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I have a text file with thousands of lines of text. Each line ends like
\\server\share\file.txt -> information
I want to remove everything following the space after the end of the file extension. So everything after " ->" (there's a space after the first quote)
How would you do this? I'd like to use vim as I am trying to understand more about it, but any program will do; I'd like to get this done soon.
718
In Bash (and ksh, zsh, dash, etc.), you can use parameter expansion with % which will remove characters from the end of the string or # which will remove characters from the beginning of the string. If you use a single one of those characters, the smallest matching string will be removed.
N command reads the next line in the pattern space. d deletes the entire pattern space which contains the current and the next line. Using the substitution command s, we delete from the newline character till the end, which effective deletes the next line after the line containing the pattern Unix.
Your description is contradictory.
Did you want to keep the arrow or not?
1 is simply accomplished with:
:%s/ ->.*/
and actually if you really did want to keep the space before the arrow, like you said, it would be:
:%s/ ->.*/ /
2 can be done with:
:%s/\( ->\).*/\1/
If you prefer to view the results of your search before the replace you can build your search first using /:
/\( ->\).*
This will highlight all results to make sure you are replacing the right thing. You can then execute the replace command with an empty search term to use the last search (the stuff thats highlighted).
:%s//\1/
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