Using a variable in a script calling curl

Question

I'm trying to replace 1.2.3.4 with the contents of variable $wanip in the following script.

wanip="4.3.2.1"
echo $wanip
content=$(curl --insecure -H "X-DNSimple-Token: foo:bar" -H "Accept: application/json" -H "Content-Type: application/json" -X PUT -d "{\"record\": {\"name\": \"foo\",\"content\": \"1.2.3.4\"}}" https://acme.com/records/123)
echo $content

If I literally replace 1.2.3.4 with $wanip*, when I run the script I'm getting a message saying: "message":"Problems parsing JSON".

Works for me. Try using set -x before the command (and set +x after). It should print ++ curl --insecure -H 'X-DNSimple-Token: foo:bar' -H 'Accept: application/json' -H 'Content-Type: application/json' -X PUT -d '{"record": {"name": "foo","content": "4.3.2.1"}}' https://acme.com/records/123 -- if it prints something else, that may point to the problem. — Gordon Davisson
– Gordon Davisson, Commented Jan 16, 2015 at 4:44
put a set -x in the line before content? If so, that is giving me an illegal operation message — WhiskerBiscuit
– WhiskerBiscuit, Commented Jan 16, 2015 at 4:55
"illegal operation", or ": invalid option"? If the latter, your script may be in DOS/Windows format rather than unix, which will cause all sorts of weird problems. Try printing the script with cat -vt scriptname, and see if it lists "^M" at the end of lines. If it does, use something like dos2unix or sed -i $'s/\r$//' scriptname to fix it, and then stay away from Windows text editors for unix scripts. — Gordon Davisson
– Gordon Davisson, Commented Jan 16, 2015 at 5:36

Nir Alfasi · Accepted Answer · 2015-01-16 07:00:06Z

1

Try adding a layer of abstraction:

#!/bin/bash
wnip="4.3.2.1"
echo $wanip
command="curl --insecure -H 'X-DNSimple-Token: foo:bar' -H 'Accept: application/json' -H 'Content-Type: application/json' -X PUT -d '{\"record\": {\"name\": \"foo\",\"content\": \"${wnip}\"}}' https://acme.com/records/123"
echo $command
content=$($command)
echo $content

edited Jan 16, 2015 at 7:00

answered Jan 16, 2015 at 4:25

Nir Alfasi

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2 Comments

WhiskerBiscuit Over a year ago

I'm getting several "couldn't resolve host" curl errors with this

Nir Alfasi Over a year ago

@WhiskerBiscuit try this

WhiskerBiscuit · Accepted Answer · 2015-01-16 05:07:40Z

0

After some hacking, I got this to work. Strange.

wanip=\"4.3.2.1\"
echo $wanip
content=$(curl --insecure -H "X-DNSimple-Token: foo:bar" -H "Accept: application/json" -H "Content-Type: application/json" -X PUT -d "{\"record\": {\"name\": \"foo\",\"content\": $wanip }}" https://acme.com/records/123)
echo $content

answered Jan 16, 2015 at 5:07

WhiskerBiscuit

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