Shell Programming and Scripting

Dear Friends,

I have a flat file which is as follows

Code:

$cat sample
123,456,1,1,1,1
sdfas,345,1,1,1,1
dfgd,234,2,3,4,1
ggffgr,234,4,3,2,1
jkhu,354.1,1,1,1
$

I want to get output of only those lines which has '1' in 3 to 5 position.

So I want output as follows

Code:

123,456,1,1,1,1
sdfas,345,1,1,1,1
jkhu,354.1,1,1,1

Kindly guide.
Anu.

Moderator's Comments:

... more code tags.

What have you tried so far?

I didn't try anything as I know way out only by using "If" statement which unfortunately I do not want to use.
Hence seeking for guidance who know grep well.

Please search forum before posting, kind of question you asked now, repeated several times on fora with different data,

you may try awk, its easy

Code:

[akshay@localhost tmp]$ cat sample
123,456,1,1,1,1
sdfas,345,1,1,1,1
dfgd,234,2,3,4,1
ggffgr,234,4,3,2,1
jkhu,354.1,1,1,1

[akshay@localhost tmp]$ awk -F, '$3 == 1 &&  $4 == 1 && $5 == 1' sample
123,456,1,1,1,1
sdfas,345,1,1,1,1
jkhu,354.1,1,1,1

[akshay@localhost tmp]$ awk  -F, '{j=1;for(i=3; i<=5; i++)j*=$i==1}j' sample
123,456,1,1,1,1
sdfas,345,1,1,1,1
jkhu,354.1,1,1,1

So if your file has several fields that you can create an expression for, then that should do it.

If the separator is , then an ignored field is .*, meaning zero or more (*) of any character (.) followed by the field separator (,)

So, to count from the beginnig of the line your expression starts as ^.*,.*, to signify start of record (^) the ignore two fields. You can then tag on 1,1,1, to specify your requirements and the rest doesn't matter if it matches or not.

I think you can end up with:-

Code:

egrep "^.*,.*,1,1,1," input_file

From your sample input, I get one less line because the one starting jkhu does not have the correct field separator between fields 2 & 3.



I hope that this helps,
Robin

Code:

awk -F, '$3==$4==$5==1' sample

Thank you friends for the help which was much needed. Special thanks to Mr. rbatte1 for taking extra efforts for step by step guiding.
Thank you.