为什么此Shell脚本不起作用？

Question

I have noticed that find -execdir is not portable, so I decided to find a portable way of using just find -exec to achieve the same effect. To do this, one must be able to determine if the path to a directory from '/' found by 'find' contains any symbolic links, and refuse to traverse it if it does. I wrote a small script to determine if a given path contains symbolic links, but it seems to always return code 1, no matter what I give it. No command that prints anything fires, except if I give it a non-directory, in which case the first printf command fires.

#!/bin/sh -e
# If any commands fail, the script should return a nonzero status
[ -d "$1" ] || printf "%s is not a directory" "$1" && exit 1  # Tests if argument is a directory
cd "$1" || echo "Could not change directory" && exit 1 # If it is a directory, goes to it
until [ "$PWD" = '/' ] # Loop until root directory reached 
do
    cd .. || echo "Could not change directory" && exit 1 # Go to parent directory
    [ -d "$PWD" ] || printf "%s is not directory" "$PWD" && exit 1 # Check that this is a directory
done
echo "Given an okay directory"
exit 0

Answer 1

In bash (unlike c-like languages) && and || have the same precedence. That means your

command || echo error && exit 1

statements are interpreted as

{ command || echo error } && exit 1

Since echo will most likely succeed even if command doesn't, the first block will succeed and the exit statement will be executed.

Answer 2

For each condition line, you should enclose the failure with () . For example:

[ -d "$1" ] || (printf "%s is not a directory" "$1" && exit 2)

I'll explaun further what @Kevin wrote: If the first statement fails (the [ -d ] ), then the second statement is executed. Since the second succeeds (only in rare cases does printf fails), then the last statement is executed. In this format, the exit statement will not get executed ONLY if BOTH the first two failed. In case it is not a directory, you get a printf and an exit. In case it is a directory, the first || becomes true, and bash doesn't bother testing the next one (printf), and goes to the && , which is the exit, again. Enclosing the failures as one will prevent this.

Answer 3

You can check if $1 is not a directory with the inverse ! -d ! -d and use if; then if; then to execute commands after it returns true.

#!/bin/sh -e
# If any commands fail, the script should return a nonzero status
if [ ! -d "$1" ]
then
    printf "%s is not a directory" "$1"
    exit 1 # Tests if argument is a directory
fi
cd "$1" # If it is a directory, goes to it
until [ "$PWD" = '/' ] # Loop until root directory reached
do
    cd .. # Go to parent directory
done
echo "Given an okay directory"
exit 0