Check if a directory exists in a shell script
在shell脚本中可以使用什么命令来检查目录是否存在？
To check if a directory exists in a shell script you can use the following:
if [ -d "$DIRECTORY" ]; then
# Control will enter here if $DIRECTORY exists.
fi
Or to check if a directory doesn't exist:
if [ ! -d "$DIRECTORY" ]; then
# Control will enter here if $DIRECTORY doesn't exist.
fi
However, as Jon Ericson points out (thanks Jon), subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check. Eg running this:
ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then
rmdir "$SYMLINK"
fi
Will produce the error message:
rmdir: failed to remove `symlink': Not a directory
So symbolic links may have to be treated differently, if subsequent commands expect directories:
if [ -d "$LINK_OR_DIR" ]; then
if [ -L "$LINK_OR_DIR" ]; then
# It is a symlink!
# Symbolic link specific commands go here.
rm "$LINK_OR_DIR"
else
# It's a directory!
# Directory command goes here.
rmdir "$LINK_OR_DIR"
fi
fi
Take particular note of the double-quotes used to wrap the variables, the reason for this is explained by 8jean in another answer.
If the variables contain spaces or other unusual characters it will probably cause the script to fail.
Remember to always wrap variables in double quotes when referencing them in a bash script. Kids these days grow up with the idea that they can have spaces and lots of other funny characters in their directory names. (Spaces! Back in my days, we didn't have no fancy spaces! ;))
One day, one of those kids will run your script with $DIRECTORY set to "My M0viez" and your script will blow up. You don't want that. So use this.
if [ -d "$DIRECTORY" ]; then
# Will enter here if $DIRECTORY exists, even if it contains spaces
fi
我发现test的双括号版本使编写逻辑测试更加自然：
if [[ -d "${DIRECTORY}" && ! -L "${DIRECTORY}" ]] ; then
echo "It's a bona-fide directory"
fi
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