Use the text of a cell from a table and use it as a string variable then use that variable as name of the bookmark, with vba word?
I have tried the following, and it says bad bookmark name. I know the constraints of a bookmark name that's why I am deleting the extra new line which ActiveDocument.Tables(1).Cell(1,1).Range.Text gets. But still why I am getting the bad bookmark name runtime error-5828
Function RemoveLastCharacter(myString As String) As String
'Remove Last Character From String
myString = Left(myString, Len(myString) - 1)
RemoveLastCharacter = myString
End Function
Sub makebookmark()
Dim kib As String
kib = ActiveDocument.Tables(1).Cell(1, 1).Range.Text
Dim kib2 As String
kib2 = RemoveLastCharacter(kib)
ActiveDocument.Tables(1).Cell(1, 3).Range.InsertAfter kib2
ActiveDocument.Bookmarks.Add Name:=kib2, _
Range:=ActiveDocument.Tables(1).Cell(1, 2).Range
End Sub
The end of cell marker is actually 2 characters. You can test this by checking the length of an empty cell.
So you could do either:
kib2 = RemoveLastCharacter(RemoveLastCharacter(kib))
or create a new function
Function RemoveEndOfCellMarker(cellText As String) As String
'Remove Last 2 Characters From String
RemoveEndOfCellMarker = Left(cellText, Len(cellText) - 2)
End Function
Related
I am trying to split data using VBA within word.
I have got the data using the following method
d = ActiveDocument.Tables(1).Cell(1, 1).Range.Text
This works and gets the correct data. Data for this example is
This
is
a
test
However, when I need to split the string into a list of strings using the delimiter as \n
Here is an example of the desired output
This,is,a,test
I am currently using
Dim dataTesting() As String
dataTesting() = Split(d, vbLf)
Debug.Print dataTesting(0)
However, this returns all the data and not just the first line.
Here is what I have tried within the Split function
\n
\n\r
\r
vbNewLine
vbLf
vbCr
vbCrLf
Word uses vbCr (ANSI 13) to write a "new" paragraph (created when you press ENTER) - represented in the Word UI by ¶ if the display of non-printing characters is activated.
In this case, the table cell content you show would look like this
This¶
is¶
a¶
test¶
The correct way to split an array delimited by a pilcro in Word is:
Dim d as String
d = ActiveDocument.Tables(1).Cell(1, 1).Range.Text
Dim dataTesting() As String
dataTesting() = Split(d, vbCr)
Debug.Print dataTesting(0) 'result is "This"
You can try this (regex splitter from this thread)
Sub fff()
Dim d As String
Dim dataTesting() As String
d = ActiveDocument.Tables(1).Cell(1, 1).Range.Text
dataTesting() = SplitRe(d, "\s+")
Debug.Print "1:" & dataTesting(0)
Debug.Print "2:" & dataTesting(1)
Debug.Print "3:" & dataTesting(2)
Debug.Print "4:" & dataTesting(3)
End Sub
Public Function SplitRe(Text As String, Pattern As String, Optional IgnoreCase As Boolean) As String()
Static re As Object
If re Is Nothing Then
Set re = CreateObject("VBScript.RegExp")
re.Global = True
re.MultiLine = True
End If
re.IgnoreCase = IgnoreCase
re.Pattern = Pattern
SplitRe = Strings.Split(re.Replace(Text, ChrW(-1)), ChrW(-1))
End Function
If this doesn't work, there may be strange unicode/Wprd characters in your Word doc. It may be soft breaks, for instance. You could try to not split with "\W+" in stead of "\s+". I cannot test this without your document.
Dim dataTesting() As String
dataTesting() = Split(d, vbLf)
Debug.Print dataTesting(0)
works fine and thank you very much for your example,
for why it have returned a whole array is because you have used 0 as index, in many programming languages 0 is the whole array, so the first element is ,
so in my case counting from 1 this perfectly split a string that I had troubles with.
To be more exact this is how it was used in my case
Dim dataTesting() As String
dataTesting() = Split(Document.LatheMachineSetup.Heads.Item(1).Comment, vbCrLf)
MsgBox (dataTesting(1))
And that comment is a multiline string.
Image
So this msg box returned exactly first line.
I have a VBA code that's designed to search a CSV String and add Carriage Returns where they should exist. I've split it up into two seperate functions - one to search the string and put the index of where the CRs should go into an array and a second function to actually add the CRs.
The issue I'm running into is that the value in the immediate window/in the watch window for the functions is correct within the function itself, but it assigns the result variable a blank string.
'*****************Import CSV**********************
'Took this straight off the internet because it was reading Jet.com files as one single line
'
Sub ImportCSVFile(filepath As String)
.....
line = SearchString(line, "SALE")
.....
End Sub
'****************Search String***************************
'This is search the string for something - It will then call a function to insert carriage returns
Function SearchString(source As String, target As String) As String
Dim i As Integer
Dim k As Integer
Dim myArray() As Variant
Dim resultString As String
Do
i = i + 1
If Mid(source, i, Len(target)) = target Then
ReDim Preserve myArray(k)
myArray(k) = i
k = k + 1
End If
DoEvents
Loop Until i = Len(source)
resultString = addCarriageReturns(source, myArray) 'resultString here is assigned a blank string
SearchString = resultString
End Function
'***************Add Carraige Returns**************************
'Cycle through the indices held in the array and place carriage returns into the string
Function addCarriageReturns(source As String, myArray As Variant) As String
Dim i As Integer
Dim resultString As String
resultString = source
For i = 0 To UBound(myArray, 1)
resultString = Left(resultString, myArray(i) + i) & Chr(13) & Right(resultString, Len(resultString) - myArray(i) + i)
Next i
addCarraigeReturns = resultString 'The value of addCarriageReturn is correct in the immediate window here
End Function
In the function the value is not blank
...but when it passes it back, it says the value is blank
I'm just curious, why do you want separate functions like this?
Can you just use:
line = Replace(line, "SALE", "SALE" & Chr(13))
What I'm trying to do
Locate the column whose header cell contains a unique string. In other words, I know the cell's text, and I know the cell is in row 1, but I don't know which column. NOTE: I want to search for the entire text, not just part of it. NOTE2: The text can vary, so I cannot hardcode the value into my code. Rather I need to use the variable in which the value is stored.
The problem
When there's no carriage return in the header text, a simple newCol = Range("1:1").Find(headerText).Column works fine. However, if there is a carriage return, this doesn't work. It throws up the error "Object variable or With block variable not set". Here's my exact header string:
Incomplete Email
(more text)
What I've already tried
I also tried using WorksheetFunction.Match(headerText, Range("1:1"), 0), but got the same issue.
Additional notes and requirements
This is part of an add-in, so I do not want to change anything in the user's excel sheet if I don't have to (i.e., I don't want to remove the carriage return).
Technically, I'm doing this in a function:
Public Function getColumn(headerText As Variant)
getColumn = Range("1:1").Find(headerText).Column
End Function
Thanks!
pls try with below code
Public Function getColumn(headerText As String)
str1 = Split(headerText, vbCrLf)
str2 = UBound(str1)
b = Range("1:1").Find(str1(0) & Chr(10) & str1(1)).Column
End Function
Here's the thing: text with and without line break is NOT the same text hence the .Find fail. What you should do is a pattern lookup. I have just tested this and it works, provided that if there is no line break there shall be a space:
Sub test()
Dim rex As RegExp, ran As Range
Dim col As Integer, headerText As String
'read you headerText here
Set rex = New RegExp
rex.Pattern = RegexIt(headerText)
For Each ran In Range("1:1")
If rex.test(ran.Text) Then
col = ran.Column
Exit For
End If
Next ran
MsgBox col
End Sub
Function RegexIt(what As String) As String
what = Replace(what, "(", "\(")
what = Replace(what, ")", "\)")
what = Replace(what, "[", "\[")
what = Replace(what, "]", "\]")
what = Replace(what, "<", "\<")
what = Replace(what, ">", "\>")
what = Replace(what, " ", "[\n ]?")
what = Replace(what, vbCrLf, "[\n ]?")
End Function
Good luck!
Edit: Reference to Microsoft VBScript Regular Expressions 5.5 required
Edit2: Edited for variable use. Explanation: Replace space in variable value with optionel space/line break, escape brackets for pattern matching.
Your code should work even if the header cell contains carriage returns:
Sub FindColumnWithTextInRowOne()
Dim headerText As String, newCol As Long
headerText = "whatever"
newCol = Range("1:1").Find(headerText).Column
MsgBox newCol
End Sub
This is because your use of Find() does not require a match to the WHOLE contents of the cell.
EDIT#1:
If the header cell was constructed using a formula, then a slightly different Find() should be used:
Sub FindColumnWithTextInRowOne()
Dim headerText As String, newCol As Long, r As Range
headerText = Range("H1").Text
newCol = Range("1:1").Find(What:=headerText, LookAt:=xlWhole, LookIn:=xlValues).Column
MsgBox newCol
End Sub
I've got a column which contains rows that have parameters in them. For example
W2 = [PROD][FO][2.0][Customer]
W3 = [PROD][GD][1.0][P3]
W4 = Issues in production for customer
I have a function that is copying other columns into another sheet, however for this column, I need to do the following
Search the cell and look for [P*]
The asterisk represents a number between 1 and 5
If it finds [P*] then copy P* to the sheet "Calculations" in column 4
Basically, remove everything from the cell except where there is a square bracket, followed by P, a number and a square bracket
Does anyone know how I can do this? Alternatively, it might be easier to copy the column across and then remove everything that doesn't meet the above criteria.
Second Edit:
I edited here to use a regular expression instead of a loop. This may be the most efficient method to achieve your goal. See below and let us know if it works for you:
Function MatchWithRegex(sInput As String) As String
Dim oReg As Object
Dim sOutput As String
Set oReg = CreateObject("VBScript.RegExp")
With oReg
.Pattern = "[[](P[1-5])[]]"
End With
If oReg.test(sInput) Then
sOutput = oReg.Execute(sInput)(0).Submatches(0)
Else
sOutput = ""
End If
MatchWithRegex = sOutput
End Function
Sub test2()
Dim a As String
a = MatchWithRegex(Range("A1").Value)
If a = vbNullString Then
MsgBox "None"
Else
MsgBox MatchWithRegex(Range("A1").Value)
End If
End Sub
First EDIT:
My solution would be something as follows. I'd write a function that first tests if the Pattern exists in the string, then if it does, I'd split it based on brackets, and choose the bracket that matches the pattern. Let me know if that works for you.
Function ExtractPNumber(sInput As String) As String
Dim aValues
Dim sOutput As String
sOutput = ""
If sInput Like "*[[]P[1-5][]]*" Then
aValues = Split(sInput, "[")
For Each aVal In aValues
If aVal Like "P[1-5][]]*" Then
sOutput = aVal
End If
Next aVal
End If
ExtractPNumber = Left(sOutput, 2)
End Function
Sub TestFunction()
Dim sPValue As String
sPValue = ExtractPNumber(Range("A2").Value)
If sPValue = vbNullString Then
'Do nothing or input whatever business logic you want
Else
Sheet2.Range("A1").Value = sPValue
End If
End Sub
OLD POST:
In VBA, you can use the Like Operator with a Pattern to represent an Open Bracket, the letter P, any number from 1-5, then a Closed Bracket using the below syntax:
Range("A1").Value LIke "*[[]P[1-5][]]*"
EDIT: Fixed faulty solution
If you're ok with blanks and don't care if *>5, I would do this and copy down column 4:
=IF(ISNUMBER(SEARCH("[P?]",FirstSheet!$W2)), FirstSheet!$W2, "")
Important things to note:
? is the wildcard symbol for a single character; you can use * if you're ok with multiple characters at that location
will display cell's original value if found, leave blank otherwise
Afterwards, you can highlight the column and remove blanks if needed. Alternatively, you can replace the blank with a placeholder string.
If * must be 1-5, use two columns, E and D, respectively:
=MID(FirstSheet!$W2,SEARCH("[P",FirstSheet!$W2)+2,1)
=IF(AND(ISNUMBER($E2),$E2>0,$E2<=5,MID($W2,SEARCH("[P",FirstSheet!$W2)+3,1))), FirstSheet!$W2, "")
where FirstSheet is the name of your initial sheet.
I am trying to take a string in a rich text box and replace them with a different string.
Now how this should work is that if two same characters are entered into the text box
e.g tt the "tt" will be replaced with "Ǿt" , it adds back one of the t's to the replaced string. Only the most recently entered string is manipulated if two same characters are entered .
I read the LAST string that is in the RichTextBox by using this method
Dim laststring As String = RichTextBox1.Text.Split(" ").Last
'hitting space bar breaks the operation so if i enter t t there will be no replacement
this is the replacement method which I use , it works correctly .
if laststring = "tt"
RichTextBox1 .Text = RichTextBox1 .Text.Replace("tt", "Ǿt")
This method is inefficient because i need to check id there are double letters for all letters and if i was to use this method it would tavke up a lot of code .
how can I accomplish this using a shorter method??
You need to put the if then section in a loop.
Dim holdstring As String
Dim doubleinstance() As String = {"bb", "tt", "uu"} ' array
Dim curstring As String = RichTextBox1.Text.Split(" ").Last
For Each item As String In doubleinstance
If RichTextBox1.Text.EndsWith(item) Then
holdstring = RichTextBox1.Text.Split(" ").Last.Length - 1 ' change to subtract 1 character from doubleinstance
RichTextBox1.Text = RichTextBox1.Text.Replace(curstring, "Ǿt" & holdstring)
MsgBox(curstring)
End If
Next item
Here's a bit of code to get you in the right direction...
There are a couple of variations of .Find, but you probably want to look at the .Select method.
With RichTextBox1
.Find("Don")
.SelectedText = "Mr. Awesome"
End With
Here is a way I came up with
Dim holdstring As String
Dim doubleinstance() As String = {"bb", "tt", "uu"} ' array
Dim curstring As String = RichTextBox1.Text.Split(" ").Last
If curstring = doubleinstance(0) And RichTextBox1.Text.EndsWith(doubleinstance(0)) Then
holdstring = RichTextBox1.Text.Split(" ").Last.Length - 1 ' change to subtract 1 character from doubleinstance
RichTextBox1.Text = RichTextBox1.Text.Replace(curstring, "Ǿt" + holdstring)
MsgBox(curstring)
End If
where i have doubleinstance(0) how do i get the if statement to not only check a single index but all of the index from 0 to 2 in this example ?