Bash: Find file with max lines count

Question

This is my try to do it

• Find all *.java files
  find . -name '*.java'
• Count lines
  wc -l
• Delete last line
  sed '$d'
• Use AWK to find max lines-count in wc output
  awk 'max=="" || data=="" || $1 > max {max=$1 ; data=$2} END{ print max " " data}'

then merge it to single line

find . -name '*.java' | xargs wc -l | sed '$d' | awk 'max=="" || data=="" || $1 > max {max=$1 ; data=$2} END{ print max " " data}'

Can I somehow implement counting just non-blank lines?

Answer 1

find . -type f -name "*.java" -exec grep -H -c '[^[:space:]]' {} \; | \
    sort -nr -t":" -k2 | awk -F: '{print $1; exit;}'

Replace the awk command with head -n1 if you also want to see the number of non-blank lines.

---

Breakdown of the command:

find . -type f -name "*.java" -exec grep -H -c '[^[:space:]]' {} \; 
'---------------------------'       '-----------------------'
             |                                   |
   for each *.java file             Use grep to count non-empty lines
                                   -H includes filenames in the output
                                 (output = ./full/path/to/file.java:count)

| sort -nr -t":" -k2  | awk -F: '{print $1; exit;}'
  '----------------'    '-------------------------'
          |                            |
  Sort the output in         Print filename of the first entry (largest count)
reverse order using the         then exit immediately
  second column (count)

Answer 2

find . -name "*.java" -type f | xargs wc -l | sort -rn | grep -v ' total$' | head -1