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Home/ Questions/Q 7128745
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T11:10:14+00:00 2026-05-28T11:10:14+00:00

My code: #!/bin/bash for i in $@; do echo $i; done; run script: #

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My code:

#!/bin/bash

for i in $@;
    do echo $i;
done;

run script:

# ./script 1 2 3

1
2
3

So, I want to skip the first argument and get:

# ./script 1 2 3

2
3
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  1. Editorial Team
    Editorial Team
    2026-05-28T11:10:14+00:00Added an answer on May 28, 2026 at 11:10 am

    Use the offset parameter expansion

    #!/bin/bash
    
    for i in "${@:2}"; do
        echo $i
    done
    

    Example

    $ func(){ for i in "${@:2}"; do echo "$i"; done;}; func one two three
    two
    three
    
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