I looked at the other links and none seem to help me out. I am writing code for a program that will count all the commas in a phrase. I am not new to programming but I am new to VBA.
Sub examp()
Dim s As String
Dim i, my_c As Integer
i = 0
s = ",jkqk;j,oiheqfjnq;ef,jwhef;ur,jwefun;jwkbnf," '<-------arbitrary, however, when I tried to make it input from a textbox it gave me error 424 as well, so I just defined it as random chars with commas
While i < Len(s)
For i = 0 To Len(s) - 1
j = s.Chars(i) <----------------------------------Error occurs here
If j = "," Then
my_c = my_c + 1
End If
Next i
Wend
Count.Text = "my_c"
End Sub
change j = s.Chars(i) to j = Mid(s,i,1)
in line Dim i, my_c As Integer only my_c is Integer, but i
is Variant. You should declare each variable explicitly: Dim i As Integer, my_c As Integer
not sure what exactly is your Count (maybe textbox), but use
Count.Text = my_c without quotes.
also I can't undersand why do you use two loops? While i < Len(s)
is odd.
For i = 0 To Len(s) - 1 should be For i = 1 To Len(s)
If you want to count commas, there is more efficient way:
Dim s As String
Dim my_c As Integer
s = ",jkqk;j,oiheqfjnq;ef,jwhef;ur,jwefun;jwkbnf,"
my_c = Len(s) - Len(Replace(s, ",", ""))
Or you can try this:
Sub test()
Dim s As String
Dim c
Dim my_c As Long
s = ",jkqk;j,oiheqfjnq;ef,jwhef;ur,jwefun;jwkbnf,"
c = Split(s, ",")
my_c = UBound(c)
Debug.Print my_c
End Sub
Related
I am writing a VBA code to add +2 to any string of numbers that are put in the function.
It works fine, until it reaches 6 and 7, then it breaks. I really have no clue why that is.
If you are wondering why I am doing this, this is part of an encryption algorithm and it is specifically looking to encrypt digits in a string.
My code is:
Sub AddNumbers()
Dim Nos As String
Dim AddNo As String
Dim Found As Boolean
Dim Split()
Nos = "0-1-2-3-4-5-6-7-8-9-10"
Sheets("Sheet1").Range("U2").Value = Nos
Length = Len(Nos)
ReDim Split(Length)
For i = 1 To Length
Found = False
Split(i) = Mid(Nos, i, 1)
For O = 48 To 55
If Split(i) = Chr(O) Then
Split(i) = Chr(O + 2)
Found = True
Exit For
End If
Next O
If Split(i) = Chr(56) Then
Split(i) = Chr(48)
ElseIf Split(i) = Chr(57) Then
Split(i) = Chr(49)
End If
Next i
AddNo = Join(Split, "")
Sheets("Sheet1").Range("U3").Value = AddNo
End Sub
I would really appreciate an insight to why it is breaking at 6 and 7.
Take me a moment, but you are double adding.
Look at your loop. When you encounter 6 (Char(54)) you add 2 and have 8 (Char(56)).
But then, after your loop you are testing again for same Split(i). Char for 6 and 7 are now accordingly 56 and 57 - so you add another 2 to them.
If Split(i) = Chr(56) And Found = False Then
Split(i) = Chr(48)
ElseIf Split(i) = Chr(57) And Found = False Then
Split(i) = Chr(49)
End If
Use the actual function Split:
Sub AddNumbers()
Dim Nos As String
Dim AddNo As String
Dim Found As Boolean
Dim SplitStr() As String
Nos = "0-1-2-3-4-5-6-7-8-9-10"
Sheets("Sheet1").Range("U2").Value = Nos
SplitStr = Split(Nos, "-")
Dim i As Long
For i = LBound(SplitStr) To UBound(SplitStr)
Dim vlue As String
vlue = StrConv(SplitStr(i), vbUnicode)
Dim substr() As String
substr = Split(Left(vlue, Len(vlue) - 1), vbNullChar)
Dim j As Long
For j = LBound(substr) To UBound(substr)
Select Case substr(j)
Case 8
substr(j) = 0
Case 9
substr(j) = 1
Case Else
substr(j) = substr(j) + 2
End Select
Next j
SplitStr(i) = Join(substr, "")
Next i
AddNo = Join(SplitStr, "-")
Sheets("Sheet1").Range("U3").Value = AddNo
End Sub
The overall problem is that you are using the Chr codes for numbers and not actual numbers. This method only returns 1 digit because a Chr() refers to a list of single characters.
You are going to need to use Split (mySplit = Split(Nos,"-")) to return each number and work with those.
The lines
If Split(i) = Chr(56) Then
Split(i) = Chr(48)
ElseIf Split(i) = Chr(57) Then
Split(i) = Chr(49)
End If
has me confused. You are saying if the value is "8" change to "0" and if it is "9" change to "1"
This is another way to do it:
Sub AddNumbers()
Dim Nos As String, Nos2 As String
Dim NumSplit As Variant
Dim Num As Variant
Dim tmp As String
Dim i As Long
Nos = "0-1-2-3-4-5-6-7-8-9-10"
Sheets("Sheet1").Range("U2").Value = Nos
NumSplit = Split(Nos, "-")
For Each Num In NumSplit
For i = 1 To Len(Num)
tmp = tmp & Mid(Num, i, 1) + 2
Next i
Nos2 = Nos2 & tmp & "-"
tmp = ""
Next Num
Nos2 = Left(Nos2, Len(Nos2) - 1)
Sheets("Sheet1").Range("U3").Value = Nos2
End Sub
It's a bit messy, but shows the basic idea of splitting the original array into the separate numbers.
The For....Next loop inside the For...Each loop takes care of any numbers with more than one digit (giving the 32).
The following code splits each lines into words and store the first words in each line into array list and the second words into another array list and so on. Then it selects the most frequent word from each list as correct word.
Module Module1
Sub Main()
Dim correctLine As String = ""
Dim line1 As String = "Canda has more than ones official language"
Dim line2 As String = "Canada has more than one oficial languages"
Dim line3 As String = "Canada has nore than one official lnguage"
Dim line4 As String = "Canada has nore than one offical language"
Dim wordsOfLine1() As String = line1.Split(" ")
Dim wordsOfLine2() As String = line2.Split(" ")
Dim wordsOfLine3() As String = line3.Split(" ")
Dim wordsOfLine4() As String = line4.Split(" ")
For i As Integer = 0 To wordsOfLine1.Length - 1
Dim wordAllLinesTemp As New List(Of String)(New String() {wordsOfLine1(i), wordsOfLine2(i), wordsOfLine3(i), wordsOfLine4(i)})
Dim counts = From n In wordAllLinesTemp
Group n By n Into Group
Order By Group.Count() Descending
Select Group.First
correctLine = correctLine & counts.First & " "
Next
correctLine = correctLine.Remove(correctLine.Length - 1)
Console.WriteLine(correctLine)
Console.ReadKey()
End Sub
End Module
My Question: How can I make it works with lines of different number of words. I mean that the length of each lines here is 7 words and the for loop works with this length (length-1). Suppose that line 3 contains 5 words.
EDIT: Accidentally had correctIndex where shortest should have been.
From what I can tell you are trying to see which line is the closest to the correctLine.
You can get the levenshtein distance using the following code:
Public Function LevDist(ByVal s As String,
ByVal t As String) As Integer
Dim n As Integer = s.Length
Dim m As Integer = t.Length
Dim d(n + 1, m + 1) As Integer
If n = 0 Then
Return m
End If
If m = 0 Then
Return n
End If
Dim i As Integer
Dim j As Integer
For i = 0 To n
d(i, 0) = i
Next
For j = 0 To m
d(0, j) = j
Next
For i = 1 To n
For j = 1 To m
Dim cost As Integer
If t(j - 1) = s(i - 1) Then
cost = 0
Else
cost = 1
End If
d(i, j) = Math.Min(Math.Min(d(i - 1, j) + 1, d(i, j - 1) + 1),
d(i - 1, j - 1) + cost)
Next
Next
Return d(n, m)
End Function
And then, this would be used to figure out which line is closest:
Dim correctLine As String = ""
Dim line1 As String = "Canda has more than ones official language"
Dim line2 As String = "Canada has more than one oficial languages"
Dim line3 As String = "Canada has nore than one official lnguage"
Dim line4 As String = "Canada has nore than one offical language"
Dim lineArray As new ArrayList
Dim countArray As new ArrayList
lineArray.Add(line1)
lineArray.Add(line2)
lineArray.Add(line3)
lineArray.Add(line4)
For i = 0 To lineArray.Count - 1
countArray.Add(LevDist(lineArray(i), correctLine))
Next
Dim shortest As Integer = Integer.MaxValue
Dim correctIndex As Integer = 0
For i = 0 To countArray.Count - 1
If countArray(i) <= shortest Then
correctIndex = i
shortest = countArray(i)
End If
Next
Console.WriteLine(lineArray(correctIndex))
Right, so using Python I would create a multidimensional list and set the values on one line of code (as per the below).
aryTitle = [["Desciption", "Value"],["Description2", "Value2"]]
print(aryTitle[0,0] + aryTitle[0,1])
I like the way I can set the values on one line. In VBA I am doing this by:
Dim aryTitle(0 To 1, 0 To 1) As String
aryTitle(0, 0) = "Description"
aryTitle(0, 1) = "Value"
aryTitle(1, 0) = "Description2"
aryTitle(1, 1) = "Value2"
MsgBox (aryTitle(0, 0) & aryTitle(0, 1))
Is there a way to set the values in one line of code?
Not natively, no. But you can write a function for it. The only reason Python can do that is someone wrote a function to do it. The difference is that they had access to the source so they could make the syntax whatever they like. You'll be limited to VBA function syntax. Here's a function to create a 2-dim array. It's not technically 'one line of code', but throw it in your MUtilities module and forget about it and it will feel like one line of code.
Public Function FillTwoDim(ParamArray KeyValue() As Variant) As Variant
Dim aReturn() As Variant
Dim i As Long
Dim lCnt As Long
ReDim aReturn(0 To ((UBound(KeyValue) + 1) \ 2) - 1, 0 To 1)
For i = LBound(KeyValue) To UBound(KeyValue) Step 2
If i + 1 <= UBound(KeyValue) Then
aReturn(lCnt, 0) = KeyValue(i)
aReturn(lCnt, 1) = KeyValue(i + 1)
lCnt = lCnt + 1
End If
Next i
FillTwoDim = aReturn
End Function
Sub test()
Dim vaArr As Variant
Dim i As Long
Dim j As Long
vaArr = FillTwoDim("Description", "Value", "Description2", "Value2")
For i = LBound(vaArr, 1) To UBound(vaArr, 1)
For j = LBound(vaArr, 2) To UBound(vaArr, 2)
Debug.Print i, j, vaArr(i, j)
Next j
Next i
End Sub
If you supply an odd number of arguments, it ignores the last one. If you use 3-dim arrays, you could write a function for that. You could also write a fancy function that could handle any dims, but I'm not sure it's worth it. And if you're using more than 3-dim arrays, you probably don't need my help writing a function.
The output from the above
0 0 Description
0 1 Value
1 0 Description2
1 1 Value2
You can write a helper function:
Function MultiSplit(s As String, Optional delim1 As String = ",", Optional delim2 As String = ";") As Variant
Dim V As Variant, W As Variant, A As Variant
Dim i As Long, j As Long, m As Long, n As Long
V = Split(s, delim2)
m = UBound(V)
n = UBound(Split(V(0), delim1))
ReDim A(0 To m, 0 To n)
For i = 0 To m
For j = 0 To n
W = Split(V(i), delim1)
A(i, j) = Trim(W(j))
Next j
Next i
MultiSplit = A
End Function
Used like this:
Sub test()
Dim A As Variant
A = MultiSplit("Desciption, Value; Description2, Value2")
Range("A1:B2").Value = A
End Sub
I am having a column which contains integer values with two special character "," and "_". I am trying to remove these character for example 1,10_2,2_3,3 should be like 1102233. Thanks in advance for your suggestions.
this function isn't foolproof but it is a good start.
Function trim(aStringToTrim As String, aElementToTrinm() As Variant) As String
Dim elementToTrim As Integer
Dim IndexInString As Integer
For elementToTrim = LBound(aElementToTrinm) To UBound(aElementToTrinm)
IndexInString = InStr(aStringToTrim, aElementToTrinm(elementToTrim))
Do While IndexInString > 0
aStringToTrim = Left(aStringToTrim, IndexInString - 1) & Right(aStringToTrim, Len(aStringToTrim) - IndexInString - Len(aElementToTrinm(elementToTrim)) + 1)
IndexInString = InStr(aStringToTrim, aElementToTrinm(elementToTrim))
Loop
Next
End Function
It can be use like this:
Sub main()
Dim myString As String
Dim caracterstoRemove As Variant
caracterstoRemove = Array(",", ".")
myString = "This, is. a, string, with. caracters to remove."
myString = trim(myString, caracterstoRemove)
End Sub
I long variable in vb.net which contains the following information,
Dim g As String = "$C:\Program Files\Cavaj Java Decompiler\cavaj.exe$C:\Users\Yoosuf\AppData\Local\Google\Chrome\Application\chrome.exe$C:\Program Files\DVD Maker\dvdmaker.exe$C:\Program Files\Adobe\Adobe Photoshop CS2\ImageReady.exe$C:\Program Files\Java\jre6\bin\javaws.exe$"
The $ symbol is used as a delimiter to separate each item from the other. I need to add the exe file name at the end of each path to a listbox. However the initial process of retrieving the variable to individual array elements is not working properly.
Dim strArr() As String = g.Split("$") 'This variable is empty
For count = 0 To strArr.Length - 1
Dim arr As String = strArr(count).Split("\")
Dim strval As String = ""
For i As Integer = 3 To arr.Length - 1
strval = arr(i)
Dim j As Integer = arr.Length - 1
strval = arr(j)
Dim result As String = strval.Substring(g.Length - 5)
result = g.Substring(g.LastIndexOf("\") + 1)
ListBox1.Items.Add(result)
Next
Next
No need to do all this work. The System.IO.Path class has methods to do this for you. You want to use either System.IO.Path.GetFileName or System.IO.Path.GetFileNameWithoutExtension. Since you've already split all the file paths, just pass those paths to either of the aforementioned methods and add the result to your listbox.
Dim strArr() As String = g.Split("$")
For Each path As String In strArr
ListBox1.Items.Add(System.IO.Path.GetFileName(path))
Next
Please refer to the code below and the associated comments. Also I have comment out some code which I feel is not required based on what you want to do.
Dim strArr() As String = g.Split("$") 'This variable is empty
For count = 0 To strArr.Length - 1
Dim arr() As String = strArr(count).Split("\") ' Split returns an array
Dim strval As String = ""
For i As Integer = 3 To arr.Length - 1
'strval = arr(i)
Dim j As Integer = arr.Length - 1
strval = arr(j)
'Dim result As String = strval.Substring(g.Length - 5)
'result = g.Substring(g.LastIndexOf("\") + 1)
ListBox1.Items.Add(strval)
Next
Next