How to find lines containing a string and then printing those specific lines and something else

Question

I use the following command to recursively search multiple files and find the line number in each file in which the string is found.

    grep -nr "the_string" /media/slowly/DATA/lots_of_files > output.txt

The output is as follows:

    /media/slowly/DATA/lots_of_files/lots_of_files/file_3.txt:3:the_string
    /media/slowly/DATA/lots_of_files/lots_of_files/file_7.txt:6:the_string is in this sentence.
    /media/slowly/DATA/lots_of_files/lots_of_files/file_7.txt:9:the_string is in this sentence too.

As shown above, the output includes the filename, line number and all the text in that line including the string.

I have also figured out how to print just the specific lines of a files containing the string using the following command:

    sed '3!d' /media/slowly/DATA/lots_of_files/lots_of_files/file_3.txt > print.txt
    sed '6!d' /media/slowly/DATA/lots_of_files/lots_of_files/file_7.txt >> print.txt
    sed '9!d' /media/slowly/DATA/lots_of_files/lots_of_files/file_7.txt >> print.txt

I created the above commands manually by reading the line numbers and filenames

Here's my question.

Q1a

Is there a way to combine both steps into one command? I'm thinking piping the line number and the filename into sed and printing the line. I'm having a problem with the order in which the grep output is generated.

Q1b

Same as above but also print the 2 lines before and 2 lines after the line containing the string (total of 5 lines)? I'm thinking piping the line number and the filename into sed and printing all the required lines somehow.

Big thanks.

Answer 1

If I am understanding the question correctly, you can accomplish this with one grep command.

For Q1a, your grep output can suppress the filename using -h, e.g.:

grep -hnr "the_string" /media/slowly/DATA/lots_of_files > output.txt

For Q1b, your grep output can include lines preceding and following matched lines using -A and -B, e.g.:

grep -hnr -A2 -B2 "the_string" /media/slowly/DATA/lots_of_files > output.txt

The output will contain a separator between matches, which you can suppress with --no-group-separator, e.g.:

grep -hnr -A2 -B2 --no-group-separator "the_string" /media/slowly/DATA/lots_of_files > output.txt

Note that the output uses a different delimiter for matching lines (:) and context lines (-).

Answer 2

Your first question as far as I know can be answered by coming at grep a different way. When you send it a list of files (or directory to recurse through with -r or -R), it will always output which file it has found a match in as well as the line number. You can get around this with a construct such as:

find /path/to/files -type f | xargs grep -n 'the_pattern'

As for your second question, if you want to see the lines before and after a match, you can use the -C (for Context) switch:

grep -C2 'pattern' /path/to/file # displays the two lines before and after a match

Related to -C are -A (for After), and -B (for Before), which only give the specified number of lines after or before a match, respectively.

You can combine the two answers thusly:

find /path/to/files -type f | xargs grep -n -C2 'the_pattern'

As for your question about sed, the example you gave only works if you already know the line numbers. You can also do something like:

sed -n '/the_pattern/p' /path/to/files/*

(but it will not recurse into subdirectories)

Answer 3

find /media/slowly/DATA/lots_of_files -type f -exec grep -h -C2 'the_pattern' {} +

This will find things which are files (as opposed to directories or links) under the /media/slowly/DATA/lots_of_files directory. It will group them up (no need for xargs this decade) and run grep on them. grep will not print the filenames (-h) but will give 2 lines of context before and after the matching lines (-C2, use -A and -B for more precise control).

The advantage of this command over the one from @cherdt is you can add additional filters into the find command, for example you can choose not to go into directories like .git