Shell Programming and Scripting

Hi All,

I have a file called test.flg.fl and the data contains in the file as below

mmddyy_list20.prn
mmddyy_list21.prn
mmddyy_list22.prn
mmddyy_list23.prn
mmddyy_list24.prn
mmddyy_list25.prn
mmddyy_list26.prn
mmddyy_list27.prn
mmddyy_list28.prn
mmddyy_list29.prn
mmddyy_list3.prn
mmddyy_list30.prn
mmddyy_list31.prn
mmddyy_list32.prn
mmddyy_list33.prn
mmddyy_list34.prn
mmddyy_list35.prn
mmddyy_list36.prn
mmddyy_list37.prn
mmddyy_list38.prn
mmddyy_list42.prn

I need to delete lines which contain list21 to list 37 (i.e. list21, list22, list23 etc etc..). So my new file test.flg.fl will be now looks like below after I delete the lines.

mmddyy_list20.prn
mmddyy_list3.prn
mmddyy_list38.prn
mmddyy_list42.prn

Can you please tell me the command how to do this to delete those lines using Unix shell scripts? Please let me know if you need more informations.

Thank you!

If you are familiar with egrep and regular expressions, then just cat the file and pipe to egrep -v to remove what you don't need. Good exercise to learn this, plenty of material to find using google.

Code:

awk -F '[t.]' ' $2 < 21 || $2 > 37' test.flg.fl

Don't cat the file. Use the filename as the final argument to grep.

However, for this exercise, awk is better:

Code:

awk -F '_list' '
 { num = $2; sub( /.prn/, "", num ); num = num + 0 }
 num < 21 || num > 37
' "$FILE"