bashshellcurl

How to pass a variable in a curl command in shell scripting


I have a curl command:

curl -u ${USER_ID}:${PASSWORD} -X GET 'http://blah.gso.woo.com:8080/rest/job-execution/job-details/${job_id}'

The variable job_id has a value in it, say, 1160. When I execute the curl command in shell it gives me the following error:

{"message":"Sorry. An unexpected error occured.", "stacktrace":"Bad Request. The request could not be understood by the server due to malformed syntax."}

If I pass the number '1160' directly in the command, as shown below, the curl command works.

curl -u ${USER_ID}:${PASSWORD} -X GET 'http://blah.gso.woo.com:8080/rest/job-execution/job-details/1160'

I want to be able to pass the value of the variable in the curl command.


Solution

  • When using variables in , you can only use doubles quotes, not single quotes : the variables inside single quotes are not expanded. Learn the difference between ' and " and `. See http://mywiki.wooledge.org/Quotes and https://web.archive.org/web/20230314111401/https://wiki.bash-hackers.org/syntax/words