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.

Answer 1

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 if ! command , which works on any command:

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

Answer 2

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 ! cmd returns zero. The [ command also accepts ! as an argument, so that [ expression ] is inverted with [ ! expression ] [ ! 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