[bash] Grep characters before and after match?

Using this:

grep -A1 -B1 "test_pattern" file

will produce one line before and after the matched pattern in the file. Is there a way to display not lines but a specified number of characters?

The lines in my file are pretty big so I am not interested in printing the entire line but rather only observe the match in context. Any suggestions on how to do this?

grep -E -o ".{0,5}test_pattern.{0,5}" test.txt 

This will match up to 5 characters before and after your pattern. The -o switch tells grep to only show the match and -E to use an extended regular expression. Make sure to put the quotes around your expression, else it might be interpreted by the shell.

With gawk , you can use match function:

    x="hey there how are you"
    echo "$x" |awk --re-interval '{match($0,/(.{4})how(.{4})/,a);print a[1],a[2]}'
    ere   are

If you are ok with perl, more flexible solution : Following will print three characters before the pattern followed by actual pattern and then 5 character after the pattern.

echo hey there how are you |perl -lne 'print "$1$2$3" if /(.{3})(there)(.{5})/'
ey there how

This can also be applied to words instead of just characters.Following will print one word before the actual matching string.

echo hey there how are you |perl -lne 'print $1 if /(\w+) there/'
hey

Following will print one word after the pattern:

echo hey there how are you |perl -lne 'print $2 if /(\w+) there (\w+)/'
how

Following will print one word before the pattern , then the actual word and then one word after the pattern:

echo hey there how are you |perl -lne 'print "$1$2$3" if /(\w+)( there )(\w+)/'
hey there how

I'll never easily remember these cryptic command modifiers so I took the top answer and turned it into a function in my ~/.bashrc file:

cgrep() {
    # For files that are arrays 10's of thousands of characters print.
    # Use cpgrep to print 30 characters before and after search patttern.
    if [ $# -eq 2 ] ; then
        # Format was 'cgrep "search string" /path/to/filename'
        grep -o -P ".{0,30}$1.{0,30}" "$2"
    else
        # Format was 'cat /path/to/filename | cgrep "search string"
        grep -o -P ".{0,30}$1.{0,30}"
    fi
} # cgrep()

Here's what it looks like in action:

$ ll /tmp/rick/scp.Mf7UdS/Mf7UdS.Source

-rw-r--r-- 1 rick rick 25780 Jul  3 19:05 /tmp/rick/scp.Mf7UdS/Mf7UdS.Source

$ cat /tmp/rick/scp.Mf7UdS/Mf7UdS.Source | cgrep "Link to iconic"

1:43:30.3540244000 /mnt/e/bin/Link to iconic S -rwxrwxrwx 777 rick 1000 ri

$ cgrep "Link to iconic" /tmp/rick/scp.Mf7UdS/Mf7UdS.Source

1:43:30.3540244000 /mnt/e/bin/Link to iconic S -rwxrwxrwx 777 rick 1000 ri

The file in question is one continuous 25K line and it is hopeless to find what you are looking for using regular grep.

Notice the two different ways you can call cgrep that parallels grep method.

There is a "niftier" way of creating the function where "$2" is only passed when set which would save 4 lines of code. I don't have it handy though. Something like ${parm2} $parm2. If I find it I'll revise the function and this answer.

You could use

awk '/test_pattern/ {
    match($0, /test_pattern/); print substr($0, RSTART - 10, RLENGTH + 20);
}' file

You can use regexp grep for finding + second grep for highlight

echo "some123_string_and_another" | grep -o -P '.{0,3}string.{0,4}' | grep string

23_string_and

You mean, like this:

grep -o '.\{0,20\}test_pattern.\{0,20\}' file

?

That will print up to twenty characters on either side of test_pattern. The \{0,20\} notation is like *, but specifies zero to twenty repetitions instead of zero or more.The -o says to show only the match itself, rather than the entire line.