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sed not working while passing the variable to print

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# 1  
Old 11-20-2012
sed not working while passing the variable to print
Hi

I have below sed to search and print particular line that matches in file and prints it, which is actually working fine as expected.

Code:
sed -n "/^2012-11-19 23:[0]*/p" filename

but if I replace date with the variable it seems it's not recognizing it and throwing out error

Code:
 v="2012-11-19"
sed -n 's/^'$v' 23:[0]*/p"  filename
 sed -n "s/^$v 23:[0]*/p"  filename

so can anyone help me resolving this??
# 2  
Old 11-20-2012
Quote:
Originally Posted by manas_ranjan
Code:
 v="2012-11-19"
sed -n 's/^'$v' 23:[0]*/p"  filename
 sed -n "s/^$v 23:[0]*/p"  filename


Try

Code:
v="2012-11-19" 
sed -n '/^'"$v"' 23:[0]*/p'  filename

Last edited by pamu; 11-20-2012 at 07:36 AM..
# 3  
Old 11-20-2012
hey pamu,

that was my mistake, actually there was (') instead of (") in my first sed with variable one you pointed out.

thanks for pointing out.

But anywas, I followed your suggestion and tried to execute as below
Code:
sed -n 's/^'"$v"' 23:[0]*/p' filename

then it pops up with
Code:
sed: -e expression #1, char 23: unterminated `s' command

error message
# 4  
Old 11-20-2012
Quote:
Originally Posted by manas_ranjan
Code:
sed -n 's/^'"$v"' 23:[0]*/p' filename

Please look at my post again. You won't find s there.Smilie
And i have highlighted s also in your previous code in red.(check post 2)

pamu
# 5  
Old 11-20-2012
thats so stupid of me, I have no idea why in the first place I even put s.
sorry for this!!

now all good..
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