How to Select or cut from the certain filed to the end of the line

Shell Programming and Scripting
1

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

2

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......

3

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......

4 
awk '{ $1=""; print }' file
5 
cut -d" " -f 2- test
6

Thankx all... Its working.........