Shell Programming and Scripting

Hi. The data file is as below:

2000922111111100232091212098324....
2123011230912832094820943684896....
3435983453409583405938453049583....
.
.
.

I need to get only the rows that match my criteria. For example: those at characters 5-10 should equal to "922111" (thus getting only the 1st row above).

As my script processes a lot of these files, I only need to check for this string first before I run succeeding codes to minimize run time. An "if-cut" condition should work but I don't want to do this per line (again, due to run time considerations).

Hope you can help me with an alternative solution.

Thanks!!!

Code:

awk 'substr($0,5,6)=="922111"' file

Sorry, but I forgot to mention that I need to filter a number range for that string. Example, I only need to show the files/rows that have chars 5-10 withiin the range of 922000 to 9229999.

Hope you can help.

Code:

awk 'substr($0,5,6)>="922000" && substr($0,5,6)<="9229999"' file

I knew it! Thanks Krishmaths!

awk 'substr($0,5,6)>="922000" && substr($0,5,6)<="9229999"' file
This will not work, since you are testing 6 characters substr($0,5,6) against a number of 7 digit number 9229999
Not sure if you like 6 or 7 digit numbers 922000 vs 9229999

Another solution:

Code:

awk 'substr($0,5,7)~/922[0-9][0-9][0-9][0-9]/' file

or:

Code:

awk --posix 'substr($0,5,7)~/922[0-9]{4}/' file

Sorry for the typo, but my earlier msg should've read 922000 to 922999 (fixed length).

It's working now. Thanks!