passing-bash-variable-to-curl quotes

Asked May 16 '17 at 20:29

Active May 16 '17 at 20:29

Viewed 20 times

I have some problems with a part of my cript, it's answering by showing the PHP code of my target page. Here is the part of my code:

curl -X POST -F 'decription='$description'' -F 'valeur=007' -F 'nom_carte=something' -F 'horodatage=something' $cible

I think it's a quoting pb. I tried possibilities but still don't get it.

asked May 16 '17 at 20:29

JLB

2

`"description='$description'"`? – chepner May 16 '17 at 20:31
More likely `description="$description"` – Barmar May 16 '17 at 21:03
1

If you're seeing the PHP source code, the problem is on the server, not your `curl` command. – Barmar May 16 '17 at 21:03
I'm with @Barmar. It looks like you may also have a quoting problem, but more fundamentally the server should *never* send the source code, no matter what malformed query comes down the wire. – Gordon Davisson May 16 '17 at 21:44

0 Answers0