In bash $@ contains all the arguments used to call the script but I am looking for a solution to remove the first one
./wrapper.sh foo bar baz ...: #!/bin/bash # call `cmd` with bar baz ... (withouyt foo one) I just want to call cmd bar baz ...
You can use shift to shift the argument array. For instance, the following code:
#!/bin/bash echo "$@" shift echo "$@" produces, when called with 1 2 3 prints 1 2 3 and then 2 3:
$ ./example.sh 1 2 3 1 2 3 2 3 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 $#.
Environment-variable-expansion! Is a very portable solution.
Remove the first argument: with $@
${@#"$1"} Remove the first argument: with $*
${*#"$1"} Remove the first and second argument: with $@
${@#"$1$2"} Both $@ or $* will work because the result of expansion is a string.
links:
Remove a fixed prefix/suffix from a string in Bash
http://www.tldp.org/LDP/abs/html/abs-guide.html#ARGLIST
Variable expansion is portable because it is defined under gnu core-utils
Search for "Environment variable expansion" at this link:
https://www.gnu.org/software/coreutils/manual/html_node/
