How to remove first bash argument and pass the others to another command?

bash shift
bash parameter expansion

In bash $@ contains all the arguments used to call the script but I am looking for a solution to remove the first one

./ foo bar baz ...:


 # call `cmd` with bar baz ... (withouyt foo one)

I just want to call cmd bar baz ...

You can use shift to shift the argument array. For instance, the following code:

echo $@
echo $@

produces, when called with 1 2 3 prints 1 2 3 and then 2 3:

$ ./ 1 2 3
1 2 3
2 3

Basically, I want to "pluck out" the first occurrence of -inf from the parameter list. (​The remaining parameters will be passed along to a different command.) The  Check arguments for specific value; Conclusion; Bash scripts are commonly used to perform a variety of tasks. These examples describe various ways you can work with command line arguments. Pass arguments through to another program. Bash scripts are often used as wrappers to launch another application. A common task is to pass the command line arguments from the script to the program being loaded.

shift removes arguments from $@.

shift [n]

Shift positional parameters.

Rename the positional parameters $N+1,$N+2 ... to $1,$2 ... If N is not given, it is assumed to be 1.

Exit Status: Returns success unless N is negative or greater than $#.

Bash - how to pass all arguments (or after shift) from one script to another Bash - local and global variables · Bash - newline and other escape character in To test above bash script we can run this on command line: #!/bin/bash #remove first argument FIRST=$1 shift echo "Removed first arg: $FIRST" . If you developed a script called that counts the words in a file, it's best to pass the file name as an argument so that the same script can be used for all the files that will be processed. For example, if the name of the file to be processed is songlist.txt , enter the following at the command line:

Environment-variable-expansion! Is a very portable solution.

Remove the first argument: with $@


Remove the first argument: with $*


Remove the first and second argument: with $@


Both $@ or $* will work because the result of expansion is a string.

links: Remove a fixed prefix/suffix from a string in Bash

Variable expansion is portable because it is defined under gnu core-utils Search for "Environment variable expansion" at this link:

Using variables to refer to data, including the results of a command. use a name or a label to refer to some other quantity: such as a value, or a command. print a sequence of numbers starting from the first argument to the second argument: the command with $( and ) , and pass it as an argument to another command: When writing a wrapper bash script, we often need to pass all arguments passed to the wrapper scrip to another script. Here is quick bash code snippet to pass all arguments to another script: Passing all arguments in bash using $@ Here is sample code to print whatever arguments are passed to it.

Use echo or printf command to display variable value: If variable $1 is not set or passed, use root as default value for u: The first syntax removes shortest part of pattern and the second syntax removes the longest part of the pattern. ${_​php_modules_enabled} php modules and delete other .ini files if  The arguments are stored in variables with a number in the order of the argument starting at 1 First Argument: $1 Second Argument: $2 Third Argument: $3 Example command: To input arguments into a Bash script, like any normal command line program, there are special variables set aside for this.

Explains how to find and remove (delete) files in a single command under a between above two syntax is that the first command remove directories as well We can also explicitly pass the -depth option to the find to process each Disables the end of file string, which is treated like any other argument. The first command and the second command2 can be any type of command from this section. Bash will create a subshell for each command and set up the first command's standard output file descriptor such that it points to the second command's standard input file descriptor. The two commands will run simultaneously and bash will wait for both of

Some are a subset of parameter substitution, and others fall under the functionality of the UNIX expr command. This results in inconsistent command syntax and  I have a command that produce a output like this: $./command1 word1 word2 word3 . I want to pass this three words as arguments to another command like this: $ command2 word1 word2 word3. How to pass command1 output as three different arguments $1 $2 $3 to command2 ?

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