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

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.

Answer 1

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

Answer 2

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

Answer 3

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'" \;