Use regular expressions for error in VBA-CodePudding

But in the regular expression through test, and the same statement using VBA in your WORD is wrong, what reason be excuse me? Code is different? Thank you

 Sub tests () 
Dim myString As the Variant, mt, mh, isr 
Application. ScreenUpdating=False 
The Set reg=CreateObject (" vbscript. Regexp ") 

MyString="me capital: RMB 123656.00, lowercase: 123654.03 intermediary capital: RMB 800.36 company certificate capital: RMB 36659.32. Japan and South Korea on the crystal" 
With reg 
The Pattern="(? <=RMB capital:) \ d + [\ d] {1, 3} 
"'the Pattern="\ d + (\. [0-9] {1, 2})" 
The Global=True 
The IgnoreCase=False 
The MultiLine=True 
The Set of mh=. The Execute (myString) 
End With 

For I=0 To mh. Count - 1 
Isr=isr & amp; Mh (I). SubMatches (0) & amp; "
"Next 

ActiveDocument. Content. The Text=isr 
Application. ScreenUpdating=True 
End Sub

. The Pattern="\ d + (\. [0-9] {1, 2})"
When test which can read the Numbers, only in the VBA. 00. 03, etc., not in front of an integer, but the test site test to read data

CodePudding user response:

 

Test (Sub) 
Dim myString As the Variant, mt, mh, isr 
Application. ScreenUpdating=False 
The Set reg=CreateObject (" vbscript. Regexp ") 

MyString="me capital: RMB 123656.00, lowercase: 123654.03 intermediary capital: RMB 800.36 company certificate capital: RMB 36659.32. Japan and South Korea on the crystal" 
With reg 
The Pattern="(RMB capital:) (\ d + \ [0-9] {1, 3})" 
The Global=True 
The IgnoreCase=False 
The MultiLine=True 
The Set of mh=. The Execute (myString) 
End With 

For I=0 To mh. Count - 1 
Isr=isr & amp; Mh (I). SubMatches (1) & amp; "
"Next 

ActiveDocument. Content. The Text=isr 
Application. ScreenUpdating=True 
End Sub

CodePudding user response:

Question, that is because the regular expression parser, you test the parser of the tool is more advanced, can explain the reverse matching, and vb and vba, VBS are older, such as the only support basic regular features, does not support & lt;=reverse match these new features,

Don't you is to get the capital behind things, actually very simple, direct write: capital: \ d +/\ d {1, 3} is good ah,

The following can also be:
Capital: [\ d \] +
(? Capital:) : [\ d \] +
(? Capital:) : (/\ \ d +)
(? Capital:) : (/\ \ d +) \.?