How to create a directory and give permission in single command

How to create a directory and give permission in single command in Linux?

I have to create lots of folder with full permission 777.

Commands

mkdir path/foldername
chmod 777 path/foldername 

I don't like to create and give permission in two commands. Can I do this in开发者_StackOverflow single command?

According to mkdir's man page...

mkdir -m 777 dirname

When the directory already exist:

mkdir -m 777 /path/to/your/dir

When the directory does not exist and you want to create the parent directories:

mkdir -m 777 -p /parent/dirs/to/create/your/dir

install -d -m 0777 /your/dir

should give you what you want. Be aware that every user has the right to write add and delete files in that directory.

IMO, it's better to use the install command in such situations. I was trying to make systemd-journald persistent across reboots.

install -d  -g systemd-journal -m 2755 -v /var/log/journal

You could write a simple shell script, for example:

#!/bin/bash
mkdir "$1"
chmod 777 "$1"

Once saved, and the executable flag enabled, you could run it instead of mkdir and chmod:

./scriptname path/foldername

However, alex's answer is much better because it spawns one process instead of three. I didn't know about the -m option.

you can use following command to create directory and give permissions at the same time

mkdir -m777 path/foldername 

Just to expand on and improve some of the above answers:

First, I'll check the mkdir man page for GNU Coreutils 8.26 -- it gives us this information about the option '-m' and '-p' (can also be given as --mode=MODE and --parents, respectively):

...set[s] file mode (as in chmod), not a=rwx - umask

...no error if existing, make parent directories as needed

The statements are vague and unclear in my opinion. But basically, it says that you can make the directory with permissions specified by "chmod numeric notation" (octals) or you can go "the other way" and use a/your umask.

Side note: I say "the other way" since the umask value is actually exactly what it sounds like -- a mask, hiding/removing permissions rather than "granting" them as with chmod's numeric octal notation.

You can execute the shell-builtin command umask to see what your 3-digit umask is; for me, it's 022. This means that when I execute mkdir yodirectory in a given folder (say, mahome) and stat it, I'll get some output resembling this:

               755                   richard:richard         /mahome/yodirectory
 #          permissions                 user:group      what I just made (yodirectory),
 # (owner,group,others--in that order)                 where I made it (i.e. in mahome)
 # 

Now, to add just a tiny bit more about those octal permissions. When you make a directory, "your system" take your default directory perms' [which applies for new directories (its value should 777)] and slaps on yo(u)mask, effectively hiding some of those perms'. My umask is 022--now if we "subtract" 022 from 777 (technically subtracting is an oversimplification and not always correct - we are actually turning off perms or masking them)...we get 755 as stated (or "statted") earlier.

We can omit the '0' in front of the 3-digit octal (so they don't have to be 4 digits) since in our case we didn't want (or rather didn't mention) any sticky bits, setuids or setgids (you might want to look into those, btw, they might be useful since you are going 777). So in other words, 0777 implies (or is equivalent to) 777 (but 777 isn't necessarily equivalent to 0777--since 777 only specifies the permissions, not the setuids, setgids, etc.)

Now, to apply this to your question in a broader sense--you have (already) got a few options. All the answers above work (at least according to my coreutils). But you may (or are pretty likely to) run into problems with the above solutions when you want to create subdirectories (nested directories) with 777 permissions all at once. Specifically, if I do the following in mahome with a umask of 022:

mkdir -m 777 -p yodirectory/yostuff/mastuffinyostuff
# OR (you can swap 777 for 0777 if you so desire, outcome will be the same)
install -d -m 777 -p yodirectory/yostuff/mastuffinyostuff

I will get perms 755 for both yodirectory and yostuff, with only 777 perms for mastuffinyostuff. So it appears that the umask is all that's slapped on yodirectory and yostuff...to get around this we can use a subshell:

( umask 000 && mkdir -p yodirectory/yostuff/mastuffinyostuff )

and that's it. 777 perms for yostuff, mastuffinyostuff, and yodirectory.

Don't do: mkdir -m 777 -p a/b/c since that will only set permission 777 on the last directory, c; a and b will be created with the default permission from your umask.

Instead to create any new directories with permission 777, run mkdir -p in a subshell where you override the umask:

(umask u=rwx,g=rwx,o=rwx && mkdir -p a/b/c)

Note that this won't change the permissions if any of a, b and c already exist though.