如何在Perl脚本中将变量传递给curl命令？

Question

I am trying to pass a variable to curl command in Perl script. But it is failing.

But when I run the same curl command from command prompt it is working.

my $id=3;

system('curl -D- -u username:password -X PUT --data {\"fields\":{\"priority\":{\"id\":\"${id}\"}}} -H "Content-Type: application/json" -k https:request);

When I execute above Perl script I am getting error as below.

{"errorMessages":[],"errors":{"priority":"The priority selected is invalid."}}after curl command 

When I run the above command from command prompt by replacing id with value it is passing.

'curl -D- -u username:password -X PUT --data {\"fields\":{\"priority\":{\"id\":\"3 \"}}} -H "Content-Type: application/json" -k https:request

Please help me here and let me know what is wrong in my code.

Answer 1

You are placing the command inside the single quote. And perl takes it as it is inside single quote. As you are passing $id variable with the command, I'll suggest you to use double quote. For example

system("curl ... $id... -k https:request");
       ^ double quote                   ^

Or you can do it like this way by using concatenation.

system('curl -D- -u ...' . $id . '... -k https:request');
                          ------ concatenation