How do you export a variable through shell script without using source command?

Question

I wanted to export variables based on the condition as:

if [[ "$1" == *"eu"* ]]; then
 export API_URL=https://xx-gateway-eu.xx.com/cccccccc
elif [[ "$1" == *"as"* ]]; then
 export API_URL=https://xx-gateway-as.xx.com/cccccccc
else
 export API_URL=https://xx-gateway-na.xx.com/cccccccc
fi

When i execute the script using "./script.sh yyy.ee.kkk.eu.dd.com", it got executed successfully but i dont see the variable got exported, echo $API_URL is empty. Could see the variable is set if i use 'source ./script.sh' and i understand the reason why its behaving like this. But, is there a way to achieve it as part of the shell script execution itself and without using source?

declension · Accepted Answer · 2022-11-28 14:01:57Z

1

Not as far as I know, because that execution happens in a new process. This is what the shebang line is all about in a script.

source (or . if you prefer) is what you want here really.

EDIT: or use `eval` if you really need to

...If you really want to avoid source, you could use eval to ensure you stay in-process:

$ 1=foo-eu- eval "$(cat script.sh)"
$ echo $API_URL
https://xx-gateway-as.xx.com/cccccccc

With the normal caveats about eval.

edited Nov 28, 2022 at 14:01

answered Nov 17, 2022 at 11:00

declension

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4 Comments

devops1 Over a year ago

Understood, but looking for a way to achieve natively on the script without using source command, as we dont have the luxury to use it.

declension Over a year ago

Can you use eval then?

devops1 Over a year ago

Could you please let me know the exact format to use eval here?

declension Over a year ago

@devops1 sorry missed that - sure, updated

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How do you export a variable through shell script without using source command?

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EDIT: or use `eval` if you really need to

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EDIT: or use eval if you really need to

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EDIT: or use `eval` if you really need to