Frequently, I type these commands in bash:
mkdir something
cd something
I almost never do this:
mkdir something
ls # something in the current directory, not ./something/
And never this:
mkdir something something2
How can I have mkdir cd into the newly-created directory? By that I mean if I mkdir something, than the command cd something is executed immediately after.
Example:
$ pwd
/home/me
$ mkdir something
$ pwd
/home/me/something
You don't have to retype the name of the directory created. Use $_ variable:
From bash 4.3 manual:
_ At shell startup, set to the absolute pathname used to invoke the shell or shell script being executed as passed in the environment or argument list. Sub‐ sequently, expands to the last argument to the pre‐ vious command, after expansion.
So usage would be as so:
bash-4.3$ mkdir mydir
bash-4.3$ cd $_
If we wanted use just one line, combine the commands into compound command { (note that leading space before { and semicolons for each command are required):
bash-4.3$ pwd
/home/xieerqi
bash-4.3$ { mkdir mydir;cd $_;}
bash-4.3$ pwd
/home/xieerqi/mydir
It's important to note that this {list;} compound command structure is chosen for a reason, because all commands within the list are executed in current shell environment.
And if we want to avoid typing everything over and over, we can make an alias out of it.
bash-4.3$ alias mkcd='{ IFS= read -r d && mkdir "$d" && cd "$d"; } <<<'
bash-4.3$ pwd
/home/xieerqi
bash-4.3$ mkcd "mydir"
bash-4.3$ pwd
/home/xieerqi/mydir
Here I used "$d" variable for both , but I could have just as equally used "$_":
bash-4.3$ alias mkcd='{ IFS= read -r d && mkdir "$d" && cd "$_"; } <<<'
I suggest creating a function:
function mydir(){ mkdir -p "$1" && cd "$1"; }
it does what you want, you can add it to your .bashrc so it will be available in all your shells.
$1 is the first argument you send to the function, the name of directory.You can use it like:
$ mydir ~/my-new-directory
$ pwd
~/my-new-directory
Note: you can change mydir with mkdir to override it, however I suggest a custom name which does not exist in system like mydir or cmkdir; You can check to see if a command exists using type command.
I usually follow mkdir something by cdAlt + . which completes the command with the last argument of the previous command, i.e. the directory name.
You can make a small function, which you can store in the file ~/.bashrc. Edit the file to add the following lines,
md () {
mkdir "$1" && cd "$1"
}
Run
source ~/.bashrc
in order to make the change work in the current terminal [window]. The new function will be there, when you open new terminals.
Using && between the commands makes the cd command run only if the mkdir command was successful.
Example: I can use the function md like this to create a test directory testdir in the current directory (in this case my home directory as seen from the prompt),
sudodus@xenial32 ~ $ md testdir
sudodus@xenial32 ~/testdir $
I will also describe my difficulties using a small bash shellscript for this purpose, because other people might try it and get confused.
You can store a shellscript in the directory ~/bin. After creating ~/bin and rebooting, it will be in PATH.
Use a name that is not used by any standard command (for example mdscript),
#!/bin/bash
mkdir "$1" && cd "$1"
Make the script executable
chmod ugo+x ~/bin/mdscript
This does not work as intended with
mdscript testdir
because the current directory is only changed in the sub-process of the shell-script, but not in the terminal [window] after finishing the shellscript.
It works when 'sourced', run with the command line
source mdscript testdir
but this is not convenient, not a good answer to the original question.
You can see how it works, if you add a pwd command into the shellscript
#!/bin/bash
mkdir "$1" && cd "$1"
pwd
and run the script mdscript
sudodus@xenial32 ~ $ mdscript testdir
/home/sudodus/testdir
sudodus@xenial32 ~ $
