I tried the following to replace all the text content in the current open document with numeric zero, but it doesn't work
Set objWdDoc = Word.Application.ActiveDocument
Set objWdRange = objWdDoc.Content
Dim re As New RegExp
re.Global = True
re.Pattern = "[a-z]"
re.IgnoreCase = True
objWdRange = re.Replace(objWdRange, "0")
Can anyone suggest a working method?
Assuming you have referenced microsoft vbscript regular expressions
objWdRange.Text = re.Replace(objWdRange, "0")
Will work, although you will of course lose any formatting.
You can also use the built-in search/replace which has limited support to find digits/characters. Record a macro of yourself doing this and you can examine the code.
Related
i'm trying to change formulas in excel, i need to change the row number of the formulas.
I'm trying do use replace regex to do this. I use an loop to iterate through the rows of the excel and need to change the formula for the row that is iterating at the time. Here is an exemple of the code:
For i = 2 To rows_aux
DoEvents
Formula_string= "=IFS(N19='Z001';'xxxxxx';N19='Z007';'xxxxxx';0=0;'xxxxxxx')"
Formula_string_new = regEx.Replace(Formula_string, "$1" & i)
wb.Cells(i, 33) = ""
wb.Cells(i, 33).Formula = Formula_string_new
.
.
.
Next i
I need to replace rows references but not the ones in quotes or double quotes. Example:
If i = 2 i want the new string to be this:
"=IFS(N2='Z001';'xxxxxx';N2='Z007';'xxxxxx';0=0;'xxxxxxx')"
I'm trying to use this regex:
([a-zA-Z]+)(\d+)
But its changing everything in quotes too. Like this:
If i = 2:
"=IFS(N2='Z2';'xxxxxx';N2='Z2';'xxxxxx';0=0;'xxxxxxx')"
If anyone can help me i will be very grateful!
Thanks in advance.
As others have written, there are probably better ways to write this code. But for a regex that will capture just the Column letter in capturing group #1, try:
\$?\b(XF[A-D]|X[A-E][A-Z]|[A-W][A-Z]{2}|[A-Z]{2}|[A-Z])\$?(?:104857[0-6]|10485[0-6]\d|1048[0-4]\d{2}|104[0-7]\d{3}|10[0-3]\d{4}|[1-9]\d{1,5}|[1-9])d?
Note that is will NOT include the $ absolute addressing token, but could be altered if that were necessary.
Note that you can avoid the loop completely with:
Formula_string = "=IFS(N19=""Z001"",""xxxxxx"",N$19=""Z007"",""xxxxxx"",0=0,""xxxxxxx"")"
Formula_string_new = regEx.Replace(Formula_string, "$1" & firstRow)
With Range(wb.Cells(firstRow, 33), wb.Cells(lastRow, 33))
.Clear
.Formula = Formula_string_new
End With
When we write a formula to a range like this, the references will automatically adjust the way you were doing in your loop.
Depending on unstated factors, you may want to use the FormulaLocal property vice the Formula property.
Edit:
To make this a little more robust, in case there happens to be, within the quote marks, a string that exactly mimics a valid address, you can try checking to be certain that a quote (single or double) neither precedes nor follows the target.
Pattern: ([^"'])\$?\b(XF[A-D]|X[A-E][A-Z]|[A-W][A-Z]{2}|[A-Z]{2}|[A-Z])\$?(?:104857[0-6]|10485[0-6]\d|1048[0-4]\d{2}|104[0-7]\d{3}|10[0-3]\d{4}|[1-9]\d{1,5}|[1-9])d?\b(?!['"])
Replace: "$1$2" & i
However, this is not "bulletproof" as various combinations of included data might match. If it is a problem, let me know and I'll come up with something more robust.
If you can identify some unique features like in the example preceding bracket ( or colon ; and trailing equal = then this might work
Sub test()
Dim s As String, sNew As String, i As Long
Dim Regex As Object
Set Regex = CreateObject("vbscript.regexp")
With Regex
.Global = True
.MultiLine = False
.IgnoreCase = True
.Pattern = "([(;][a-zA-Z]{1,3})(\d+)="
End With
i = 1
s = "=IFS(NANA19='Z001';'xxxxxx';NA19='Z007';'xxxxxx';0=0;'xxxxxxx')"
sNew = Regex.Replace(s, "$1" & i & "=")
Debug.Print s & vbCr & sNew
End Sub
What I want to do is to figure out how you use a regular expression to extract the inner most item from a HTML tag collection. That is:
TARGET TEXT
Function FindInnerHtml(Work As String) As String
Dim Results As String, myRegExp, myMatches As Object, thisMatch As Object
Let myRegExp = New RegExp
myRegExp.IgnoreCase = True
myRegExp.Global = True
myRegExp.Pattern = ">(.*?)<"
Set myMatches = myRegExp.Execute(Work)
If (myMatches.Count = 0) Then
Results = myMatches(0)
Results = Replace$(Replace$(Results, ">", ""), "<", "")
End If
FindInnerHtml = Results
End Function
What I do get from the function is the inner HTML, that is the target text, what I would rather be able to do is to ensure I'm not in need of adding that double replace$() to clean up the results.
It's crude and fails miserably for edge cases but something like this could work:
<[a-zA-Z]{1}[a-zA-Z\d]*>([^><]*)</[a-zA-Z]{1}[a-zA-Z\d]*>
$1 will contain the inner text
https://regex101.com/r/iuLdJV/3
I wish to select and add comments after certain words, e.g. “not”, “never”, “don’t” in sentences in a Word document with VBA. The Find/Replace with wildcards works fine, but “Use wildcards” cannot be selected with “Match case”. The RegEx can “IgnoreCase=True”, but the selection of the word is not reliable when there are more than one comments in a sentence. The Range.start seems to be getting modified in a way that I cannot understand.
A similar question was asked in June 2010. https://social.msdn.microsoft.com/Forums/office/en-US/f73ca32d-0af9-47cf-81fe-ce93b13ebc4d/regex-selecting-a-match-within-the-document?forum=worddev
Is there a new/different way of solving this problem?
Any suggestion will be appreciated.
The code using RegEx follows:
Function zRegExCommentor(zPhrase As String, tComment As String) As Long
Dim sTheseSentences As Sentences
Dim rThisSentenceToSearch As Word.Range, rThisSentenceResult As Word.Range
Dim myRegExp As RegExp
Dim myMatches As MatchCollection
Options.CommentsColor = wdByAuthor
Set myRegExp = New RegExp
With myRegExp
.IgnoreCase = True
.Global = False
.Pattern = zPhrase
End With
Set sTheseSentences = ActiveDocument.Sentences
For Each rThisSentenceToSearch In sTheseSentences
Set rThisSentenceResult = rThisSentenceToSearch.Duplicate
rThisSentenceResult.Select
Do
DoEvents
Set myMatches = myRegExp.Execute(rThisSentenceResult)
If myMatches.Count > 0 Then
rThisSentenceResult.Start = rThisSentenceResult.Start + myMatches(0).FirstIndex
rThisSentenceResult.End = rThisSentenceResult.Start + myMatches(0).Length
rThisSentenceResult.Select
Selection.Comments.Add Range:=Selection.Range
Selection.TypeText Text:=tComment & "{" & zPhrase & "}"
rThisSentenceResult.Start = rThisSentenceResult.Start + 1 'so as not to find the same phrase again and again
rThisSentenceResult.End = rThisSentenceToSearch.End
rThisSentenceResult.Select
End If 'If myMatches.Count > 0 Then
Loop While myMatches.Count > 0
Next 'For Each rThisSentenceToSearch In sTheseSentences
End Function
Relying on Range.Start or Range.End for position in a Word document is not reliable due to how Word stores non-printing information in the text flow. For some kinds of things you can work around it using Range.TextRetrievalMode, but the non-printing characters inserted by Comments aren't affected by these settings.
I must admit I don't understand why Word's built-in Find with wildcards won't work for you - no case matching shouldn't be a problem. For instance, based on the example: "Never has there been, never, NEVER, a total drought.":
FindText:="[n,N][e,E][v,V][e,E][r,R]"
Will find all instances of n-e-v-e-r regardless of the capitalization. The brackets let you define a range of values, in this case the combination of lower and upper case for each letter in the search term.
The workarounds described in my MSDN post you link to are pretty much all you can if you insist on RegEx:
Using the Office Open XML (or possibly Word 2003 XML) file format will let you use RegEx and standard XML processing tools to find the information, add comment "tags" into the Word XML, close it all up... And when the user sees the document it will all be there.
If you need to be doing this in the Word UI a slightly different approach should work (assuming you're targeting Word 2003 or later): Work through the document on a range-by-range basis (by paragraph, perhaps). Read the XML representation of the text into memory using the Range.WordOpenXML property, perform the RegEx search, add comments as WordOpenXML, then write the WordOpenXML back into the document using the InserXml method, replacing the original range (paragraph). Since you'd be working with the Paragraph object Range.Start won't be a factor.
I have a data column that has a heading value with multiple levels, where I only want the first three levels, but I cannot figure out how to get the parsed value?
I was reading this and it shows how to use create a function to return a boolean for the condition, but how would I create a function that would return a parsed value?
This is the Regular Expression that I think I need.
^(\d.\d.\d)
I'm looking for something that would change 1.2.3.4.5. to 1.2.3 and similar for any other header I have that has more than three levels.
Ideally, I'd like to be able to put it into my Query Design as a Field Expression, but I'm not sure how I would do that.
I assumed your input values could have more than one digit between the dots. In other words, I think you want this ...
? RegExpGetMatch("1.2.3.4.5.", "^(\d+\.\d+\.\d+).*", 1)
1.2.3
? RegExpGetMatch("1.27.3.4.5.", "^(\d+\.\d+\.\d+).*", 1)
1.27.3
If that is the correct behavior, here is the function I used.
Public Function RegExpGetMatch(ByVal pSource As String, _
ByVal pPattern As String, _
ByVal pGroup As Long) As String
'requires reference to Microsoft VBScript Regular Expressions
'Dim re As RegExp
'Set re = New RegExp
'late binding; no reference needed
Dim re As Object
Set re = CreateObject("VBScript.RegExp")
re.Global = True
re.Pattern = pPattern
RegExpGetMatch = re.Replace(pSource, "$" & pGroup)
Set re = Nothing
End Function
See also this answer by KazJaw. His answer taught me how to select the match group with RegExp.Replace.
In a query run within an Access session, you could use the function like this:
SELECT
RegExpGetMatch([Data Column], "^(\d+\.\d+\.\d+).*", 1) AS parsed_value
FROM YourTable;
Note however a custom VBA function is not usable for queries run from outside an Access session.
Try changing your RegEx to ^(\d\.\d\.\d). You need to escape the . since it has a special meaning in RegExp.
I have been having trouble with the Access key term LIKE and it's use. I want to use the following RegEx (Regular Expression) in query form as a sort of "verfication rule" where the LIKE operator filters my results:
"^[0]{1}[0-9]{8,9}$"
How can this be accomplished?
I know you were not asking about the VBA, but it maybe you will give it a chance
If you open a VBA project, insert new module, then pick Tools -> References and add a reference to Microsoft VBScript Regular Expressions 5.5. Given that pate the code below to the newly inserted module.
Function my_regexp(ByRef sIn As String, ByVal mypattern As String) As String
Dim r As New RegExp
Dim colMatches As MatchCollection
With r
.Pattern = mypattern
.IgnoreCase = True
.Global = False
.MultiLine = False
Set colMatches = .Execute(sIn)
End With
If colMatches.Count > 0 Then
my_regexp = colMatches(0).Value
Else
my_regexp = ""
End If
End Function
Now you may use the function above in your SQL queries. So your question would be now solved by invoking
SELECT my_regexp(some_variable, "^[0]{1}[0-9]{8,9}$") FROM some_table
if will return empty string if nothing is matched.
Hope you liked it.
I don't think Access allows regex matches (except in VBA, but that's not what you're asking). The LIKE operator doesn't even support alternation.
Therefore you need to split it up into two expressions.
... WHERE (Blah LIKE "0#########") OR (Blah LIKE "0########")
(# means "a single digit" in Access).