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I'm calling a shell script from another shell script and the called script requires some input (command line) parameters.
I'm have below mentioned code, but thats not working. I don't know why the argument values are not passed to the called script.
script1.sh
=======================================
#!/bin/bash
ARG1="val1"
ARG2="val2"
ARG3="val3"
. /home/admin/script2.sh "$ARG1" "$ARG2" "$ARG3"
script2.sh
=======================================
#!/bin/bash
echo "arg1 value is: $1 ....."
echo "arg2 value is: $2 ....."
echo "arg3 value is: $3 ....."
But when I execute script1.sh I get following result:
arg1 value is: .....
arg2 value is: .....
arg3 value is: .....
What am I missing?
538
Using arguments Inside the script, we can use the $ symbol followed by the integer to access the arguments passed. For example, $1 , $2 , and so on. The $0 will contain the script name.
Arguments can be passed to the script when it is executed, by writing them as a space-delimited list following the script file name. Inside the script, the $1 variable references the first argument in the command line, $2 the second argument and so forth. The variable $0 references to the current script.
By sourcing the second script with . /home/admin/script2.sh, you're effectively including it in the first script, so you get the command line arguments to the original script in $@. If you really want to call the other script with arguments, then do
/home/admin/script2.sh "$ARG1" "$ARG2" "$ARG3"
(make sure it's executable).
106
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