How to list files and match first line in bash script?

Question

I would like to check for (only) python files for those which do not have the #!/usr/bin/env python in the first line. So, I write a bash script that does the following:

#!/bin/bash

#list all of python files

for file in `find . -name "*.py"`
do
        if [ `head -1 $file` != "#!/usr/bin/env python"] then;
                 echo "no match in file $file"
        else then;
                 echo "match!"
        fi
done

However, for some reason I cannot get the if statement correct! I've looked at many questions, but I cannot find one that succinctly describes the issue. Here is the error message:

./run_test.sh: line 9: syntax error near unexpected token `else'
./run_test.sh: line 9: `    else then;'

where am I going awry? Thank you.

correct syntax is : `if [[ \`head -1 $file\` != "#!/usr/bin/env python" ]]; then`. also delete the then; after else — amdixon, Sep 05 '15 at 02:05
Why is the double brackets (`[[` and `]]`) necessary? It doesn't work without them - I just checked. — makansij, Sep 05 '15 at 03:37
Either you need to surround command in double quotes or use [[. If output of head -1 $file is more than one word or null, then it will break [ syntax. — Mahesh Kharvi, Sep 05 '15 at 05:09

score 2 · Answer 1 · edited May 23 '17 at 12:29

You can do something like

find . -type f -name '*.py' -exec \
  awk 'NR==1 && /#!\/usr\/bin\/env python/ \
  { print "Match in file " FILENAME; exit } \
  { print "No match in file " FILENAME; exit }' \
{} \;

If you are going to loop over it, don't use a for loop

#!/bin/bash
find . -type f -name '*.py' -print0 | while IFS= read -r -d $'\0' file; do
  if [[ $(head -n1 "$file") == "#!/usr/bin/env python" ]]; then
      echo "Match in file [$file]"
  else
      echo "No match in file [$file]"
  fi
done

score 1 · Answer 2 · answered Sep 05 '15 at 02:10

Things to notice:

The [] after your if statement needs correct spacing
The ';' (if you enter a new line is not necessary) goes after the if and not after the then

You added an extra then after the else.

#!/bin/bash

#list all of python files

for file in `find . -name "*.py"` 
do
     if [ `head -1 $file` != "#!/usr/bin/env python" ]; 
     then
         echo "no match in file $file"
     else 
         echo "match!"
     fi
done

Sanjay Manohar · Answer 3 · 2015-09-06T16:46:30.870

-1

can you use -exec option by any chance? I find it easier.

find . -name "*.py" -exec head -1 {} | grep -H '#!/usr/bin/env python' \;

You can control the output using grep options.

edit

Thanks to @chepner - To avoid the pipe being swallowed too early:

-exec sh -c "head -1 {} | grep -H '#!/usr/bin/env python'" \;

edited Sep 06 '15 at 16:46

answered Sep 05 '15 at 02:05

Sanjay Manohar

6,920
3
35
58

1

I like this since this most definitely seems like a one-off kind of task. – lintuxvi Sep 05 '15 at 02:40
Yeah, I like this....what does the `-H` option do? I read the man page, but it doesn't make sense to me: ` -H Always print filename headers with output lines.` – makansij Sep 05 '15 at 03:22
oh, it should print the name of the file, before the matching text. I think `-lHZ` may actually be best for you, giving just zero-terminated filenames. – Sanjay Manohar Sep 05 '15 at 03:45
1

You can't have a pipeline in the argument to `-exec`. – chepner Sep 05 '15 at 04:03
Oops I didn't test it, my bad. Anyone suggest the fix? Can I enclose in `$()` or quotes for example? – Sanjay Manohar Sep 05 '15 at 09:30
1

`-exec sh -c "head -1 {} | grep -H '#!/usr/bin/env python'" \;` – chepner Sep 06 '15 at 15:03
@SanjayManohar `-exec sh -c '...'` creates a subshell. Fine in morre complicated cases, but just piping as in op's original edit is better/faster here. – Reinstate Monica Please Sep 06 '15 at 23:59

How to list files and match first line in bash script?

3 Answers3

edit