How can I grep a directory based on the contents of two successive lines?

Question

How can I grep a directory for lines that contain "Foo", but only get matches when the next line also contains "Bar"?

Answer 1

Grep itself doesn't seem to support it, use pcregrep instead:

Foo
Bar
Foo
abc

pcregrep -M "Foo\nBar" file

Got:

Foo
Bar

Answer 2

Using grep only, you could construct the following pipe:

grep -A1 'Foo' input_file | grep -B1 'Bar' | grep 'Foo'

The first grep will get all the lines that contain Foo as well as the line after the match. Then we get lines that contain Bar as well as the line before the match, and finally extract the lines from this output that contain Foo.

EDIT: As manatwork pointed out, there are some problematic cases to be observant of. Although an interesting challenge, due to grep's line oriented functionality, any solution with it is likely to be a 'hack' and you are probably better off using something like pcregrep which is more suited to the task at hand.

Answer 3

@warl0ck pointed me in the right direction with pcregrep, but I said "contains", not "is", and I asked about a directory, not a file.

This seems to work for me.

pcregrep -rMi 'Foo(.*)\n(.*)Bar' .

Answer 4

With a sed script :

#!/bin/sed -nf

/^Foo/{
    h         # put the matching line in the hold buffer
    n         # going to nextline
    /^Bar/{   # matching pattern in newline
        H     # add the line to the hold buffer
        x     # return the entire paragraph into the pattern space
        p     # print the pattern space
        q     # quit the script now
    }
}

To use it :

chmod +x script.sed
printf '%s\n' * | ./script.sed

The printf here display all files in the current directory on one line each, and pass it to sed.

Note : this is sorted by alphabetical order.

More infos of useful pattern space and hold space HERE.

grymoire.com have really good stuff about shell programming.

Answer 5

With awk:

awk '/bar/ && prev != "" {print FILENAME ": " prev "\n" FILENAME ": " $0}
     /foo/ {prev=$0; next}
     {prev=""}' file1...

(general note about awk limitation: beware that if some file names may contain "=" characters, you'll need to pass them as ./filename instead of filename to awk)

Answer 6

While I prefer Nathan's solution using pcregrep, here is solution using only grep

grep -o -z -P  'Foo(.*)\n(.*)Bar' file

Options explanation:

• -o print only matched part. Necessary since inclusion of -z will printout the whole file (unless there is a \0 somewhere)
• -z Treat the input as a set of lines, each terminated by a zero byte (the ASCII NUL character) instead of a newline.
• -P perl regex syntax

EDIT : This version prints out entire matched lines

    grep -o -P -z  '(.*)Foo(.*)\n(.*)Bar(.*)' file