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How to delete all character before certain word

 
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# 1  
Old 01-30-2009
How to delete all character before certain word
Hi,

For example, i have a string

"123 456 789 abc 111 222 333"

and I would like to delete all the characters before abc so that it becomes

"abc 111 222 333"

how can i do that in unix? using sed?

note: I actually don't know how many words/charachters before "abc", so the "cut" command should not work.

Thanks.
Andrew
Last edited by phandy; 01-30-2009 at 01:24 AM.. Reason: Use better title
# 2  
Old 01-30-2009
Code:
$ echo "123 456 789 abc 111 222 333" | sed 's/^.*abc/abc/'
abc 111 222 333
$

# 3  
Old 01-30-2009
Yes, it works! Many thanks Corona688Smilie
# 4  
Old 01-30-2009
Oh! I found that this command will delete until the last occurance instead of first, meaning if

Code:
$ echo "123 456 789 abc 111 abc 222 333" | sed 's/^.*abc/abc/'

will return "abc 222 333".

Is it possible to return "abc 111 abc 222 333" instead? Many thanks.
# 5  
Old 01-30-2009
Any idea?Smilie
# 6  
Old 01-31-2009
sed can't do this, but perl can:

Code:
$ echo "123 456 789 abc 111 abc 222 333" | perl -p -e 's/^.*?abc/abc/'
abc 111 abc 222 333
$

# 7  
Old 01-31-2009
Quote:
Originally Posted by Corona688
sed can't do this, but perl can:

Code:
$ echo "123 456 789 abc 111 abc 222 333" | perl -p -e 's/^.*?abc/abc/'
abc 111 abc 222 333
$

With sed you do something like:

Code:
$
$ echo "123 456 abc 111 abc 222 abc 333" | sed 's/[^abc]*\(abc.*\)/\1/'
abc 111 abc 222 abc 333
$
$

Regards
 
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