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vi part-pattern matching and deletion

 
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# 1  
Old 03-29-2005
vi part-pattern matching and deletion
Hi,

I have a log file which shows the files which has been changed over the last week.

They follow this pattern:

old_file_version_number@@new_file_version_number

Now I need to know how to delete from each line parts starting from @@.
I would be issuing the command inside vi(m).

So in the end the log file will contain just:

old_file_version_number

Thanks,
Vino
# 2  
Old 03-29-2005
Usually I would suggest you search the forums but I found that it took me over an hour to find the correct (hopefully correct) solution.

To effect the complete file:
:1,$s/@@.*$//g

I'm sure those that know this better than I do (which would be most) can correct it if this is the wrong answer. But I tested and I believe it works for what you asked for.
# 3  
Old 03-29-2005
That will work or

:%s/@@.*$//g
will do the same
# 4  
Old 03-29-2005
Quote:
Originally Posted by reborg
That will work or

:%s/@@.*$//g
will do the same
actually there's no need to the trailing 'g' as there's ONLY one '$' [end of line] in a line:

:%s/@@.*$//
# 5  
Old 03-29-2005
Quote:
Originally Posted by vgersh99
actually there's no need to the trailing 'g' as there's ONLY one '$' [end of line] in a line:

:%s/@@.*$//
Very true, force of habit.
# 6  
Old 04-04-2005
Kool...thanks to all.
That worked.

Vino
 
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