UNIX for Dummies Questions & Answers

I have a fixed length file in the following format

<date><product_code><other data>

The file size is huge and I have to extract only the lines that match a certain product code which is of 2 bytes length. I cannot use normal grep since that may give undesirable results. When I search for prod cd 20, if I give grep 20, it matches with date 2008... Since this is a fixed length file, I know the position where to search. Is there any command to search at particular position and extract the matching lines?

use the following command :

awk 'substr($0,9,2)=="02" {print}' <filename>

That works fine. Much appreciated.

I am curious to know. Are there any alternate ways to do it other than using awk? Like sed, grep or cut. Do any other unix utility provide the option of searching at a particular position and length?

Code:

sed -n -e 's/^[0-9]\{8\}20*/&/p' file

Is there a generic syntax of this?

sed -n -e 's/^[0-9]\{8\}20*/&/p' file

I understand as you are skipping until 8th position and do a pattern search with 20*. I do not understand how the [0-9] and & help the search.

Code:

> cat parts
01010602grape
02020701apple
02020602banana
05250607pear
12250702melon
01010805banana02
10130202kiwi

Code:

> cat parts | grep "^......02"
01010602grape
02020602banana
12250702melon
10130202kiwi

Note the
. to wildcard for a character position
^ to wildcard starting at left column