How to replace last 8 digits?


 
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# 1  
Old 06-03-2013
How to replace last 8 digits?

Hi,
How I can replace last 8 ZEROS with 22991231?
Code:
  
19523479811841494432A2013052700000000
19523479811730333980A2013052700000000
19523479811417044397A2013052700000000
19523479811205895810C2013010120130131
A9523479811205895810A2013020120130228
19523479811205895810I2013030120130331
19523479811689774317A2013052700000000
19523479811659473312C2013050120130531
19523479811659473312C2013060120130630

Thanks!

Last edited by Scott; 06-03-2013 at 01:30 PM.. Reason: Fixed code tags
# 2  
Old 06-03-2013
Code:
awk '{sub(/00000000$/, "22991231"); print}' oldfile > newfile

try that.
This User Gave Thanks to jim mcnamara For This Post:
# 3  
Old 06-03-2013
Code:
sed 's/........$/22991231/' file

This User Gave Thanks to Yoda For This Post:
# 4  
Old 06-03-2013
Thanks. It worked. Can I use column number with sed command?

Last edited by Scott; 06-03-2013 at 01:30 PM.. Reason: Removed code tags
# 5  
Old 06-03-2013
Nope. sed has no concept of 'column', doing so in a plain regex while possible would be a terrible mess.

awk does have columns. Use awk.

Code:
awk '{sub(/00000000$/, "22991231", $5); print}' oldfile > newfile

...for fifth column.
# 6  
Old 06-20-2013
How can you print the 5th column of that file using awk if there is no FS ?
# 7  
Old 06-21-2013
Code:
sed 's/^\(.\{29\}\)00000000/\122991231/' oldfile > newfile

The first 29 characters in \( \) are restored by \1 so the replacement starts at position 30.
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