grepping a specific column ???

[HP System Handle Owner]

Can anyone advise on a command or script which will read through a large file line by line extracting lines which contain a specific strings of characters which appear between certain places on a line.

e.g grep abcde filename1 > filename2 works, but I only want not look in characters 5-10 of each line and no where else on the line.

Many thanks

[John Poff]

Try this:

cut -c5-10 filename1 | grep findthis

[Kevin Wright · Accepted solution]

you need to use awk for this, if I understand you question correctly.
columns start at $0, $1, $2
so to print the second column
awk '{print $1}'
whitespace is the default deliminator, to change this use -F .

[Vincenzo Restuccia]

awk '{ print $5,$10 }'filename1|grep file_find

[MANOJ SRIVASTAVA]

Hi

try this
cat filename | awk '{ print $5 } | cut -c5-10

Manoj Srivastava

[Rodney Hills]

How about

awk 'substr($0,5,5)="datum"{print $0}' your_file

awk can compare a substring and then print only thos lines
or

perl -nie 'print if substr($_,4,5) eq "datum"'

perl can also compare a substring and then print

[Carlos Fernandez Riera]

awk ' $2 == "data" { print $0 }' file.

[Bruce Regittko_1]

Hi,

Try using a regular expression:

grep ".....abcde" file1 > file2

The . wil match any single character. You could probably get away with the following, which would guard against the wrong number of "dots":

grep ".\{5\}abcdef" file1 > file2

See the regexp(5) and the grep(1) man pages for more information.

--Bruce