Shell Programming and Scripting

Hi guys,

I need to search for a particular pattern within the first 5 chars of the line. If the pattern is found then output the whole line. Unfortunately grep only searches the whole line

For example if i am searching for aaaaa in these lines

aaaaabbbbbccccc
bbbbbcccccaaaaa

grep would return both lines but i only want the first line. Also the search pattern (aaaaa) needs to be read from an input file as there could be more than one search patern

Any tips!!!

Code:

egrep '^.{0,4}aaaaa' file

... assuming you mean it can start in any position from 0 through 5. If the first five characters need to be precisely aaaaa then you don't even need egrep.

Code:

 grep '^aaaaa' file

To read the patterns from a file, something like

Code:

sed -e 's%^%^.{0,4}%' patternfile | egrep -f - file

If your egrep doesn't permit -f - or doesn't have the -f option, there are workarounds, of course.

Grep can evaluate based on regular expressions as well. See the -E flag.

Code:

grep -E "^aaaaa" <file.to.search>

I think you can combine the -f flag as well to read inputs from a file.

Using sed

Code:

sed -n "/^aaaaa/p" <file.to.search>

So if i wanted to read in the input file (ie file full of patterns) it would be something like this i assume :-

cat input file || grep '^aaaaa' file

What does the ^ do exactly ?

sorry being dum vino - how do i pass in the pattern file ? Would they below work ?

would grep -f {pattern.file} -E "^" {file.to.search}

Code:

while read -r line
do
 case $line in
    aaaaa* ) echo $line;;
 esac
done < file

The sed script I posted above modifies each line in patternfile suitably, and feeds the result to egrep -f, so it should be a complete solution to your problem. The ^ in a regular expression is an "anchor" which matches beginning of line. Read an introduction to regular expressions to learn more. Here's that script again, adjusted to my current understanding of your requirements.

Code:

sed 's/^/^/' patternfile | grep -f - file