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04-10-2008
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How to Select or cut from the certain filed to the end of the line
Hi:
I have few rows in file..Like suppose... 9063C0 44 00051363603253033253347 3333 070248 06 9063C0 5G PAN00013 9063C0 44 00061030305040404250724 0506 100248 08 9063C0 43 01 00000089 I need to cut the row starting after "9063C0" till the end of the line and store in another file... My o/P is: 44 00051363603253033253347 3333 070248 06 5G PAN00013 44 00061030305040404250724 0506 100248 08 43 01 00000089 |
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04-10-2008
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Use grep and sed combination
Hi Rajesh,
Use grep and sed combinations to do the need. grep ^9063C0 test | sed 's/^9063C0//' where test is the file name. OUTPUT : =========== $> grep ^9063C0 test 9063C0 44 00051363603253033253347 3333 070248 06 9063C0 5G PAN00013 9063C0 44 00061030305040404250724 0506 100248 08 9063C0 43 01 00000089 $> grep ^9063C0 test | sed 's/^9063C0//' 44 00051363603253033253347 3333 070248 06 5G PAN00013 44 00061030305040404250724 0506 100248 08 43 01 00000089 Hope this is what u wanted...... Last edited by helper; 04-10-2008 at 07:20 AM.. |
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04-10-2008
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Hi Rajesh,
Use grep and sed combinations to do the need. grep "9063C0" test | sed 's/9063C0//' where test is the file name. OUTPUT : =========== $> grep ^9063C0 test 9063C0 44 00051363603253033253347 3333 070248 06 9063C0 5G PAN00013 9063C0 44 00061030305040404250724 0506 100248 08 9063C0 43 01 00000089 $> grep ^9063C0 test | sed 's/9063C0//' 44 00051363603253033253347 3333 070248 06 5G PAN00013 44 00061030305040404250724 0506 100248 08 43 01 00000089 Hope this is what u wanted...... |
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04-10-2008
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Code:
awk '{ $1=""; print }' file
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04-10-2008
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Code:
cut -d" " -f 2- test |
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04-10-2008
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Thankx all... Its working.........
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