How to find patterns across multiple lines using grep?

Question

I want to find files that have \"abc\" AND \"efg\" in that order, and those two strings are on different lines in that file. Eg: a file with content:

blah bl         


        

Answer 1

As an alternative to Balu Mohan's answer, it is possible to enforce the order of the patterns using only grep, head and tail:

for f in FILEGLOB; do tail $f -n +$(grep -n "pattern1" $f | head -n1 | cut -d : -f 1) 2>/dev/null | grep "pattern2" &>/dev/null && echo $f; done

This one isn't very pretty, though. Formatted more readably:

for f in FILEGLOB; do
    tail $f -n +$(grep -n "pattern1" $f | head -n1 | cut -d : -f 1) 2>/dev/null \
    | grep -q "pattern2" \
    && echo $f
done

This will print the names of all files where "pattern2" appears after "pattern1", or where both appear on the same line:

$ echo "abc
def" > a.txt
$ echo "def
abc" > b.txt
$ echo "abcdef" > c.txt; echo "defabc" > d.txt
$ for f in *.txt; do tail $f -n +$(grep -n "abc" $f | head -n1 | cut -d : -f 1) 2>/dev/null | grep -q "def" && echo $f; done
a.txt
c.txt
d.txt

Explanation

• tail -n +i - print all lines after the ith, inclusive
• grep -n - prepend matching lines with their line numbers
• head -n1 - print only the first row
• cut -d : -f 1 - print the first cut column using : as the delimiter
• 2>/dev/null - silence tail error output that occurs if the $() expression returns empty
• grep -q - silence grep and return immediately if a match is found, since we are only interested in the exit code