Why is eval not working with command substitution?

Question

I have this command

$("${DIR}/test.py" "$1")

the output of which is

export SOME_VAR="hello world" export ANOTHER_VAR="hello"

But eval $("${DIR}/test.py" "$1")

seems to do nothing. What am I doing wrong?

I want to export those variables in my shell script.

did you mean that the output of `${DIR}/test.py" "$1"` is "export..."? — niry, Nov 18 '18 at 04:01
neither would work. Without eval, i get the output i posted above. With $() I wont get command substitution and hence just get the same output i posted above. I want to "evaluate" that output too in my script. — masterpiece, Nov 18 '18 at 18:15
I tried using both `#!/bin/bash` and `#!/bin/sh` same result. i am on bash3.2 — masterpiece, Nov 18 '18 at 18:49
trying to export to current shell so other programs can use these vars — masterpiece, Nov 18 '18 at 18:58
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/85941/discussion-between-niry-and-masterpiece). — niry, Nov 18 '18 at 19:10

score 1 · Accepted Answer · answered Nov 18 '18 at 22:03

1

After discussion, we determined that the problem was not using source to allow exporting vars to the current shell.

answered Nov 18 '18 at 22:03

niry

1 Answers1