i + 1 & " are overlapping!")
' Now Lets see if we can fix it:
wasfixed = False
For k = i To 1 Step -1
If eb(k, 2) = eb(i + 1, 2) Then
' okay, the row k to i seem to be contained in i+1 and following. Delete k to i ...
For j = k To ebcnt - (i - k + 1) ' ... by copying upwards all later rows...
eb(j, 1) = eb(j + (i - k + 1), 1)
eb(j, 2) = eb(j + (i - k + 1), 2)
eb(j, 3) = eb(j + (i - k + 1), 3)
eb(j, 4) = eb(j + (i - k + 1), 4)
Next j
ebcnt = ebcnt - (i - k + 1) ' and correcting the total rowcount
wasfixed = True
Exit For
End If
Next k
If Not (wasfixed) Then
MsgBox ("The input text analysis failed - entering debug mode...")
Debug.Assert False
End If
End If
i = i + 1
Wend

I have noticed this same issue with certain character being a trigger for this problem when it comes after something. In your case, it looks like something after "DE" Since I didn't write the code I haven't thoroughly search in the code why this would trigger a repeat but I'm guessing that some of the hex conversion in the function causes this problem. In my case, I avoided the issue by having a space in front of the entire string being input to the generator. For some reasons, having the space at the beginning somehow prevent the triggering of the repeat. The reader program that reads the barcode in my case would remove the space from the string anyway so it doesn't matter.
I don't know whether it's a problem for your application, but try putting a blank space (" ") in front of the problematic string (DE86672500200000123456 )and see whether that works.
BCD
001
1
SCT
SOLADES1HDB
Recipient First and Last Name
DE86672500200000123456
EUR123.45

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Format of long strings in VBA IDE

I'm writing a macro which takes a VBA code module, exports it, reads the exported file into a 64 bit string and saves this as a constant in another VBA module (by using VBComponent.codeModule.InsertLines). For some reason, my 64 bit strings have linebreaks every 72 characters in the IDE (see image)
For some reason the strings generated from exporting files do this, but not a string generated with String(500,"a"). I wonder whether anyone can provide any insight into this behaviour, I would like the entire expression to be on a single line.
Sub test() 'in a module named "testModule"
exampleString = String(500, "a")
ThisWorkbook.VBProject.VBComponents.Item("testModule").codeModule.InsertLines _
2, "Const str As String = """ & exampleString & """"
End Sub
gives
Sub test() 'in a module named "testModule"
Const str As String = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
exampleString = String(500, "a")
ThisWorkbook.VBProject.VBComponents.Item("testModule").codeModule.InsertLines _
2, "Const str As String = """ & exampleString & """"
End Sub
while
exampleString = Base64EncodedModule()
results in that bizarre behaviour in the image. Maybe there's something really obvious I'm missing, but the base64 string looks exactly like something I could type, so I don't know why it's being split over multiple lines when I write it to the codeModule programmatically.
A minimum bit of code
Function Base64EncodedModule() As String
'export module
Dim exportPath As String: exportPath = Environ("temp") & "\" & "tempModule.bas"
ThisWorkbook.VBProject.VBComponents("Module1").Export exportPath
'read file as bytes
Dim inStream As Object: Set inStream = CreateObject("ADODB.Stream")
inStream.Open
inStream.Type = 1 'Binary file
inStream.LoadFromFile exportPath
'encode as base 64
Dim objXML As MSXML2.DOMDocument
Dim objNode As MSXML2.IXMLDOMElement
Set objXML = New MSXML2.DOMDocument
Set objNode = objXML.createElement("b64")
objNode.DataType = "bin.base64"
objNode.nodeTypedValue = inStream.Read() 'read bytes from file
Base64EncodedModule = objNode.text
Kill exportPath 'remove temp file
End Function
The string is base64 encoded and can contain line separators every n blocks of 4 chars depending on the encoding algorithm. It seems to be the case here.
So either remove the line breaks with Replace or use a decoding/encoding algorithm without line break like this one:
'
' Base 64 encoding '
'
Public Sub FromBase64(Text As String, Out() As Byte)
Dim b64(0 To 255) As Byte, str() As Byte, i&, j&, v&, b0&, b1&, b2&, b3&
Out = ""
If Len(Text) Then Else Exit Sub
str = " ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
For i = 2 To UBound(str) Step 2
b64(str(i)) = i \ 2
Next
ReDim Out(0 To ((Len(Text) + 3) \ 4) * 3 - 1)
str = Text & String$(2, 0)
For i = 0 To UBound(str) - 7 Step 2
b0 = b64(str(i))
If b0 Then
b1 = b64(str(i + 2))
b2 = b64(str(i + 4))
b3 = b64(str(i + 6))
v = b0 * 262144 + b1 * 4096& + b2 * 64& + b3 - 266305
Out(j) = v \ 65536
Out(j + 1) = (v \ 256&) Mod 256
Out(j + 2) = v Mod 256
j = j + 3
i = i + 6
End If
Next
If b2 = 0 Then
Out(j - 3) = (v + 65) \ 65536
j = j - 2
ElseIf b3 = 0 Then
Out(j - 3) = (v + 1) \ 65536
Out(j - 2) = ((v + 1) \ 256&) Mod 256
j = j - 1
End If
ReDim Preserve Out(j - 1)
End Sub
Public Function ToBase64(data() As Byte) As String
Dim b64(0 To 63) As Byte, str() As Byte, i&, j&, v&, n&
n = UBound(data) - LBound(data) + 1
If n Then Else Exit Function
str = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
For i = 0 To 127 Step 2
b64(i \ 2) = str(i)
Next
ReDim str(0 To ((n + 2) \ 3) * 8 - 1)
For i = LBound(data) To UBound(data) - (n Mod 3) Step 3
v = data(i) * 65536 + data(i + 1) * 256& + data(i + 2)
str(j) = b64(v \ 262144)
str(j + 2) = b64((v \ 4096) Mod 64)
str(j + 4) = b64((v \ 64) Mod 64)
str(j + 6) = b64(v Mod 64)
j = j + 8
Next
If n Mod 3 = 2 Then
v = data(n - 2) * 256& + data(n - 1)
str(j) = b64((v \ 1024&) Mod 64)
str(j + 2) = b64((v \ 16) Mod 64)
str(j + 4) = b64((v * 4) Mod 64)
str(j + 6) = 61 ' = '
ElseIf n Mod 3 = 1 Then
v = data(n - 1)
str(j) = b64(v \ 4 Mod 64)
str(j + 2) = b64(v * 16 Mod 64)
str(j + 4) = 61 ' = '
str(j + 6) = 61 ' = '
End If
ToBase64 = str
End Function

Damerau-Levenshtein algorithm isn't working on short strings

I have a for loop that takes a user's input and one of the keys in my dictionary and passes them to a Damerau-Levenshtein function and based on the distance, overwrites the user's input with the dictionary key (The for loop is to cycle through each dictionary key). This works fine enough for strings larger than three characters, but if the string is three or fewer characters the algorithm returns with the wrong key. Here's the for loop:
1950 For j = 0 To dict.Count - 1
1960 distance = DamerauLevenshtein(SplitStr(i), dict.Keys(j))
1970 'MsgBox dict.Keys(j) & vbCrLf & distance ' used for debugging
1980 If distance < 4 Then
1990 If distance < leastDist Then
2000 leastDist = distance
2010 SplitStr(i) = dict.Keys(j)
2020 End If
2030 End If
2040 Next
2050 MsgBox "The distance is: " & leastDist & vbCrLf & "The entered text was " & tempStr & vbCrLf & "The replaced word is " & SplitStr(i)
SplitStr(i) holds the user's input, which comes from a split function. I arbitrarily picked 4 for a good distance
I stole the algorithm from a bytes.com forum post. Algorithm below:
Function DamerauLevenshtein(str1, str2, Optional intSize = 256)
Dim intTotalLen, arrDistance, intLen1, intLen2, i, j, arrStr1, arrStr2, arrDA, intMini
Dim intDB, intI1, intJ1, intD
str1 = UCase(str1)
str2 = UCase(str2)
intLen1 = Len(str1)
intLen2 = Len(str2)
intTotalLen = intLen1 + intLen2
ReDim arrStr1(intLen1)
ReDim arrStr2(intLen2)
ReDim arrDA(intSize)
ReDim arrDistance(intLen1 + 2, intLen2 + 2)
arrDistance(0, 0) = intTotalLen
For i = 0 To intSize - 1
arrDA(i) = 0
Next
For i = 0 To intLen1
arrDistance(i + 1, 1) = i
arrDistance(i + 1, 0) = intTotalLen
Next
For i = 1 To intLen1
arrStr1(i - 1) = Asc(Mid(str1, i, 1))
Next
For j = 0 To intLen2
arrDistance(1, j + 1) = j
arrDistance(0, j + 1) = intTotalLen
Next
For j = 1 To intLen2
arrStr2(j - 1) = Asc(Mid(str2, j, 1))
Next
For i = 1 To intLen1
intDB = 0
For j = 1 To intLen2
intI1 = arrDA(arrStr2(j - 1))
intJ1 = intDB
If arrStr1(i - 1) = arrStr2(j - 1) Then
intD = 0
Else
intD = 1
End If
If intD = 0 Then intDB = j
intMini = arrDistance(i, j) + intD
If intMini > arrDistance(i + 1, j) + 1 Then intMini = arrDistance(i + 1, j) + 1
If intMini > arrDistance(i, j + 1) + 1 Then intMini = arrDistance(i, j + 1) + 1
If intMini > arrDistance(intI1, intJ1) + i - intI1 + j - intJ1 - 1 Then intMini = arrDistance(intI1, intJ1) + i - intI1 + j - intJ1 - 1
arrDistance(i + 1, j + 1) = intMini
Next
arrDA(arrStr1(i - 1)) = i
Next
DamerauLevenshtein = arrDistance(intLen1 + 1, intLen2 + 1)
End Function
If I type in "Cire" the algorithm correctly returns "CORE".
"Raman" returns "REMAN"
"Cosnigned" returns "CONSIGNED
However, "Now" should return "New" but returns "OCM".
"New" also returns "OCM" (so distance should be 0, but is 2.)
"FP" should be "FP" but returns "OCM", distance is 2
"DPF" Should be "DPF" but returns "OCM", distance is 2
I just learned about the algorithm, so I'm sure I'm missing something important, but I just can't see it. Thoughts?
I figured it out. After much searching I found a post saying that an edit distance is commonly 2. (They didn't specify any merits on why 2 is common)
I switched my if statement to 2 from 4 and now all of the problem terms are being corrected as they should be.

VBA IF statement not finding indicated marker

There is a moment where the program has t = 1 but my if statement wont find it.
what gives?
Most of the question is code to fully experiment with the issue
What i am trying to do with my if statement is to find when t = whole number integers example 1,2,3,4,5 then do stuff to return other results but i cant find the moments when t= 1 so im stuck
Dim neq As Double
neq = 2
Dim e As Double
e = Exp(1)
Dim t_int As Integer
t_int = 5
'''''COUNTERS
Dim i As Integer
Dim j As Integer
Dim colOf As Integer
'''''EQUATION CONTROL
Dim h(3) As Double
Dim n As Double
'''''EQUATION CONTROL
Dim u() As Double
Dim uStar() As Double
Dim uOld() As Double
Dim uEx As Double
'''''EQUATION CONTROL
Dim f() As Double
Dim fOld() As Double
'''''EQUATION CONTROL
Dim t As Double
Dim tOld As Double
Dim tNew As Double
'''''SIZING ARRAY
ReDim u(neq)
ReDim uOld(neq)
ReDim uStar(neq)
ReDim f(neq)
ReDim fOld(neq)
'''''INITAL VAULES
h(1) = 0.1
h(2) = 0.05
h(3) = 0.025
u(1) = 2
u(2) = 0
colOf = 12
For j = 1 To 1
Cells(1, 1 + colOf) = "h(" & j & ") = " & h(j)
Cells(2, 1 + colOf) = "t"
Cells(2, 2 + colOf) = "u(1)"
Cells(2, 3 + colOf) = "u(2)"
Cells(2, 4 + colOf) = "uEx"
For n = 1 To (t_int / h(j))
tOld = t
t = tOld + h(j)
For i = 1 To neq
uOld(i) = u(i)
Next i
For i = 1 To neq
fOld(i) = fDeriv(uOld, tOld, i)
uStar(i) = uOld(i) + h(j) * fOld(i)
Next i
For i = 1 To neq
f(i) = fDeriv(uStar, t, i)
u(i) = uOld(i) + (h(j) * (fOld(i) + f(i))) / 2
Next i
i = i - 1
uEx = 2 * e ^ -t * (Cos((3 ^ 0.5) * t) + ((3 ^ 0.5) ^ -1) * Sin((3 ^ 0.5) * t))
Cells(n + 2, 1 + colOf) = t
Cells(n + 2, 2 + colOf) = u(1)
Cells(n + 2, 3 + colOf) = u(2)
Cells(n + 2, 4 + colOf) = uEx
**If t = 1 Then Debug.Print t**
Next n
colOf = colOf + 5
Next j

VBA array adding

For j = 1 To 8
Sheet5.Cells(j + 1, 2) = 480
Next
t = 0
c = 0
For j = LBound(arrayTime) + 1 To UBound(arrayTime)
MsgBox "j " & j
'MsgBox (t)
numMins = Sheet5.Cells((j + 1) - (8 * c), 2) - arrayTime(j)
If numMins < 0 Then
t = t + 1
ReDim Preserve arrayTime(numrows - 1 + t)
arrayTime(numrows - 1 + t) = arrayTime(j)
MsgBox (arrayTime(numrows - 1 + t))
Else
Sheet5.Cells((j + 1) - (8 * c), 2) = numMins
End If
If j = 8 * (c + 1) Then
c = c + 1
End If
MsgBox ("end " & t)
Next
Im trying to add an value to arrayTime if the condition is true. I successfully added it but the for loop will not re-dimension to loop through the added element. The array originally contains 12 elements then I add a 13th but the loop does to recognize the 13th element and only loops 12 times. Any suggestions on how to get the for loop to loop 13 times?
Add a loop counter, say i, and set it to LBound(arrayTime) + 1 then use a Do Until (i = UBound(arrayTime)). This forces VBA to recalculate the upper bound before each loop.

Excel VBA - Run-time error 1004

When I run the code below I get the annoying 1004 error.
Sub TEST_MACRO()
Dim shSource As Worksheet
Dim shDest As Worksheet
Dim DateRange As Range
Dim i As Integer, nullcounter As Integer, nullcounterov As Integer, tablelength As Integer, tablelengthov As Integer, DateRangeSize As Integer
Dim q As Integer
Set shSource = ThisWorkbook.Sheets("Sheet1")
Set shDest = ThisWorkbook.Sheets("Sheet2")
Set DateRange = shSource.Application.InputBox("Select date", Type:=8)
DateRangeSize = DateRange.Rows.Count
nullcounter = 0
nullcounterov = 0
tablelength = 3
tablelengthov = 3
For q = 0 To DateRangeSize - 1
shDest.Range("C4:I17").ClearContents
'THIS IS THE CODE FOR ABC
For i = 0 To 3
If IsEmpty(shSource.Cells(DateRange.Row + q, 2 + i)) = True Or shSource.Cells(DateRange.Row + q, 2 + i) = 0 Then
nullcounter = nullcounter + 1
Else
shDest.Cells(4 + i - nullcounter, 4) = shSource.Cells(DateRange.Row + q, 2 + i)
shDest.Cells(4 + i - nullcounter, 5) = shSource.Cells(DateRange.Row + q, 6 + i)
shDest.Cells(4 + i - nullcounter, 3) = shSource.Cells(1, 2 + i)
tablelength = tablelength + 1
End If
Next
'THIS IS THE CODE FOR XYZ
For i = 0 To 6
If IsEmpty(shSource.Cells(DateRange.Row + q, 10 + i)) = True Or shSource.Cells(DateRange.Row + q, 10 + i) = 0 Then
nullcounterov = nullcounterov + 1
Else
shDest.Cells(4 + i - nullcounterov, 8) = shSource.Cells(DateRange.Row + q, 10 + i)
shDest.Cells(4 + i - nullcounterov, 9) = shSource.Cells(DateRange.Row + q, 17 + i)
shDest.Cells(4 + i - nullcounterov, 7) = shSource.Cells(1, 10 + i)
tablelengthov = tablelengthov + 1
End If
Next
Next
End Sub
The excel sheet I run this on looks like this:
http://i.imgur.com/V7tWTKq.png
The code works for ABC but it doesn't for XYZ. I'm guessing the 0 value cells are messing it up but I don't understand why.
The goal of the code is:
User is prompted to select a range of size DateRangeSize.
For each row in the range the code copies the values of ABC, ABC-D, XYZ and XYZ-D if they aren't 0 and writes them to sheet 2.
If the number of rows in the range is 1, the code works fine. But if the number of rows is greater than 1, I get the 1004 error where this part of the code is highlight:
shDest.Cells(4 + i - nullcounterov, 8) = shSource.Cells(DateRange.Row + q, 10 + i)
I appreciate the help.
EDIT: I just want to add that the code ALWAYS works for ABC. If the number of rows is 2 then the ABC values for the second row are printed in sheet2 but the code breaks when it attempts to do the same for XYZ.
EDIT 2: I added 0 values to the ABC portion but the code still works for ABC! This is so frustrating.
I figured it out. I defined
nullcounter = 0
nullcounterov = 0
tablelength = 3
tablelengthov = 3
Outside of the loop so they kept on increasing.
shDest.Cells(4 + i - nullcounterov, 8) = shSource.Cells(DateRange.Row + q, 10 + i)
Eventually 4 + i << nullcounterov and excel tried to write to a cell that didn't exist.
The fixed code:
Sub TEST_MACRO()
Dim shSource As Worksheet
Dim shDest As Worksheet
Dim DateRange As Range
Dim i As Integer, nullcounter As Integer, nullcounterov As Integer, tablelength As Integer, tablelengthov As Integer, DateRangeSize As Integer
Dim q As Integer
Set shSource = ThisWorkbook.Sheets("Sheet1")
Set shDest = ThisWorkbook.Sheets("Sheet2")
Set DateRange = shSource.Application.InputBox("Select date", Type:=8)
DateRangeSize = DateRange.Rows.Count
For q = 0 To DateRangeSize - 1
nullcounter = 0
nullcounterov = 0
tablelength = 3
tablelengthov = 3
shDest.Range("C4:I17").ClearContents
'THIS IS THE CODE FOR ABC
For i = 0 To 3
If IsEmpty(shSource.Cells(DateRange.Row + q, 2 + i)) = True Or shSource.Cells(DateRange.Row + q, 2 + i) = 0 Then
nullcounter = nullcounter + 1
Else
shDest.Cells(4 + i - nullcounter, 4) = shSource.Cells(DateRange.Row + q, 2 + i)
shDest.Cells(4 + i - nullcounter, 5) = shSource.Cells(DateRange.Row + q, 6 + i)
shDest.Cells(4 + i - nullcounter, 3) = shSource.Cells(1, 2 + i)
tablelength = tablelength + 1
End If
Next
'THIS IS THE CODE FOR XYZ
For i = 0 To 6
If IsEmpty(shSource.Cells(DateRange.Row + q, 10 + i)) = True Or shSource.Cells(DateRange.Row + q, 10 + i) = 0 Then
nullcounterov = nullcounterov + 1
Else
shDest.Cells(4 + i - nullcounterov, 8) = shSource.Cells(DateRange.Row + q, 10 + i)
shDest.Cells(4 + i - nullcounterov, 9) = shSource.Cells(DateRange.Row + q, 17 + i)
shDest.Cells(4 + i - nullcounterov, 7) = shSource.Cells(1, 10 + i)
tablelengthov = tablelengthov + 1
End If
Next
Next
End Sub