grep or cat using sed

Shell Programming and Scripting
1

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.

2

It might be a bit easier with awk - if by smaller, you mean numerically

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

If that's not what you were asking, please provide some input data. Thanks.

3

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!!!!!:smiley:

4 
sed '/^[4-9][0-9]\{8\}/d' urfile