How to check if a string is a valid JSON string in JavaScript without using Try/Catch

Question

Something like:

var jsonString = \'{ \"Id\": 1, \"Name\": \"Coke\" }\';

//should be true
IsJsonString(jsonString);

//should be false
IsJsonString(\"foo\");         


        

Answer 1

I know i'm 3 years late to this question, but I felt like chiming in.

While Gumbo's solution works great, it doesn't handle a few cases where no exception is raised for JSON.parse({something that isn't JSON})

I also prefer to return the parsed JSON at the same time, so the calling code doesn't have to call JSON.parse(jsonString) a second time.

This seems to work well for my needs:

function tryParseJSON (jsonString){
    try {
        var o = JSON.parse(jsonString);

        // Handle non-exception-throwing cases:
        // Neither JSON.parse(false) or JSON.parse(1234) throw errors, hence the type-checking,
        // but... JSON.parse(null) returns null, and typeof null === "object", 
        // so we must check for that, too. Thankfully, null is falsey, so this suffices:
        if (o && typeof o === "object") {
            return o;
        }
    }
    catch (e) { }

    return false;
};