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I have to remove 03,reocd,49 which contain 000000000000000000 ,

 
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Top Forums UNIX for Dummies Questions & Answers I have to remove 03,reocd,49 which contain 000000000000000000 ,
# 1  
Old 07-04-2011
I have to remove 03,reocd,49 which contain 000000000000000000 ,
Hi

This is my inuput file
Code:
01,011600033,011600033,110516,0801,3,90,,2/
02,011600033,011103093,1,110317,0801,,2/
03,000000000000000000,,056,+00000000000,,,703,+00000000000,,,904,+00000000000,,/
49,+00000000000000000,00002/
03,004000100015860001,,056,+00029000000,,,703,+00150000000,,,904,+00000054375,,/
49,+00000000179054375,00002/

I have to remove 03,reocd,49 which conatain 000000000000000000 ,

My out put like this
Code:
01,011600033,011600033,110516,0801,3,90,,2/
02,011600033,011103093,1,110317,0801,,2/
03,004000100015860001,,056,+00029000000,,,703,+00150000000,,,904,+00000054375,,/
49,+00000000179054375,00002/

Last edited by pludi; 07-04-2011 at 07:08 AM..
# 2  
Old 07-04-2011
well, your 49 record actually conatins one less zero than you stated you're looking for, but with that caveat (and using the shorter string, would :

Code:
#  egrep -v "^(03|49).*00000000000000000" infile
01,011600033,011600033,110516,0801,3,90,,2/
02,011600033,011103093,1,110317,0801,,2/
03,004000100015860001,,056,+00029000000,,,703,+00150000000,,,904,+00000054375,,/
49,+00000000179054375,00002/

suffice for your needs ?
# 3  
Old 07-04-2011
Sed way..(based on your input file)
Code:
sed '/00000000000*/d' inputfile > outfile

# 4  
Old 07-05-2011
Thanks for u r reply
i have a one question insted of give outputfile that will reflect on input file
# 5  
Old 07-05-2011
If you are using GNU sed then you can use -i switch to make the change directly.
Otherwise you can use perl.

Or with sed you can use formulation like :
Code:
sed 's/0000000000*/d' inputfile > outfile.tmp && mv outfile.tmp inputfile

# 6  
Old 07-05-2011
use

Code:
 
sed -i

 
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