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MS Office Forum / Word / Programming / February 2005A bit of A challenge for me to do.
Hello From Steved 4.29x5x Band Of Colours Using the below Macro will return Band Of Colours Which is what I want 4.Band Of Colours Using the below Macro will return Of Colors missing the word Band Please what do I need to do to the macro to allways give me Band Of Colours Dim s As String Dim p As Integer Selection.Paragraphs(1).Range.Select s = Selection.Text p = InStr(1, s, " ") p = InStr(p + 1, s, " ") s = Right(s, Len(s) - p) Selection.Paragraphs(1).Range.Text = s Thankyou. Answered last week in another forum. > Hello From Steved > [quoted text clipped - 19 lines] > > Thankyou. It might look similar but it is not the sme question. Thankyou. >-----Original Message----- >Answered last week in another forum. [quoted text clipped - 24 lines] > >. You'll need to explain things a little more clearly. Is this a generic task, or does the sentence always contain "band of colours" or some other oblique reference to Sherlock Holmes? > It might look similar but it is not the sme question. > [quoted text clipped - 32 lines] >> >>. Hello Jezebel from Steved Every second line is what I trying to do with the macro 1. 17138 Drovers Call (6) 2g b Drovers Call (6) 2g b 2. 4349 Coup Hogan (4) 2g ch Coup Hogan (4) 2g ch 3. 3. The Guttersnipe (7) 2g gr The Guttersnipe (7) 2g gr 1. 23 Aim To Score (6) 4g ch Aim To Score (6) 4g ch 3. 29x5x The Oracle (4) 5g br The Oracle (4) 5g br Thankyou. >-----Original Message----- >You'll need to explain things a little more clearly. Is this a generic task, [quoted text clipped - 39 lines] > >. Appears to be exactly the same as the problem you posted last week: delete everything up to and including the second space in each line. As suggested then, use a regular expression. > Hello Jezebel from Steved > [quoted text clipped - 65 lines] >> >>. Below Jezebel is what you done for me and I am very grateful, but using the below Deletes the word Band. On all others I use your find and replace it is perfect. Find: [! ]@ [! ]@ ([!^013]@[^013]) Replace with: \1 4. Band Of Colours (AUS) (2) 2g b Of Colours (AUS) (2) 2g b Hopefully you now see my issue. >-----Original Message----- >Appears to be exactly the same as the problem you posted last week: delete [quoted text clipped - 72 lines] > >. Of course. The spec was to delete up to and including the second space, which in this case comes after the word 'Band' -- that's the issue with regular expressions ... the text you're processing has to be, by definition, regular in some specifiable sense. Hence my subsequent question ... what is the regular element in the strings you are checking? > Below Jezebel is what you done for me and I am very > grateful, but using the below Deletes the word Band. [quoted text clipped - 90 lines] >> >>. Hello Jezebe from Steved The criteria I require is delete everything up to the first word, as in examples below. 1. 17138 Drovers Call (6) 2g b Drovers Call (6) 2g b 3. 29x5x The Oracle (4) 5g br The Oracle (4) 5g br 4. Band Of Colours (AUS) (2) 2g b Band Of Colours (AUS) (2) 2g b Thankyou. >-----Original Message----- >Of course. The spec was to delete up to and including the second space, [quoted text clipped - 96 lines] > >. Please forgive me for jumping in here... As Jezebel said, in order to delete material from a string you need to establish what the rules are for deletion. In this case, there don't seem to be any consistent string-related rules (you can't delete up to the second <space> for example because some strings only have one <space> before the material to be retained). If it's correct that the material you want to keep ALWAYS begins with a capital letter A-Z, then you could use VBA (not wildcard search) to increment through the string from left to right to establish the position of the first instance of a capital letter, then trim off everything to the left of that position. There may be other solutions but that's the only one that strikes me having read this thread. > Hello Jezebe from Steved > [quoted text clipped - 124 lines] > > > >. Hello Chuck from Steved Thankyou for your reply So Is it not possible to write a VBA to search for the first word in a String or if that is possible I would be happy. Cheers. >-----Original Message----- >Please forgive me for jumping in here... [quoted text clipped - 144 lines] >> >. You need to define 'word' in this context. If you mean a string beginning with a capital letter, or comprising letters only, then Find and Replace will do it, as will VBA. Here's another VBA -- Dim pWords() as string pWords = Split(Range.Text, " ") For i = 0 to ubound(pWords) if pWords(i) like "[a-zA-Z]{1,}" then pStartWord = i exit for end if Next Output = pWords(pStartWord) For i = pStartWord + 1 to ubound(pWords) Output = Output & " " & pWords(i) Next > Hello Chuck from Steved > [quoted text clipped - 174 lines] >>> >>. Hello Jezebel from Steved Jezebel as you would have quessed and that is that I am not very good at explaning my issue here. I would like to thankyou for your patience. Cheers. >-----Original Message----- >You need to define 'word' in this context. If you mean a string beginning [quoted text clipped - 195 lines] > >. Hi Jezebel When I tested your code it worked fine except that it kept returning the original string as Output. However if I amended the line containing the Like operator it worked fine, eg: Replace: if pWords(i) like "[a-zA-Z]{1,}" then with: if pWords(i) like "[a-zA-Z]*" then Unless I'm missing something? I also took the liberty of making a Function from your code: Function StripStringToWords(sString As String) Dim pWords() As String Dim Output As String Dim i As Long Dim pStartWord As Long pWords = Split(sString, " ") For i = 0 To UBound(pWords) If pWords(i) Like "[a-zA-Z]*" Then pStartWord = i Exit For End If Next Output = pWords(pStartWord) For i = pStartWord + 1 To UBound(pWords) Output = Output & " " & pWords(i) Next StripStringToWords = Output End Function Chuck > You need to define 'word' in this context. If you mean a string beginning > with a capital letter, or comprising letters only, then Find and Replace [quoted text clipped - 192 lines] > >>> > >>.
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