How to determine if a file is NOT-executable in bash script

Question

I am looping through a directory and need to find all files that are not executable. I know

if [ -x $dir ]; then 
    echo "$dir is an executable file"
fi

shows that it is executable but how do I do the opposite of this? I have tried

if [ !-x $dir ]; then 
    echo "$dir is not-executable"
fi

however that does not work.

Spacing is crucial in shell scripts. There's a vast difference between [ !-x $dir ] and [ ! -x $dir ] (and you should use [ ! -x "$dir" ] to avoid some surprises). — Jonathan Leffler
– Jonathan Leffler, Commented Mar 9, 2021 at 20:30

John Kugelman · Accepted Answer · 2021-03-09 20:18:04Z

4

Running the line through Shell Check shows:

if [ !-x $dir ]; then
      ^-- SC1035: You are missing a required space here.
         ^-- SC2086: Double quote to prevent globbing and word splitting.

Adding the missing space and quotes results in:

if [ ! -x "$dir" ]; then

You can also put the ! outside the brackets using the generic syntax if ! command, which works on any command:

if ! [ -x "$dir" ]; then

answered Mar 9, 2021 at 20:18

John Kugelman

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William Pursell · Accepted Answer · 2021-03-09 20:25:10Z

4

Either:

if ! [ -x "$dir" ]; then 
    echo "$dir is not an executable file"
fi

or:

if [ ! -x "$dir" ]; then 
    echo "$dir is not an executable file"
fi

will work. In general, any command can be negated by !. So if cmd returns non-zero, ! cmd returns zero. The [ command also accepts ! as an argument, so that [ expression ] is inverted with [ ! expression ]. Which you choose is pretty much a stylistic choice and makes little difference.

Of course, you can also just do:

if [ -x "$dir" ]; then
    :
else
    echo "$dir is not an executable file"
fi

edited Mar 9, 2021 at 20:25

answered Mar 9, 2021 at 20:19

William Pursell

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