I'm trying to make a simple text converter program for my first Visual Basic program. I've already written this in Python but I'm not sure how to do it in Visual Basic.
I need the program to run through the characters of a string and once it encounters a ( to then remove the bracket and ignore the rest of the text until it encounters a ).
For Example,
"This is a (not so good) sentence",
becomes
"This is a Sentence".
Previously I did this with a for loop what looked at a character in the string, then checked if it was open. If it wasn't it would append the character to an output string to the next character or if it was it would then trigger a Boolean to true what would stop the character being appended. It would then continue to stop the future characters from being appended until it found a close bracket. At that point, it would then make the Boolean false, stop the closed bracket from being appended and move back to appending characters, unless it was another bracket.
Sorry if this description is a bit rough as I'm not the best at describing things and I'm very new to visual basic. Thanks for any help given
You can achieve this by several different methods. Here is one method using Instr. Instr returns the character position (index) of a given string in another string. You can use this to determine the bounds of where to include/exclude the string chunk.
You didn't specify if there could be multiple sections encapsulated in () so I assumed there wouldn't be. However, this is a relatively easy tweak by adding either a Do...Loop or a While... loop in the Function.
Hope it helps:
Option Explicit
Public Function removeBrackets(Source As String, Optional RemoveDoubleSpaces As Boolean = False)
Dim FirstBracket As Long
Dim SecondBracket As Long
FirstBracket = InStr(1, Source, "(")
SecondBracket = InStr(1, Source, ")")
If FirstBracket >= SecondBracket Or FirstBracket = 0 Or SecondBracket = 0 Then Exit Function
removeBrackets = Left$(Source, FirstBracket - 1) & Right$(Source, Len(Source) - SecondBracket)
If RemoveDoubleSpaces Then removeBrackets = Replace$(removeBrackets, " ", " ")
End Function
'Run this
Sub Test()
Debug.Print "The value returned is: " & removeBrackets("This is a (not so good) sentence") ' Example given
Debug.Print "The value returned is: " & removeBrackets("This is a (not so good) sentence", True) ' Example given, slight revision. Remove double spaces
Debug.Print "The value returned is: " & removeBrackets("This is a (not so good sentence") ' missing ending bracket
Debug.Print "The value returned is: " & removeBrackets("This is a not so good) sentence") ' missing starting bracket
Debug.Print "The value returned is: " & removeBrackets("This is a not so good sentence") ' No brackets
End Sub
Related
I have MS Access form where the user pastes a string into a field {Vars}, and I want to reformat that string into a new field so that (a) it retains whole words, and (b) "fits" within 70 columns.
Specifically, the user will be cutting/pasting variable names from SPSS. So the string will go into the field as whole names---no spaces allowed---with line breaks between each variable. So the first bit of VBA code looks like this:
Vars = Replace(Vars, vbCrLf, " ")
which removes the line breaks. But from there, I'm stumped---ultimately I want the long string that is pasted in the Vars field to be put on consecutive multiple lines that each are no longer than 70 columns.
Any help is appreciated!
Okay, for posterity, here is a solution:
The field name on the form that captures the user input is VarList. The call to the SPSS_Syntax function below returns the list of variable names (in "Vars") that can then be used elsewhere:
Vars = SPSS_Syntax(me.VarList)
Recall that user input into Varlist comes in as each variable (word) with a line break in between each. The problem is that we want the list to be on one line (horizontal, not vertical) AND a line can be no more than 256 characters in length (I'm setting it to 70 characters below). Here's the function:
Public Function SPSS_Syntax(InputString As String)
InputString = Replace(InputString, vbNewLine, " ") 'Puts the string into one line, separated by a space.
MyLength = Len(InputString) 'Computes length of the string
If MyLength < 70 Then 'if the string is already short enough, just returns it as is.
SPSS_Syntax = InputString
Exit Function
End If
MyArray = Split(InputString, " ") 'Creates the array
Dim i As Long
For i = LBound(MyArray) To UBound(MyArray) 'for each element in the array
MyString = MyString & " " & MyArray(i) 'combines the string with a blank space in between
If Len(MyString) > 70 Then 'when the string gets to be more than 70 characters
Syntax = Syntax & " " & vbNewLine & MyString 'saves the string as a new line
MyString = "" 'erases string value for next iteration
End If
Next
SPSS_Syntax = Syntax
End Function
There's probably a better way to do it but this works. Cheers.
I have a declaration like number= InputBox("Number for:", "Number:"), number is declared as Dim number As Double but when I enter a double number, for example 5.4, into the Inputbox and transmit it into a cell, the cell shows me 54, it deletes the point.
How can I fix this?
THX
If you want to detect which settings your Excel uses for the Decimal seperator, try the code below:
MsgBox "Excel uses " & Chr(34) & Application.DecimalSeparator & Chr(34) & " as a decimal seperator"
if you want to change it to ., then use the line below:
Application.DecimalSeparator = "."
Unfortunately, VBA is horrible at handling differences in decimal seprators. In your case, you should probably use a comma (,), instead of a punctuation/dot (.).
Edit: Using the Application.DecimalSeparator method, it now works regardless of regional settings. Be aware though, it seems to cause some issues if you change the comma separator settings for Excel (it seems that VBA somewhat ignores this setting). If you do not change that however, the example should work in all other cas
Sub GetNumberFromInputBox()
Dim val As String
Dim num As Double
'Get input
val = InputBox("Number for:", "Number:")
Debug.Print Application.DecimalSeparator
If IsNumeric(val) Then
'Try to convert to double as usual
num = CDbl(val)
'If the dot is removed automatically, then
'you will se a difference in the length. In
'those cases, replace the dot with a comma,
'before converting to double
If Len(val) <> Len(num) Then
If Application.DecimalSeparator = "," Then
num = CDbl(Replace(val, ".", ","))
Else
num = CDbl(Replace(val, ",", "."))
End If
End If
'Pring the number
Debug.Print "You selected number: " & num
Else
'If its not a number at all, throw an error
Debug.Print "You typed " & val & ", which is not a number"
End If
End Sub
I wrote a custom function that parses an HL7 interface message. These are messages sent between healthcare information systems, but basically it's just a long text string, delimited with various characters to indicate different fields, that I paste into a cell in Excel. The function I created searches and counts to find the fields specified in the arguments.
DISCLAIMER: I am new to VBA. I've been teaching myself via online research and trial-and-error over the past 3-4 weeks, so I'm no VBA expert. I'd prefer NOT to use arrays because when I tried that, the code got too complex for me to troubleshoot. So, I'm OK with the code being easy-to-follow, as opposed to being the fastest/most-efficient.
Anyhow, I've got it working pretty well to do what I want, but I'm stuck on adding in logic for an OPTIONAL argument.
So, this is how I WANT it to work:
Formula =KWHL7(A1, "MSH", 8)
NOTE only 3 arguments
Result I Want ADT^A08
Result I Get ADT
NOTE I know I told it to stop at the next instance of "HL7_SUBFIELD_DELIMITER" which is " ^ "
Formula =KWHL7(A1, "MSH", 8,1)
NOTE the optional 4th argument
Result I Want ADT
Formula =KWHL7(A1, "MSH", 8,2)
NOTE the optional 4th argument
Result I Want A08
The contents "value" of cell A1:
<11>MSH|^~\&|OPS|384|RISIC|384|20160923093012||ADT^A08|Q1230569238T1410271390|P|2.3|||*PM_ALLERGY*|||8859/1<13>
EVN||20160923<13>
PID|1||000000808^^^SCH MRN^MRN^SC||ZZTEST^LEANN||20160706|F|||459 CORPORATION ST.^^BEAVER^PA^15009^USA||(724)775-7418^PRN|||S||000000008082^^^SCH Account Number^FIN NBR|||||||0<13>
PV1|1|I|SCH Periop^^^^^^||||08888^Bullian^Leann~08888^Naylor^Daniel|||10|||||||08888^Nguyen-potter^Rose~00187^TEST^STCHRISRES^L^MD^^MD^^SCH Doc Number|1|1287593^^^TEMP FIN^VISITID||||||||||||||||||||384||A|||20160707131900<13>
PV2|||PA^<13>
OBX|1||Dosing Weight^WEIGHT||5|kg<13>
OBX|2||Height^HEIGHT||25|cm<13>
AL1|1|Drug|d00308^morphine^Multum Drug||66382015<13>
ZAL|||16655315|16655315||Active|66382015^Anaphylaxis^673967||||20160923093008|^Naylor^Daniel|0<13>
AL1|3|Drug|d00012^codeine^Multum Drug||103576018<13>
ZAL|||16655323|16655307||Active|103576018^Diarrhea^673967||||20160923093008|^Naylor^Daniel|0<13>
<28><13>
My VBA code (sorry for all the comments, I'm just learning!):
Public Function KWHL7(KW_Cell_With_HL7_Message As Variant, KW_HL7_Segment_Name As String, KW_HL7_Field_Number As Integer)
'KW_Cell_With_HL7_Message = KW_Cell_With_HL7_Message.Value
'KW_Cell_With_HL7_Message = ActiveCell.Value
'KW_HL7_Segment_Name = "PID"
'KW_HL7_Field_Number = 18
Const HL7_SEGMENT_DELIMITER = vbLf 'using "<13>" did not work due to carriage return
Const HL7_FIELD_DELIMITER = "|" ' Pipe means next field
Const HL7_SUBFIELD_DELIMITER = "^"
'Various carriage returns and line breaks: vbLf, vbCr, vbCrLf, vbNewLine, Chr(10), Chr(13)
KWSegmentStringToSearchFor = HL7_SEGMENT_DELIMITER & KW_HL7_Segment_Name 'Using the segment delimiter ("<13>" or "vbLf" / carriage return) before segment name implies that the segment / line STARTS with this text
KWSegmentCharacterPosition = InStr(1, KW_Cell_With_HL7_Message, KWSegmentStringToSearchFor)
'** FOR TESTING ** MsgBox ("Segment Character Position: " & KWSegmentCharacterPosition & ", 5 Characters starting there = " & Mid(KW_Cell_With_HL7_Message, KWSegmentCharacterPosition, 5))
'Now we have the character position of the start of the proper SEGMENT / line
'Now we have to find the Proper Field in that segment
'So we'll use this position + the length of the end of the Segment Delimiter as the start
'***WARNING***: Still must add logic to make sure we stop if we encounter another Segment Delimiter
KWFieldCharacterPosition = KWSegmentCharacterPosition + Len(HL7_SEGMENT_DELIMITER) 'instead of starting at character 0, start at the beginning of the segment found previously
' ** FOR TESTING ** MsgBox ("Length of Segment Delimiter = " & Len(HL7_SEGMENT_DELIMITER))
' ** FOR TESTING ** MsgBox ("Field Character Position: " & KWFieldCharacterPosition & ", 5 Characters starting there = " & Mid(KW_Cell_With_HL7_Message, KWFieldCharacterPosition, 5))
For J = 1 To KW_HL7_Field_Number
KWFieldCharacterPosition = InStr(KWFieldCharacterPosition + 1, KW_Cell_With_HL7_Message, HL7_FIELD_DELIMITER)
If KWFieldCharacterPosition = 0 Then Exit For
Next
' ** FOR TESTING ** MsgBox ("Field Character Position: " & KWFieldCharacterPosition & ", 5 Characters starting there = " & Mid(KW_Cell_With_HL7_Message, KWFieldCharacterPosition, 5))
'Determine the number of characters to return after the start position
'Want to pull text UNTIL the next Segment Delimiter or Field Delimiter or Subfield Delimiter
'Find the position of the next Segment Delimiter or Field Delimiter or Subfield Delimiter
'Since the InStr function does not accept multiple substrings to search for, and does not allow OR statements inside...
Next_HL7_Segment_Delimiter = InStr(KWFieldCharacterPosition + 1, KW_Cell_With_HL7_Message, HL7_SEGMENT_DELIMITER)
Next_HL7_Field_Delimiter = InStr(KWFieldCharacterPosition + 1, KW_Cell_With_HL7_Message, HL7_FIELD_DELIMITER)
Next_HL7_Subfield_Delimiter = InStr(KWFieldCharacterPosition + 1, KW_Cell_With_HL7_Message, HL7_SUBFIELD_DELIMITER)
'Added logic to handle issue where the next delimiter was not found, making result 0, making it the lowest value in the next lines of code
If Next_HL7_Segment_Delimiter = 0 Then Next_HL7_Segment_Delimiter = 99999
If Next_HL7_Field_Delimiter = 0 Then Next_HL7_Field_Delimiter = 99999
If Next_HL7_Subfield_Delimiter = 0 Then Next_HL7_Subfield_Delimiter = 99999
'Set the Last Character Position to whichever Next Delimiter is the lowest / minimum number - Segment or Field or Subfield
KWLastCharacterPosition = WorksheetFunction.Min(Next_HL7_Segment_Delimiter, Next_HL7_Field_Delimiter, Next_HL7_Subfield_Delimiter)
' ** FOR TESTING ** MsgBox ("Last Character Position: " & KWLastCharacterPosition & ", 5 Characters starting there = " & Mid(KW_Cell_With_HL7_Message, KWLastCharacterPosition, 5))
'Determine the number of characters to return in the MID function by subtracting the first character position from the last character position
KWNumberOfCharactersToReturn = KWLastCharacterPosition - KWFieldCharacterPosition - 1
' ** FOR TESTING ** MsgBox ("Number of characters to return: " & KWNumberOfCharactersToReturn)
KWResult = Mid(KW_Cell_With_HL7_Message, KWFieldCharacterPosition + 1, KWNumberOfCharactersToReturn)
'MsgBox ("Result: Segment " & KW_HL7_Segment_Name & ":" & KW_HL7_Field_Number & " is " & KWResult)
KWHL7 = KWResult
End Function
The problem I had with using the split function was that it put everything into arrays. And since I needed to search FIRST for the KWSegmentStringToSearchFor (i.e. "MSH" or "PV1"), before couting the pipe (|) characters, I would need the array to have separate nested arrays and it got way too confusing for me.
So I abandoned the split function, and my initial plans to use arrays, and just wrote everything to find things sequentially. So it searches for the KWSegmentStringToSearchFor (i.e. "MSH" or "PV1") with InStr() and then counts the pipe (|) characters from there to determine which number field to return.
Since the strings are of variable length, but delimited with special characters, next I have to determine how many characters to return with the MID function. So I search for the next delimiter FROM THERE / using the field I found as the starting point and call that the end of my field.
The issue:
The logic considers ANY of the 3 possible delimiters the end of the field.
If I take that out, the code wouldn't know where the end of the string is.
Even if I add some sort of IF statement that IF the optional 4th argument exists (which I'm not sure how to do yet), THEN ignore the ^ as a delimiter... that would always return the full field (ADT^A08). It wouldn't return just the sub-field / component I want.
Thanks!
A simple answer, would be to split on the LineFeed then it may need tweaking would be, split(range("a1").value,"|")(intFieldNumber)
i.e.
split("11>MSH|^~\&|OPS|384|RISIC|384|20160923093012||ADT^A08|Q1230569238T1410271390","|")(8)
gives the result ADT^A08
Take a look at this property(Given you have a table on the first worksheet):
Application.Sheets(1).ListObjects(1).name
How many characters can this property contain? Well, after testing out a few strings I've come to the conclusion that its 255, any string with more than 255 characters causes an error to be thrown:
Run-Time Error 5 - Invalid procedure call or arguement
Take a look at this property:
Application.Sheets(1).ListObjects(1).Summary
How many characters can this property contain? Again, test several strings and you'll come out with a number that's around 50,000, You set it any higher and you get the same error, except in this case excel will sometimes crash or spit out a different error(after multiple attempts):
Dim i As Integer
Dim a As String
For i = 1 To 5001
a = a & "abcdefghih"
Next i
Application.Sheets(1).ListObjects(1).Summary = a
Method "Summary" of object 'ListObject' failed
This sort of "hidden" character limit comes up all over the place(here, here, less specifically here, and so classically here), and it doesn't seem like they're documented anywhere, for example take a look at the page for ListObject.Name, its not noted how many characters you can store in that variable...
So is there a better way to determine this? Are the strings you are setting in properties being stored in a fixed length string somewhere that can be accessed to determine what their maximum length is, or is there some other form of documentation that can be leveraged in order to obtain this information?
It strikes me as odd these character limits that are set on most strings within standard VBA objects, I wonder what their purpose is, why the designers choose to limit "ListObjects.name" to 255 characters and whether that was an arbitrary default limit or whether that was a conscious decision that was made. I believe that the standard string length is this, I wonder why the deviation from this standard.
To summarize the points I've made above and to condense this question into one sentence:
Is there a generic way to determine the maximum length of a string that can be set within an object's property, without first testing that string's property by giving it another value and ignoring errors/checking for character truncation?
First of all, if your intention is to store meta information about objects, you could maybe make use of CustomDocumentProperties. You can find examples on their usage here and here and some nice wrappers by Chip Pearson here.
Since they are still very limited (255 chars) in length (thanks for pointing that out!), the best solution might be to use CustomXMLParts like described here. The hard part would then be building correct XML using VBA, but maybe not impossible, if you add a reference to Microsoft XML.
But to provide some help with your question concerning maximum lengths for string properties, too, here is a test setup you can use to (relatively) quickly find these limits for arbitrary properties.
Just replace the ActiveWorkbook.Sheets(1).Name on line 19 with the property you want to test and run TestMaxStringLengthOfProperty():
Option Explicit
Const PRINT_STEPS = True ' If True, calculation steps will be written to Debug.Print
Private Function LengthWorks(ByVal iLengthToTest As Long) As Boolean
Dim testString As String
testString = String(iLengthToTest, "#") ' Build string with desired length
' Note: The String() method failed for different maximum string lengths possibly
' depending on available memory or other factors. You can test the current
' limit for your setup by putting the string assignment in the test space.
' In my tests I found maximum values around 1073311725 to still work.
On Error Resume Next
' ---------------------------------------------------------------------------------
' Start of the Test Space - put the method/property you want to test below here
ActiveWorkbook.Sheets(1).Name = testString
' End of the Test Space - put the method/property you want to test above here
' ---------------------------------------------------------------------------------
LengthWorks = Err.Number = 0
On Error GoTo 0
End Function
Private Sub TestMaxStringLengthOfProperty()
Const MAX_LENGTH As Long = 1000000000 ' Default: 1000000000
Const MAXIMUM_STEPS = 100 ' Exit loop after this many tries, at most
' Initialize variables for check loop
Dim currentLength As Long
Dim lowerBoundary As Long: lowerBoundary = 0
Dim upperBoundary As Long: upperBoundary = MAX_LENGTH
Dim currentStep As Long: currentStep = 0
While True ' Infinite loop, will exit sub directly
currentStep = currentStep + 1
If currentStep > MAXIMUM_STEPS Then
Debug.Print "Exiting because maximum number of steps (" & _
CStr(MAXIMUM_STEPS) & _
") was reached. Last working length was: " & _
CStr(lowerBoundary)
Exit Sub
End If
' Test the upper boundary first, if this succeeds we don't need to continue search
If LengthWorks(upperBoundary) Then
' We have a winner! :)
Debug.Print "Method/property works with the following maximum length: " & _
upperBoundary & vbCrLf & _
"(If this matches MAX_LENGTH (" & _
MAX_LENGTH & "), " & _
"consider increasing it to find the actual limit.)" & _
vbCrLf & vbCrLf & _
"Computation took " & currentStep & " steps"
Exit Sub
Else
' Upper boundary must be at least one less
upperBoundary = upperBoundary - 1
PrintStep upperBoundary + 1, "failed", lowerBoundary, upperBoundary, MAX_LENGTH
End If
' Approximately halve test length
currentLength = lowerBoundary + ((upperBoundary - lowerBoundary) \ 2)
' "\" is integer division (http://mathworld.wolfram.com/IntegerDivision.html)
' Using `left + ((right - left) \ 2)` is the default way to avoid overflows
' when calculating the midpoint for our binary search
' (see: https://en.wikipedia.org/w/index.php?title=Binary_search_algorithm&
' oldid=809435933#Implementation_issues)
If LengthWorks(currentLength) Then
' If test was successful, increase lower boundary for next step
lowerBoundary = currentLength + 1
PrintStep currentLength, "worked", lowerBoundary, upperBoundary, MAX_LENGTH
Else
' If not, set new upper boundary
upperBoundary = currentLength - 1
PrintStep currentLength, "failed", lowerBoundary, upperBoundary, MAX_LENGTH
End If
Wend
End Sub
Private Sub PrintStep(ByVal iCurrentValue As Long, _
ByVal iWorkedFailed As String, _
ByVal iNewLowerBoundary As Long, _
ByVal iNewUpperBoundary As Long, _
ByVal iMaximumTestValue As Long)
If PRINT_STEPS Then
Debug.Print Format(iCurrentValue, String(Len(CStr(iMaximumTestValue)), "0")) & _
" " & iWorkedFailed & " - New boundaries: l: " & _
iNewLowerBoundary & " u: " & iNewUpperBoundary
End If
End Sub
The short answer is no.
Regards, Zack Barresse
I am attempting to program a loop into a DDEPoke call to a VBA-supported function known as OPC. This will enable me to write to a PLC (RSLogix 500) database from an excel spreadsheet.
This is the code:
Private Function Open_RsLinx()
On Error Resume Next
Open_RsLinx = DDEInitiate(RsLinx, C1)
If Err.Number <> 0 Then
MsgBox "Error Connecting to topic", vbExclamation, "Error"
OpenRSLinx = 0 'Return false if there was an error
End If
End Function
Sub CommandButton1_Click()
RsLinx = Open_RsLinx()
For i = 0 To 255
DDEPoke RsLinx, "N16:0", Cells(1 + i, 2)
Next i
DDETerminate RsLinx
End Sub
This code works and will, if there is a link set up with an OPC server (in this case through RSLinx) write data to the PLC.
The problem is that I can't get the part DDEPoke RsLinx, "N16:0", Cells(1 + i, 2) to write data, sequentially, from one excel cell to one element of the PLC's data array.
I tried to do DDEPoke RsLinx, "N16:i", Cells(1 + i, 2) and DDEPoke RsLinx, "N16:0+i", Cells(1 + i, 2) but neither has any effect and the program doesn't write anything at all.
How can I set up the code to get N16:0 to increment all the way up to N16:255 and then stop?
Break the variable i out of the string. Be careful for the implicit type conversion though, depending on which (Str() or CStr()), you'll wind up with a leading space. Thus, convert the number Str(i), then wrap with Trim() to make sure there's no extra spaces, and concatenate that result back to your "N" string:
RsLinx = Open_RsLinx()
For i = 0 To 255
DDEPoke RsLinx, "N16:" & Trim(Str(i)), Cells(1 + i, 2)
Next i
The reason the i didn't work when it's inside the string is because that in VBA, anything within a set of quotes is considered a literal string. Unlike some other languages (PHP comes to mind) where variables can be resolved within a string like that, VBA must have variables concatenated. Consider the following:
Dim s As String
s = "world"
Debug.Print "Hello s!"
This outputs the literal of Hello s! to the immediate window, because s is treated not as a variable, but as part of the literal string. The correct way is through concatenation:
Dim s As String
s = "world"
Debug.Print "Hello " & s & "!"
That outputs the expected Hello World! to the immediate window, because s is now treated as a variable and is resolved and concatenated.
If that were not the case, the following might be difficult to deal with:
Dim i As Integer
For i = 0 to 9
Debug.Print "this" & i
Next i
You would then have:
th0s0
th1s1
th2s2
th3s3
th4s4
'etc
That'd make things pretty difficult to manage in a lot of cases.
With all that said, there are some languages - notably PHP - where, when using a certain set of quotes (either "" or '' - I don't recall which offhand), in fact does resolve the variable when embedded into the string itself:
$i = 5;
echo "this is number $i";
VBA does not have this feature.
Hope it helps...