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Shell Programming and Scripting

grep or cat using sed

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# 1  
Old 07-13-2010
grep or cat using sed
Is there a way using grep or cat a file to create a new file based on whether the first 9 positions of each record is less than 399999999?

This is a fixed file format.
Last edited by ski; 07-13-2010 at 02:03 PM..
# 2  
Old 07-13-2010
It might be a bit easier with awk - if by smaller, you mean numerically

Code:
awk 'substr($0,1,9) < 399999999' file > new_file

If that's not what you were asking, please provide some input data. Thanks.
# 3  
Old 07-13-2010
Thanks so much....will try that now. I knew there had to be an easier way then reading the file.

---------- Post updated at 12:13 PM ---------- Previous update was at 12:07 PM ----------

Perfect! Perfect! Thanks again!!!!!Smilie
# 4  
Old 07-14-2010
Code:
sed '/^[4-9][0-9]\{8\}/d' urfile

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