Calling shell script inside shell script based on no of arguments passed to first script

Question

We are doing automation to databases where we have one script is calling another script in the shell. Forex :-

first-script.sh arg1 agr2 arg3 [Suppose 3 arguments are passed, then we have to call second-script.sh 3 times]
#!/bin/bash 
./second-script.sh $1 
./second-script.sh $2 
./second-script.sh $3

How can we remove this dependency? We want something like if 2 arguments are passed then automatically first script should run like

first-script.sh arg1 agr2
#!/bin/bash 
./second-script.sh $1 
./second-script.sh $2

and if 5 arguments are passed then, it should run like

first-script.sh arg1 agr2 arg3 arg4 arg5 
#!/bin/bash 
./second-script.sh $1 
./second-script.sh $2 
./second-script.sh $3
./second-script.sh $4 
./second-script.sh $5

fpmurphy · Accepted Answer · 2021-08-06 05:11:30Z

2

Use "$@" (or possibly $*) to represent all the arguments:

#!/bin/bash

for arg in "$@"
do
   ./second-script.sh "$arg"
done

answered Aug 6, 2021 at 5:11

fpmurphy

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Comments

Keith · Accepted Answer · 2021-08-06 05:44:34Z

2

You just need to loop over the arguments.

for arg in $@  ; do 
    second-script.sh "$arg"
done

edited Aug 6, 2021 at 5:44

answered Aug 6, 2021 at 5:07

Keith

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1 Comment

user1934428 Over a year ago

You for got to quote the variables.

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