[BASH] passing a variable in CURL

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I'm trying to pass a defined variable in a BASH CURL command. I can't seem to get the single quotes or double quotes correct and it results in an error:

line 22: --user: command not found

Hardcoding the value is no problem.

What I have tried:

My code example:

password=MyPS@!

curl "https://somewebsite.com" \
--request POST \
--header "Accept:application/json" \
--header "Content-Type:application/json" \
--data "{\"header1\":\"text1\",\"header2\":\"text2\",\"header3\":\"text3\"}" \

#--user 'Administrator':'MyPS@!'  #< --- This works
--user 'Administrator':'$password'  #< --- Does not work

Posted 1-Nov-21 6:43am

wifinut

Updated 1-Nov-21 7:07am

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Richard MacCutchan · Answer 1 · 2021-11-01T07:07:00

Solution 1

It is because of the uncontinued blank line and the single quotes around $password. The following works*:

Shell

curl "https://somewebsite.com" \
--request POST \
--header "Accept:application/json" \
--header "Content-Type:application/json" \
--data "{\"header1\":\"text1\",\"header2\":\"text2\",\"header3\":\"text3\"}" \
--user 'Administrator':"$password"

*my test did not use curl, but echo, just to check the generated command.

Posted 1-Nov-21 7:07am

Richard MacCutchan

Comments

wifinut 1-Nov-21 13:21pm

👍Thanks so much! Now I learned that I cannot have a blank line in a CURL command.
(facepalm)

Richard MacCutchan 1-Nov-21 13:33pm

The issue was that the blank line did not have a continuation character at the end, so the command string stopped there. The following line was then passed to the shell as if it was a command.

[BASH] passing a variable in CURL

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[BASH] passing a variable in CURL

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