I am trying to learn the basics of shell. I used vim editro for creating my own list of commands to be executed. Here is the way I created the code
vi mycommands
then inside this file I wrote
cd Documents
I am using macOS Catalina which has zsh by default but switched to bash
So when I write the following command in the terminal:
$ sh +x mycommands
It shows
+cd Documents
The Documents has some files and directories but it is not changing directory.Where am I going wrong?
Any help will be greatly appreciated.
Scripts run like sh myscript execute in a separate sub-shell, not the current shell. Changing directory inside a script will not cause your shell to change directory. If you want to change directory in your shell, you need to run the commands in your shell.
To do that, run:
. ./myscript (sh, bash) or source ./myscript (bash).
See this question.
Related
I have a shell script with some functionalities. I want to convert it to an executable file. Any idea if this is possible?
Thanks
Add the following line at the very top of your script:
#!/bin/sh
This is known as a shebang. It indicates which program to invoke the shell script with when you execute it. You could change it to anything. Eg, to run a zsh script you would use #!/bin/zsh, which is the path to the zsh interpreter on my machine.
Then you need to mark the file as executable using chmod. You can do this as follows:
chmod +x myscript
You can now run the script like this:
/full/path/to/myscript
Or, if you're in the directory the script is in:
./myscript
The '.' expands to the path of your current working directory.
If you want to be able to run the script from anywhere, stick it somewhere in your path. Eg.
mv myscript /usr/bin
You can now run the script from anywhere by typing in just the name.
I am trying to get a bash script to run in git bash while specifying a different .bashrc than the one in my home directory (or none at all) however it is proving an impossible task.
To my understanding this should work:
"C:\Program Files (x86)\Git\bin\sh.exe " --rcfile .bashrc --login -i C:/Scripts/myscript.sh
However no matter what I try either the --rcfile file flag will be completely ignored or the script will get parse errors because it is not parsed by bash.
The following are my findings:
--login flag is needed to get the script to be parsed by bash rather than windows command prompt
--rcfile and also --norc are completely ignored if flags --login is used
I have tried every possible combination I think of, including calling the script within my .bashrc file, swapping the flags around, using the -c flag to run the script command and swapping my .bashrc files around to try using the --norc flag instead.
Is this just a result of shitty bash implementation for windows or am I doing something wrong?
Any help on the matter is appreciated.
You can try sourcing your .bashrc inside the script myscript.sh.
source .bashrc
Or
. .bashrc
As far as I can tell, the -i flag is overridden by the fact that you provide a script for bash to run. Your shell isn't actually interactive, so --rcfile is ignored. The only way I can tell to both run a script and source an additional file is to use a non-interactive login shell; however, in that case, you are restricted to using .bash_profile, .bash_login, or .profile, whichever is found first:
bash --login myscript.sh
There is no --loginfile to override the choice of file sourced prior to myscript.sh.
UPDATE: I forgot about BASH_ENV.
export BASH_ENV=.bashrc
bash myscript.sh
I do not know how you would go about adding BASH_ENV to your environment in Git bash.
In the following script (saved as script.sh):
#!/bin/sh
cd $MY_PYTHON_WORKING_DIRECTORY
python script1.py
python script2.py
Then, when I try to run the command script.sh in my bash shell, I got the error bash: script.sh: command not found. Why does this not work as expected? If the first line of any scripts start by #! prefix, then the following path on the line is interpreted as a command, right? For your information, even if I changed my first line to #!/bin/bash, the same error still occurred. If I run the script as either sh script.sh or bash script.sh, then the script ran as expected.
Is there any way to run the script by just hitting script.sh?
One more question, between sh and bash, which should I use? I'm on OS X 10.8 and my default shell is currently set bash, but I wonder which one to use going forward.
Thanks.
First, make the script executable:
chmod u+x script.sh
Second, your current directory is not in your $PATH. Therefore, you have to run the script with a path (relative is enough):
./script.sh
I am not a Solaris expert and I am trying to create a shell script that will change my prompt to PWD and the ksh to bash and I have this:
PS1='$PWD $ ' exec bash --noprofile --rcfile /dev/null
or
PS1='\w $' exec bash --noprofile --rcfile /dev/null
Both of them dont work from a sh. if i add them from the command line then the first time my bash appears on prompt and the second time the PS1='$PWD $' kicks in and my prompt changes.
Firstly, why is PS1='$PWD $' not working from shell script . and why do i have to run the command from command line twice to acheive my results.
Also, in my export/home/syed/ directory there are three files local.login, local.profile, and local.cshrc. is there any way i can use them that when ever i log in i dont need to run my shell script and upon login i get bash shell and my prompt as i want it
(am i asking too much, i dont like the ksh as it does not have any features like up arrow recall last commands and tab auto complete features)
thanks
Syed...
When you exec from within a script, the script is what is replaced, not the parent shell.
Try sourcing the script rather than running it.
Also, in Solaris, you can use passwd -e to change your login shell.
You may be able to symlink ~/.profile to your existing ~/local.profile (or similar). Note that .cshrc is for the C Shell and is not compatible with ksh or Bash.
If you want that your default shell will be bash, change it in /etc/passwd
When you exec bash it sets up its own environment from scratch. Pass it an --rcfile containing the settings you would like for it to inherit.
I tried activating a VirtualEnv through a shell script like the one below but it doesn't seem to work,
#!/bin/sh
source ~/.virtualenvs/pinax-env/bin/activate
I get the following error
$ sh virtualenv_activate.sh
virtualenv_activate.sh: 2: source: not found
but if I enter the same command on terminal it seems to work
$ source ~/.virtualenvs/pinax-env/bin/activate
(pinax-env)gautam#Aspirebuntu:$
So I changed the shell script to
#!/bin/bash
source ~/.virtualenvs/pinax-env/bin/activate
as suggested and used
$ bash virtualenv_activate.sh
gautam#Aspirebuntu:$
to run the script .
That doesn't throw an error but neither does that activate the virtual env
So any suggestion on how to solve this problem ?
PS : I am using Ubuntu 11.04
TLDR
Must run the .sh script with source instead of the script solely
source your-script.sh
and not
your-script.sh
Details
sh is not the same as bash (although some systems simply link sh to bash, so running sh actually runs bash). You can think of sh as a watered down version of bash. One thing that bash has that sh does not is the "source" command. This is why you're getting that error... source runs fine in your bash shell. But when you start your script using sh, you run the script in an shell in a subprocess. Since that script is running in sh, "source" is not found.
The solution is to run the script in bash instead. Change the first line to...
#!/bin/bash
Then run with...
./virtualenv_activate.sh
...or...
/bin/bash virtualenv_activate.sh
Edit:
If you want the activation of the virtualenv to change the shell that you call the script from, you need to use the "source" or "dot operator". This ensures that the script is run in the current shell (and therefore changes the current environment)...
source virtualenv_activate.sh
...or...
. virtualenv_activate.sh
As a side note, this is why virtualenv always says you need to use "source" to run it's activate script.
source is an builtin shell command in bash, and is not available in sh. If i remember correctly then virtual env does a lot of path and environment variables manipulation. Even running it as bash virtualenv_blah.sh wont work since this will simply create the environment inside the sub-shell.
Try . virtualenv_activate.sh or source virtualenv_activate.sh this basically gets the script to run in your current environment and all the environment variables modified by virtualenv's activate will be available.
HTH.
Edit: Here is a link that might help - http://ss64.com/bash/period.html
On Mac OS X your proposals seems not working.
I have done it this way. I'am not very happy with solution, but share it anyway here and hope, that maybe somebody will suggest the better one:
In activate.sh I have
echo 'source /Users/andi/.virtualenvs/data_science/bin/activate'
I give execution permissions by: chmod +x activate.sh
And I execute this way:
`./activate.sh`
Notice that there are paranthesis in form of ASCII code 96 = ` ( Grave accent )
For me best way work as below.
Create start-my-py-software.sh and pest below code
#!/bin/bash
source "/home/snippetbucket.com/source/AIML-Server-CloudPlatform/bin/activate"
python --version
python /home/snippetbucket.com/source/AIML-Server-CloudPlatform/main.py
Give file permission to run like below.
chmod +x start-my-py-software.sh
Now run like below
.start-my-py-software.sh
and that's it, start my python based server or any other code.
ubuntu #18.0
In my case, Ubuntu 16.04, the methods above didn't worked well or it needs much works.
I just made a link of 'activate' script file and copy it to home folder(or $PATH accessible folder) and renamed it simple one like 'actai'.
Then in a terminal, just call 'source actai'. It worked!