[bash] grep for multiple strings in file on different lines (ie. whole file, not line based search)?

I want to grep for files containing the words Dansk, Svenska or Norsk on any line, with a usable returncode (as I really only like to have the info that the strings are contained, my one-liner goes a little further then this).

I have many files with lines in them like this:

Disc Title: unknown
Title: 01, Length: 01:33:37.000 Chapters: 33, Cells: 31, Audio streams: 04, Subpictures: 20
        Subtitle: 01, Language: ar - Arabic, Content: Undefined, Stream id: 0x20, 
        Subtitle: 02, Language: bg - Bulgarian, Content: Undefined, Stream id: 0x21, 
        Subtitle: 03, Language: cs - Czech, Content: Undefined, Stream id: 0x22, 
        Subtitle: 04, Language: da - Dansk, Content: Undefined, Stream id: 0x23, 
        Subtitle: 05, Language: de - Deutsch, Content: Undefined, Stream id: 0x24, 
(...)

Here is the pseudocode of what I want:

for all files in directory;
 if file contains "Dansk" AND "Norsk" AND "Svenska" then
 then echo the filename
end

What is the best way to do this? Can it be done on one line?

awk '/Dansk/{a=1}/Norsk/{b=1}/Svenska/{c=1}END{ if (a && b && c) print "0" }' 

you can then catch the return value with the shell

if you have Ruby(1.9+)

ruby -0777 -ne 'print if /Dansk/ and /Norsk/ and /Svenka/' file

This is a blending of glenn jackman's and kurumi's answers which allows an arbitrary number of regexes instead of an arbitrary number of fixed words or a fixed set of regexes.

#!/usr/bin/awk -f
# by Dennis Williamson - 2011-01-25

BEGIN {
    for (i=ARGC-2; i>=1; i--) {
        patterns[ARGV[i]] = 0;
        delete ARGV[i];
    }
}

{
    for (p in patterns)
        if ($0 ~ p)
            matches[p] = 1
            # print    # the matching line could be printed
}

END {
    for (p in patterns) {
        if (matches[p] != 1)
            exit 1
    }
}

Run it like this:

./multigrep.awk Dansk Norsk Svenska 'Language: .. - A.*c' dvdfile.dat

I did that with two steps. Make a list of csv files in one file With a help of this page comments I made two scriptless steps to get what I needed. Just type into terminal:

$ find /csv/file/dir -name '*.csv' > csv_list.txt
$ grep -q Svenska `cat csv_list.txt` && grep -q Norsk `cat csv_list.txt` && grep -l Dansk `cat csv_list.txt`

it did exactly what I needed - print file names containing all three words.

Also mind the symbols like `' "

Here's what worked well for me:

find . -path '*/.svn' -prune -o -type f -exec gawk '/Dansk/{a=1}/Norsk/{b=1}/Svenska/{c=1}END{ if (a && b && c) print FILENAME }' {} \;
./path/to/file1.sh
./another/path/to/file2.txt
./blah/foo.php

If I just wanted to find .sh files with these three, then I could have used:

find . -path '*/.svn' -prune -o -type f -name "*.sh" -exec gawk '/Dansk/{a=1}/Norsk/{b=1}/Svenska/{c=1}END{ if (a && b && c) print FILENAME }' {} \;
./path/to/file1.sh

How to grep for multiple strings in file on different lines (Use the pipe symbol):

for file in *;do 
   test $(grep -E 'Dansk|Norsk|Svenska' $file | wc -l) -ge 3 && echo $file
done

Notes:

1. If you use double quotes "" with your grep, you will have to escape the pipe like this: \| to search for Dansk, Norsk and Svenska.

2. Assumes that one line has only one language.

Walkthrough: http://www.cyberciti.biz/faq/howto-use-grep-command-in-linux-unix/

Yet another way using just bash and grep:

For a single file 'test.txt':

  grep -q Dansk test.txt && grep -q Norsk test.txt && grep -l Svenska test.txt

Will print test.txt iff the file contains all three (in any combination). The first two greps don't print anything (-q) and the last only prints the file if the other two have passed.

If you want to do it for every file in the directory:

   for f in *; do grep -q Dansk $f && grep -q Norsk $f && grep -l Svenska $f; done

You can use:

grep -l Dansk * | xargs grep -l Norsk | xargs grep -l Svenska

If you want also to find in hidden files:

grep -l Dansk .* | xargs grep -l Norsk | xargs grep -l Svenska

You can do this really easily with ack:

ack -l 'cats' | ack -xl 'dogs'

• -l: return a list of files
• -x: take the files from STDIN (the previous search) and only search those files

And you can just keep piping until you get just the files you want.

I had this problem today, and all one-liners here failed to me because the files contained spaces in the names.

This is what I came up with that worked:

grep -ril <WORD1> | sed 's/.*/"&"/' | xargs grep -il <WORD2>

If you have git installed

git grep -l --all-match --no-index -e Dansk -e Norsk -e Svenska

The --no-index searches files in the current directory that is not managed by Git. So this command will work in any directory irrespective of whether it is a git repository or not.

Expanding on @kurumi's awk answer, here's a bash function:

all_word_search() {
    gawk '
        BEGIN {
            for (i=ARGC-2; i>=1; i--) {
                search_terms[ARGV[i]] = 0;
                ARGV[i] = ARGV[i+1];
                delete ARGV[i+1];
            }
        }
        {
            for (i=1;i<=NF; i++) 
                if ($i in search_terms) 
                    search_terms[$1] = 1
        }
        END {
            for (word in search_terms) 
                if (search_terms[word] == 0) 
                    exit 1
        }
    ' "$@"
    return $?
}

Usage:

if all_word_search Dansk Norsk Svenska filename; then
    echo "all words found"
else
    echo "not all words found"
fi

Simply:

grep 'word1\|word2\|word3' *

see this post for more info

If you only need two search terms, arguably the most readable approach is to run each search and intersect the results:

 comm -12 <(grep -rl word1 . | sort) <(grep -rl word2 . | sort)

This searches multiple words in multiple files:

egrep 'abc|xyz' file1 file2 ..filen 

grep –irl word1 * | grep –il word2 `cat -` | grep –il word3 `cat -`

• -i makes search case insensitive
• -r makes file search recursive through folders
• -l pipes the list of files with the word found
• cat - causes the next grep to look through the files passed to it list.