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regex to remove commentaries and blanks

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# 1  
Old 06-18-2009
regex to remove commentaries and blanks at the end of the line
Hi all,

I need to prune a var's content as follows:

VAR='blah blah # seew seew'

NEWVAR='blah blah'

(without blanks)

I need also to perform this change by using variable substitution within bash shell. I've tried it with the following subst:

VAR2=${VAR/ \#*/}

but the blanks still remain... Any ideas are welcome.

Thanks in advance
Last edited by yomaya; 06-18-2009 at 07:34 AM..
# 2  
Old 06-18-2009
do you want something like this...
Code:
var2=`echo $var | tr -d ' '`

# 3  
Old 06-18-2009
Thanks, but the text to preserve has also blanks... Maybe I should have been more precise...

Code:
VAR='blah blah blah      # seew seew'
         ^    ^
        
NEWVAR='blah blah blah' 
            ^    ^     (<- should get preserved)

Last edited by yomaya; 06-18-2009 at 07:15 AM..
# 4  
Old 06-18-2009
I dont get you. Give some meaningful example.
Last edited by rakeshawasthi; 06-18-2009 at 07:30 AM..
# 5  
Old 06-18-2009
Quote:
Originally Posted by rakeshawasthi
I dont get you. Give some meanigful example.
Code:
VAR='blah blah blah      # seew seew'
         ^    ^
        
NEWVAR='blah blah blah' 
            ^    ^     (<- should get preserved)

The subchain that should be removed is:
Code:
'      # seew seew'

# 6  
Old 06-18-2009
something like this ??

Code:
echo "'blah blah blah      # seew seew'" | sed 's/\(.*\)#.*/\1/g' | sed 's/.[ ]*$//g'

# 7  
Old 06-18-2009
Quote:
Originally Posted by panyam
something like this ??

Code:
echo "'blah blah blah      # seew seew'" | sed 's/\(.*\)#.*/\1/g' | sed 's/.[ ]*$//g'

Thank you very much. It does the job properly!

Do you know if it's possible to do this without invoking 'sed' (for instance by parameter substitution)? The reason why it would be nicer doing it this way is that VAR is in fact an array (which I didn't mention for clarity) with a few hundred entries.
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