iterate over argument list in linux shell

Question

I want to iterate over the argument list in shell, I know how to do this with

for var in $@

But I want to do this with

for ((i=3; i<=$#; i++))

I need this because the first two arguments won't enter into the loop. Anyone knows how to do this? Looking forward to you help.

cheng

reader_1000 · Accepted Answer · 2011-10-02 08:30:51Z

11

This might help:

for var in "${@:3}"

for more information you can look at:

answered Oct 1, 2011 at 13:30

reader_1000

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Comments

Community · Accepted Answer · 2017-05-23 12:32:21Z

5

reader_1000 provides a nice bash incantation, but if you are using an older (or simpler) Bourne shell you can use the creaking ancient (and therefore highly portable)

VAR1=$1
VAR2=$2
shift 2
for arg in "$@"
...

CommunityBot

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answered Oct 1, 2011 at 13:41

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shady · Accepted Answer · 2015-10-05 17:03:15Z

1

Though this is an old question, there is another way you can do this. And, maybe this is what you are asking for.

for(( i=3; i<=$#; i++ )); do  
    echo "parameter: ${!i}"  #Notice the exclamation here, not the $ dollar sign.
done

answered Oct 5, 2015 at 17:03

shady

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