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?

Eastlake

Your solution as is will probably fall over when encountering unusual file names. Use -print0 in find in conjunction with -0 option in xargs, something like this - find . -name '*.java' -print0 | xargs -0 wc -l | sort -n | tail -2 | head -1 – Rune

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)

Cusec

Great, I like this more, cause it revealed find -exec option, which is more useful than looping – Eastlake

It'd fail for file names that contain colons or newlines. – Distinguishing

@EdMorton your filenames contain newlines? – Eastlake

MarekSebera I personally don't deliberately create file names containing newlines but I do come across them on various systems I work on and assuming your software will never have to work when the file names contains newlines is one of the ways your code can fail and be exploited. Using -print0 and xargs -0 as @Rune suggested and I used in my answer is one way to help you avoid such problems. – Distinguishing

So the 3 answers to your filenames contain newlines? are - a) no, not the filenames I manually create, b) maybe, e.g. filenames I create as a result of being required by a customer to write a tool to create file names from input that itself can contain newlines, e.g. fields in a CSV exported from Excel and c) yes, the filenames I don't create but that can exist on machines my software runs on. – Distinguishing

Answer 2

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

Modular

Not bad, but needs edit to show only file with most lines of code, now it shows all files with their counts – Eastlake

yeah ..you are right.just forgot to add one more pipe.added now – Modular

Super helpful to get a "top 10" of files with most lines in it, by changing head -1 to head -10 – Stiffler

Answer 3

Something like this might work:

find . -name '*.java'|while read filename; do
    nlines=`grep -v -E '^[[:space:]]*$' "$filename"|wc -l`
    echo $nlines $filename
done|sort -nr|head -1

(edited as per Ed Morton's comment. I must have had too much coffee :-) )

Francophile

Answer 4

To get the size of all of your files using awk is just:

$ find . -name '*.java' -print0 | xargs -0 awk '
BEGIN { for (i=1;i<ARGC;i++) size[ARGV[i]]=0 }
{ size[FILENAME]++ }
END { for (file in size) print size[file], file }
'

To get the count of the non-empty lines, simply make the line where you increment the size[] conditional:

$ find . -name '*.java' -print0 | xargs -0 awk '
BEGIN { for (i=1;i<ARGC;i++) size[ARGV[i]]=0 }
NF { size[FILENAME]++ }
END { for (file in size) print size[file], file }
'

(If you want to consider lines that contain only blanks as "empty" then replace NF with /^./.)

To get only the file with the most non-empty lines just tweak again:

$ find . -name '*.java' -print0 | xargs -0 awk '
BEGIN { for (i=1;i<ARGC;i++) size[ARGV[i]]=0 }
NF { size[FILENAME]++ }
END {
   for (file in size) {
      if (size[file] >= maxSize) {
         maxSize = size[file]
         maxFile = file
      }
   }
   print maxSize, maxFile
}
'

Distinguishing