Passing Bash Variable to CURL

Question

when i give the values like this it works :

curl --silent \ --insecure \ -X POST \ -d '{"Name" : "Vikram"}' \ -H "Content-Type: application/json" \ $restUrl

but when i give it like this :

post="'{\"Name\" : \"Vikram\"}'"    
echo $post   // Prints '{"Name" : "Vikram"}'
echo ${post} // Prints '{"Name" : "Vikram"}'

but the following does not work and throws an 400 error:

curl --silent \ --insecure \ -X POST \ -d ${post} \ -H "Content-Type: application/json" \ $restUrl

Solution 1

It seems there are multiple issues.

1) To address the question asked by your title, when you use $post as an argument, the white space in its value causes it to be treated as multiple arguments by Bash. Try putting quotes around it so that it's treated as one argument.

2) I'm guessing you added single quotes to $post, in an attempt to have Bash treat it as a single parameter. Try removing the single quotes.

3) For me at least, all of the backslashes in the curl command were causing it to fail. Maybe you had it split across multiple lines and the copy/paste didn't translate that. I've removed them in my example below.

Putting all together:

post="{\"Name\" : \"Vikram\"}"
curl --silent --insecure -X POST -d "${post}" -H "Content-Type: application/json" $restUrl

Solution 2

Don't put single quotes inside the $post variable, put them around the value to quote it. Then you don't need to escape the double quotes inside the value.

post='{"Name" : "Vikram"}'

Then quote the variable when you use it in the curl command line, to prevent it from being split into multiple arguments.

You also shouldn't have all those backslashes in the curl command line. They're escaping the space after them, so they'll be treated as literal arguments rather than delimiters. The usual time to use backslash in a command is if you're splitting it across multiple lines, then you need to escape the newline so it doesn't end the command.

curl --silent --insecure -X POST -d "${post}" -H "Content-Type: application/json" "$restUrl"

Comments

[Vikram]

But still am getting the same error i tried what you suggested but i could not resolve the issue.

[twm]

@Vikram I've updated my answer. I think you probably also have single quotes that you want to remove. It's hard to say without knowing what's on the server side for $restUrl.

[Vikram]

contentType="Content-Type: application/json" How to include it with header like -H ${contentType} does not work. Any workaround

[Barmar]

Did you read my answer? What does it say about quoting variables?

[Barmar]

@Vikram 99% of the time you want to quote variables when you use them.

[Barmar]

It should work. Update the question to show exactly what you're doing.

[Barmar]

Or post a new question, since you've already marked this question as solved.

[Vikram]

contentType="Content-Type: application/json" curl --silent --insecure -X POST -d "${post}" -H "${contentType}" $restUrl but here it returns 415

[Vikram]

Let us continue this discussion in chat.

[Barmar]

You can't put the variable assignment on the same line that you use the variable.

[Barmar]

All variables are expanded before executing the command, so it doesn't execute the assignment until after the variable is expanded.

[Vikram]

Can i use the curl command like this : curl --silent \ --insecure \ -X POST \ -d {"Name" : "Vikram"}, '{"Country" : "India"} \ -H "Content-Type: application/json" \ $restUrl Hi i have the requirement to pass multiple parameters and the rest method is : @RequestMapping(method = RequestMethod.POST, value = "/save") public void save( @RequestParam(name = "country") String country, @RequestBody Name name)