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Return all characters to the left of the last delimeter of each line

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# 1  
Old 06-28-2012
Return all characters to the left of the last delimeter of each line
Hello,

Working on a ksh script and a little stumped... how can I return all characters to the left of the last delimeter per line in a file, skipping any lines without that delimeter?

ie, sample.txt:

Code:
Once_upon-a-Midnight_dreary_while
I pondered, weak_and weary
over many a quaint
and curious volume_of_forgotten lore,

while I nodded, nearly_napping, suddenly_
there came _ a tapping
_as of someone gently rapping, rapping_at my
_chamber door.

Tis some visitor! I muttered_rapping at my chamber door.
Only this, and nothing more.

If delimeter is "_", output should be:

Code:
Once_upon-a-Midnight_dreary
I pondered, weak
and curious volume_of
while I nodded, nearly_napping, suddenly
there came 
_as of someone gently rapping, rapping

Tis some visitor! I muttered

Thanks for any pointers.
# 2  
Old 06-28-2012
Code:
sed -n 's/_[^_]*$//p' infile

This User Gave Thanks to radoulov For This Post:
MoreCowbell
# 3  
Old 06-28-2012
Thank you - would you also use sed if performing the same string manipulation on a variable in a script?
# 4  
Old 06-28-2012
I wouldn't. It's more efficient to use parameter expansion itself.
Code:
${var%_*}

Regards,
Alister
This User Gave Thanks to alister For This Post:
MoreCowbell
# 5  
Old 06-28-2012
neat. Thanks.
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