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| Thread | Thread Starter | Forum | Replies | Last Post |
| search and replace | a.suryakumar | Shell Programming and Scripting | 1 | 04-28-2008 08:30 AM |
| Perl: Search for string on line then search and replace text | Crypto | Shell Programming and Scripting | 4 | 01-04-2008 07:24 AM |
| search and replace | dannyd | UNIX for Dummies Questions & Answers | 18 | 04-17-2007 04:42 PM |
| sed search and replace | d__browne | UNIX for Dummies Questions & Answers | 7 | 04-26-2006 06:46 AM |
| search and Replace | mukeshannamalai | UNIX for Advanced & Expert Users | 4 | 09-14-2001 03:21 AM |
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#1
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vi search/replace using a set
Hi,
I'm trying to do a global search/replace in vi using a set - I want to find every occurance of a carriage return followed by a character and replace it with a space. I've tried the following: :%s/\n[a-z][A-Z]/ [a-z][A-Z]/g It does the search ok, but it replaces the characters with the literal value "[a-z][A-Z]" How do i escape the replace condition to use the set? Thanks a million! -Hud |
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#2
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One point: your search string, as you have it, would find \n followed by a lowercase letter followed by an uppercase letter. To match any single character, whether upper- or lowercase, use only a single pair of square brackets.
As for what you want to do, try the following which tagged the "character" subexpression: Code:
:%s/\n\([a-zA-Z]\)/ \1/g |
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#3
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You are my hero
You rule!
That totally worked! I did end up changing it to this though to expand it even more: :%s/\n\([^0-9]\)/ \1/g Thanks a million! -Hud |
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