Bash scripting, cannot substitute a command line variable

Question

This is a small command line script that is used to post a json body in an http server. I am finding difficulties to pass the first command line argument $1 to the json body.

#!/bin/bash
curl  -X POST -d '{  "game": 16, "id": $(($1)) }' http://localhost:10000/

The command does not fail, however the http body contains exactly

{ "game": 16, "id": $(($1)) }

I want to run the script ./script 123 and send the json body

{  "game": 16, "id": 3 }

How can I do this using bash?

Yes, everything is needed. It is going to be a number actually, eg 12345 so I need all chars — cateof
– cateof, Commented Sep 12, 2017 at 11:51
In general, you should not be creating JSON using string interpolation, because the contents of $1 may need to be correctly escaped to produce valid JSON. Something like $(jq --arg val "$1" '{game: 16, id: $val}') is more appropriate. — chepner
– chepner, Commented Sep 12, 2017 at 14:32

kometonja · Accepted Answer · 2017-09-12 11:48:35Z

2

You can also use single quotes so you don't have to escape double quotes like this:

#!/bin/bash
curl  -X POST -d '{  "game": 16, "id": '$1' }' http://localhost:10000/

answered Sep 12, 2017 at 11:48

kometonja

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arco444 · Accepted Answer · 2017-09-12 11:42:42Z

1

Using single quotes will print literal characters. You need to use double quotes for string interpolation. Try:

curl  -X POST -d "{  \"game\": 16, \"id\": $1 }" http://localhost:10000/

answered Sep 12, 2017 at 11:42

arco444

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