Bash: Execute command from variable without eval

Question

I have a problem with a script which should execute a curl command composed from the given arguments. It works if I use eval in the subshell, however, I would like to avoid that as it seems to be bad style according to this

Here is the relevant part of the script:

#!/bin/bash
COMMAND="curl -s -o /dev/null -w '%{http_code}' -X POST -H \"Content-Type:application/json\" -H \"Accept:application/json\" https://www.example.com/login -d '{\"username\":\"$1\",\"password\":\"$2\"}'"
echo $(eval $COMMAND)

With eval, I get 201 or 401 depending on the validity of the given credentials as expected. If I remove eval and use echo $($COMMAND) I get 400 which indicates a wrongly constructed data parameter.

Can someone explain to me, how I have to construct the command such that I can directly pass it to a subshell?

Please see [BashFAQ/050](http://mywiki.wooledge.org/BashFAQ/050). — Dennis Williamson, Sep 22 '14 at 17:37

Pierre-Yves Gillier · Answer 1 · 2014-09-17T07:59:14.233

3

You may use backticks or nesting to avoid using eval.

Backticks

$ var=\`curl -s -o /dev/null -w '%{http_code}' -X POST -H \"Content-Type:application/json\" -H \"Accept:application/json\" https://www.example.com/login -d '{\"username\":\"$1\",\"password\":\"$2\"}'"\`

Nesting

$ var = $(curl -s -o /dev/null -w '%{http_code}' -X POST -H \"Content-Type:application/json\" -H \"Accept:application/json\" https://www.example.com/login -d '{\"username\":\"$1\",\"password\":\"$2\"}'")

Nesting is supported on the vast majority of shells and more readable than backticks.

edited Sep 17 '14 at 07:59

answered Sep 16 '14 at 10:17

Pierre-Yves Gillier

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Although since you are explicitly using bash, I would suggest you replace the backticks with nest-friendly `var=$( curl ... )` – Cameron Kerr Sep 16 '14 at 11:17
Right. Answer updated ! – Pierre-Yves Gillier Sep 16 '14 at 13:23

score 0 · Accepted Answer · answered Sep 16 '14 at 12:23

Thanks Pierre.

After some more attempts of trial and error about where to place which kind of quotes I finally got this working:

$(curl -s -o /dev/null -w '%{http_code}' -X POST -H "Content-Type:application/json" -H "Accept:application/json" https://example.com/login -d "{\"username\":"\"$1\"",\"password\":"\"$2\""}")

Bash: Execute command from variable without eval

2 Answers2