Get directory when last folder in path ends in given string (sed in ifelse)-CodePudding

I am attempting to find multiple files with .py extension and grep to see if any of these files contain the string nn. Then return only the directory name (uniques), afterwards, if the last folder of the path ends in nn, then select this.

For example:

find `pwd` -iname '*.py' | xargs grep -l 'nn' | xargs dirname | sort -u | while read files; do if [[ sed 's|[\/](.*)*[\/]||g' == 'nn' ]]; then echo $files; fi; done

However, I cannot use sed in if-else expression, how can I use it for this case?

CodePudding user response：

[[ ]] is not bracket syntax for an if statement like in other languages such as C or Java. It's a special command for evaluating a conditional expression. Depending on your intentions you need to either exclude it or use it correctly.

If you're trying to test a command for success or failure just call the command:

if command ; then
  :
fi

If you want to test the output of the command is equal to some value, you need to use a command substitution:

if [[ $( command ) = some_value ]] ; then
  :
fi

In your case though, a simple parameter expansion will be easier:

# if $files does not contain a trailing slash
if [[ "${files: -2}" = "nn" ]] ; then
  echo "${files}"
fi

# if $files does contain a trailing slash
if [[ "${files: -3}" = "nn/" ]] ; then
  echo "${files%/}"
fi