grep multiple strings, count lines, echo output for each string

Question

I have a directory of files that is structured like this:

data/directory1/file1.xml
data/directory2/file2.xml
data/directory3/file3.xml
...

There are thousands of sub directories within data/, each with one xml file.

I would like to recursively scan data/ and look for three patterns, then echo the count of the pattern match (ideally) with a label before each. So, for my output, I'd like to see

Pattern 1 = 20
Pattern 2 = 100
Pattern 3 = 432

I've been able to use this to run these one at a time using:

grep -E -r 'Pattern 1' data/ | wc -l
grep -E -r 'Pattern 2' data/ | wc -l
grep -E -r 'Pattern 3' data/ | wc -l 

Now I'm trying to put this all into a shell script with one argument, which would be parent directory of data/. Then output the count to each pattern to either standard output or count.txt.

Your help is appreciated!

Answer 1

You should check grep -c, it does exactly what you describe. If you need more flexibility, you could probably use awk. Try something like this (typed directly in the browser and not tested):

dir=$1
# add tests to check if $1 exists (-d) and/or is non null (-n)

find "${dir}" | awk <<\EOF

BEGIN {
   pat1 = 0;
   pat2 = 0;
   ...
}

/pattern/ { pat1 += 1;}
/otherpattern/ {pat2 += 2;}
...

END {
   print "pattern", pat1;
   print "otherpattern", pat2;
   ...
}
EOF

I'll leave it up to you to work out the details of the find command or the awk script.

Answer 2

The following script loops over the pattern and prints the total count of matches for each pattern. It searches under the directory whose name is passed as an argument, or under the current directory if you don't pass an argument.

#!/bin/sh
for p in 'Pattern 1' 'Pattern 2' 'Pattern 3; do
  printf '%s = ' "$p"; grep -E -r -e "$p" "${1:-.}/data" | wc -l
done