bash, compare the output of a command

Question

I am trying to do something like this:

  if [ $(wc -l $f) -lt 2 ]

where $f is a file. When I run this, I get the error message:

  [: too many arguments

Does anybody know how to fix this line in my command?

The full script is:

for f in *.csv
do
  if [ $(wc -l $f) -lt 2 ]
      then
      echo $f
      fi
done

Answer 1

at least in my case wc -l filename does output 32 filename being 32 the number of lines. so you must strip of the filename after the line count. You could change your code from

if [ $(wc -l $f) -lt 2 ]

to

if [ $(wc -l $f | cut -f1 -d' ') -lt 2 ]

or

if [ $(wc -l < $f) -lt 2 ]

If does not solve your problem please add the output of wc -l filename to your question or as a comment.

Answer 2

Try this :

  if (( $(wc -l < "$f") < 2 ))

or if you want to keep your syntax :

  if [ $(wc -l < "$f") -lt 2 ]

Note

((...)) is an arithmetic command, which returns an exit status of 0 if the expression is nonzero, or 1 if the expression is zero. Also used as a synonym for "let", if side effects (assignments) are needed. See http://mywiki.wooledge.org/ArithmeticExpression