I have a folder which contains some folders, these folder are moved very often so I made a script to see if they exist, and if not then create them. This is what I did to (which I though would) achieve it:
if [ ! -f "$DIR/0folder" ]
then
mkdir "$DIR/0folder"
fi
But, even if 0folder already exists, it still tries to make it which mkdir tells me. Like here;
mkdir: /Allfoldersgoeshere/subfolder/0folder: File exists
Why? It should just ignore it because it already exists?
The -f in your test is checking if FILE exists and is a regular file.
What you need is -d to test if FILE exists and is a directory.
if [ ! -d "$DIR/0folder" ]
then
mkdir "$DIR/0folder"
fi
It is not mandatory to check if a directory exists though. According to the man page of mkdir we see the following
man mkdir | grep -A1 -- -p
-p, --parents
no error if existing, make parent directories as needed
However, if FILE exists and is a regular file mkdir -p will fail with
mkdir: /Allfoldersgoeshere/subfolder/0folder': Not a directory.
In this scenario handling the file that is expected to be a directory will be necessary before directory creation.
-f is for files and -d for directories?
man test will show the test options.
-e.
mkdir -p info.
From the manual (man bash) under CONDITIONAL EXPRESSIONS:
-f file
True if file exists and is a regular file.
-d file
True if file exists and is a directory.
So, to check for the existence of a directory (not a file)...
if [[ ! -d "$DIR/0folder" ]] ; then
mkdir "$DIR/0folder"
fi
Neither -d or -f is useful here for robust tests - because they're just opposite sides of the same coin.
touch file; [ ! -d file ] && mkdir file
mkdir: cannot create directory ‘file’: File exists
See? Better would be [ -e ... ] || mkdir ... to check for existence, which would at least save on the error above, but even that has its issues - like race conditions or not creating parents as needed or unresolved symlinks.
if [ ! -e "$DIR/0folder" ] &&
[ ! -L "$DIR/0folder" ]
then mkdir -- "$DIR/0folder"
fi
...is closer but still racy and still assumes all parents exist as well.
Probably you could work out a sure test if you needed to, but you really don't. You can get atomic directory creation w/ mkdir -p and test its return. There aren't too many standard shell tools which offer atomic fs ops - it's best we take advantage of those that do.
Read the manual. The -p option will make a directory and intermediate directories as required if it doesn't exist or silently fail. Assuming that you are not processing thousands of files then you could just do mkdir -p "${DIR}/0folder or if you feel the need to test then
[ -d "${DIR}/0folder" ] || mkdir -p "${DIR}/0folder"
-d test before mkdir -p is rather useless. If it is useful to know whether or not a directory was created, just test mkdir exit status: 0 means a directory was created. That said. +1 for mkdir -p... to nullify the notification when exit status is non-zero, mkdir -p dir/dir 2>/dev/null
You're doing: if the $DIR/0folder doesn't exist, make a directory with that name.
Just change -f (file) by -d (directory).
That should solve your problem, now you could be interested on some tips that people had given here but that is apart.
-fby-d?mkdir -p "$dir/0folder"It will create if it doesn't exist otherwise nothing happens.