SyntaxError: Unexpected token o in JSON at position 1

Question

I'm parsing some data using a type class in my controller. I'm getting data as follows:

{  
   "data":{  
      "userList":[  
         {  
            "id":1,
            "name":"soni"
         }
      ]
   },
   "status":200,
   "config":{  
      "method":"POST",
      "transformRequest":[  
         null
      ],
      "transformResponse":[  
         null
      ],
      "url":"/home/main/module/userlist",
      "headers":{  
         "rt":"ajax",
         "Tenant":"Id:null",
         "Access-Handler":"Authorization:null",
         "Accept":"application/json, text/plain, */*"
      }
   },
   "statusText":"OK"
}

I tried to store the data like this

var userData = _data;
var newData = JSON.parse(userData).data.userList;

How can I extract the user list to a new variable?

You may need not use `JSON.parse`. Try using `userData` directly as an object. — Mohit Bhardwaj, Jul 14 '16 at 17:25
If `console.log(typeof userData)` shows `object` then you already have an javascript object and not a JSON string you need to parse. — t.niese, Jul 14 '16 at 17:27
Usually whenever you get this error - `Unexpected token o in JSON`, most probably you are trying to parse an object which is already in parsed form. — Mohit Bhardwaj, Jul 15 '16 at 06:48
Data is already in JSON format, No need to covert it or parse it. — akshay_sushir, Aug 05 '21 at 08:56

score 553 · Accepted Answer · answered Jul 14 '16 at 17:37

553

The JSON you posted looks fine, however in your code, it is most likely not a JSON string anymore, but already a JavaScript object. This means, no more parsing is necessary.

You can test this yourself, e.g. in Chrome's console:

new Object().toString()
// "[object Object]"

JSON.parse(new Object())
// Uncaught SyntaxError: Unexpected token o in JSON at position 1

JSON.parse("[object Object]")
// Uncaught SyntaxError: Unexpected token o in JSON at position 1

JSON.parse() converts the input into a string. The toString() method of JavaScript objects by default returns [object Object], resulting in the observed behavior.

Try the following instead:

var newData = userData.data.userList;

answered Jul 14 '16 at 17:37

TimoStaudinger

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So it this applicable for all kinds of browsers ? – AdityaKapreShrewsburyBoston Jun 12 '17 at 18:51
@Timo Could you look this link. – Varun Sharma Nov 15 '17 at 05:38
Sometimes this error can surface when the api url endpoint that serves the request has an error in its code or some other included or used file and it throws a error which is not handled or noted. Mostly you can get this if you look at the network tab in Browser developer tools or you can hit the endpoint using post man and see what comes up. – MuturiAlex May 23 '20 at 16:54
Thank you for the explanation on the JSON String vs JS Obj distinction, easier to remember how to do it right when you know why it happened. – BrettJ Oct 26 '20 at 04:49
Saved me a few minutes, thanks. I was using require. – Robert Moskal Dec 14 '21 at 19:22
1

In express I was doing JSON.parse(response.data) which I had forgot was already parsed into json object hence causing this error. – Emile Jan 22 '22 at 10:52

score 211 · Answer 2 · edited Sep 18 '21 at 10:46

211

The first parameter of the JSON.parse function is expected to be a string, and your data is a JavaScript object, so it will coerce it to the string "[object Object]". You should use JSON.stringify before passing the data:

JSON.parse(JSON.stringify(userData))

edited Sep 18 '21 at 10:46

Sebastian Simon

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answered Apr 04 '18 at 19:30

huruji

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this will not work when the string contains a double quote do you have any solution – Mr Coder Jun 08 '20 at 13:29
17

Maybe I'm missing something obvious, but why would you turn an object into JSON (which is what `JSON.stringify()` does), only to parse the JSON back into an object? Why not use the object directly? This seems completely superfluous. – Ivar Aug 25 '20 at 13:44
Probably not the example from the OP - but sometimes you need the JSON structure and not JavaScript one for some libraries. This helped. – G. Urikh Sep 24 '21 at 13:20
1

If you need JSON then you **wouldn't** use `JSON.parse` since that turns the JSON back into JS. – Quentin Jan 24 '22 at 14:50

score 51 · Answer 3 · edited Aug 04 '20 at 17:29

51

Don't ever use JSON.parse without wrapping it in try-catch block:

// payload 
let userData = null;

try {
    // Parse a JSON
    userData = JSON.parse(payload); 
} catch (e) {
    // You can read e for more info
    // Let's assume the error is that we already have parsed the payload
    // So just return that
    userData = payload;
}

// Now userData is the parsed result

edited Aug 04 '20 at 17:29

calvin11

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answered Jul 14 '16 at 17:42

Kousha

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This doesn't actually answer the question, but comments on lack of exception handling. – Richard Duerr Oct 26 '16 at 21:48
3

@RichardDuerr, but this will helps to fix error of head topic. i.e. SyntaxError: Unexpected token o in JSON at position 1 – Niko Jojo May 29 '19 at 07:02
12

That merely hides the error, but not actually solve it. – Richard Duerr May 29 '19 at 23:44

score 28 · Answer 4 · edited Aug 14 '17 at 13:34

28

Just above JSON.parse, use:

var newData = JSON.stringify(userData)

edited Aug 14 '17 at 13:34

bruntime

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answered Apr 06 '17 at 10:26

Sukhchain

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`JSON.stringify()` converts a JavaScript object to a string representation of it, which is the opposite of what `JSON.parse()` does. You were getting the `SyntaxError` because you were trying to parse something that was already an object. In @Sukhchain's solution, it gets converted to a string to avoid that. – Hubert Oct 03 '17 at 18:54
1

The downside of that is, in the end you're using `JSON.parse()` too much redundantly. Even though it's a pretty fast process, parsing JSON is done synchronously and can potentially block your UI, so I'd advise against using that. Instead, you could check if your variable is an object, for example using `typeof(userData) === 'object'` before attempting to parse it. – Hubert Oct 03 '17 at 18:54

score 10 · Answer 5 · answered Jul 08 '20 at 10:36

10

We can also add checks like this:

function parseData(data) {
    if (!data) return {};
    if (typeof data === 'object') return data;
    if (typeof data === 'string') return JSON.parse(data);

    return {};
}

answered Jul 08 '20 at 10:36

Pulkit chadha

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score 6 · Answer 6 · answered Feb 10 '21 at 08:43

You can simply check the typeof userData & JSON.parse() it only if it's string:

var userData = _data;
var newData;
if (typeof userData === 'object')
  newData = userData.data.userList; // dont parse if its object
else if (typeof userData === 'string')
  newData = JSON.parse(userData).data.userList; // parse if its string

Yergalem · Answer 7 · 2017-06-12T16:10:18.410

Unexpected 'O' error is thrown when JSON data or String happens to get parsed.

If it's string, it's already stringfied. Parsing ends up with Unexpected 'O' error.

I faced similar( although in different context), I solved the following error by removing JSON Producer.

    @POST
    @Produces({ **MediaType.APPLICATION_JSON**})
    public Response login(@QueryParam("agentID") String agentID , Officer aOffcr ) {
      return Response.status(200).entity("OK").build();

  }

The response contains "OK" string return. The annotation marked as @Produces({ **MediaType.APPLICATION_JSON})** tries to parse the string to JSON format which results in Unexpected 'O'.

Removing @Produces({ MediaType.APPLICATION_JSON}) works fine. Output : OK

Beware: Also, on client side, if you make ajax request and use JSON.parse("OK"), it throws Unexpected token 'O'

O is the first letter of the string

JSON.parse(object) compares with jQuery.parseJSON(object);

JSON.parse('{ "name":"Yergalem", "city":"Dover"}'); --- Works Fine

score 3 · Answer 8 · answered Sep 07 '22 at 11:25

3

The reason for the error is Object is not in the form of string, it is {} and it should be string Object('{}')

answered Sep 07 '22 at 11:25

KARTHIKEYAN.A

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score 0 · Answer 9 · answered Sep 24 '20 at 14:05

0

Give a try catch like this, this will parse it if its stringified or else will take the default value

let example;
   try {
   example  = JSON.parse(data)
  } catch(e) {
    example = data
  }

answered Sep 24 '20 at 14:05

Pran R.V

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score -1 · Answer 10 · edited Dec 09 '21 at 13:19

-1

First set request value in variable like:

let reqData = req.body.reqData;
if (reqData) {
  let reqDataValue = JSON.parse(JSON.stringify(reqData));
}

edited Dec 09 '21 at 13:19

Mario Petrovic

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answered Dec 09 '21 at 07:32

Ajay Prajapati

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SyntaxError: Unexpected token o in JSON at position 1

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