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How to find the last argument passed to a shell script?
=> $@ is all of them. => $0 is script name. I want thisfilename.txt stored in a shell variable called $last. How do I find the last argument passed to a shell script written in bash or ksh under Unix like operating systems?
How to pass a command to a zsh script?
In an sh script, you could call eval, but don’t do this in zsh, because users wouldn’t expect to have to write zsh syntax which is a little different from sh syntax. 0 11 * * * my_wrapper.sh “task_name” “command arg1 arg2 arg3 …”
What does it mean to pass a script to a shell?
A shell script is a string that you pass to the sh program for execution. This allows the user to write any shell snippet without having to explicitly invoke a shell, and is simpler to parse since it’s a single string, hence a single argument to your program.
Is it possible to pass named arguments in Bash?
Is this possible in Bash or Zsh? There is a helpful Small getopts tutorial or you can type help getopts at the shell prompt. The only caveat is that you have to use the syntax my_script –p_out=/some/path –arg_1=5. STEPS and REPOSITORY_NAME are ready to use in the script.
How to return the last argument of a command?
To return the last argument of the most recently used command use the special parameter: In this instance it will work if it is used within the script before another command has been invoked. Just use !$. Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question.
How can I recall the argument of the previous Bash command?
The ! prefix is used to access previous commands. In the command line you can press esc -. or alt +. It cycles through the last argument of your previous commands. If you know the number given in the history for a particular command, you can pretty much take any argument in that command using following terms.