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

04/13/2020 15:00:01

Something like:

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

//should be true

//should be false

The solution should not contain try/catch. Some of us turn on "break on all errors" and they don't like the debugger breaking on those invalid JSON strings.

Verified Answer (168 Votes)

09/14/2010 20:45:57

A comment first. The question was about not using try/catch.
If you do not mind to use it, read the answer below. Here we just check a JSON string using a regexp, and it will work in most cases, not all cases.

Have a look around the line 450 in

There is a regexp that check for a valid JSON, something like:

if (/^[\],:{}\s]*$/.test(text.replace(/\\["\\\/bfnrtu]/g, '@').
replace(/"[^"\\\n\r]*"|true|false|null|-?\d+(?:\.\d*)?(?:[eE][+\-]?\d+)?/g, ']').
replace(/(?:^|:|,)(?:\s*\[)+/g, ''))) {

  //the json is ok


  //the json is not ok


EDIT: The new version of json2.js makes a more advanced parsing than above, but still based on a regexp replace ( from the comment of @Mrchief )


Answer #2 (880 Votes)

09/14/2010 20:11:15

Use a JSON parser like JSON.parse:

function IsJsonString(str) {
    try {
    } catch (e) {
        return false;
    return true;

Answer #3 (439 Votes)

12/05/2013 10:46:35

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;
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