JavaScript JSON Syntax Error: Unrecognized Expression

Question

I am trying to use jQuery to iterate through the objects in this JSON expression:

[
    {
        "value": 609,
        "label": "Wyandotte, MI"
    },
    {
        "value": 3141,
        "label": "Wilmington, NC"
    }
]

Such that each object consists of two properties, value and label.

This is what I have so far:

$(data).each(function () {
     alert('value: ' + this.value + ' label: ' + this.label);
});

But I keep getting an error: Uncaught Error: Syntax error, unrecognized expression

Can anyone help?

Edit:
The data variable in the example was arbitrary. What is actually happening is I'm getting an array of objects something like this:

data =
[
    locations: "[{"value":5626,"label":"Bensenville, IL"}]",
    notes: "Sample note"
]

Then I call data = JSON.parse(data);

And then I perform the iteration on data.locations, which is producing the error.

Edit (again):
https://jsfiddle.net/e2p7gdod/
I keep trying to reproduce what I'm seeing.
This is how I'm producing my return result:

public JsonResult Foo()
{
    var data = new JObject();
    data["locations"] = "[{'value': 609,'label': 'Wyandotte, MI'},{'value': 3141,'label': 'Wilmington, NC'}]";
    data["supervisor"] = "John Doe";
    data["notes"] = "Sample note";

    return new JsonResult()
    {
        Data = JsonConvert.SerializeObject(data),
        JsonRequestBehavior = JsonRequestBehavior.AllowGet
    };
}

Your code as posted does not produce the error. Exactly what happens before your .each() call? Where does data get its value? — Pointy
– Pointy, Commented Jul 10, 2015 at 20:53
Works fine as-is: jsfiddle.net/yk4wb7dv (assuming your array is data) — iCollect.it Ltd
– iCollect.it Ltd, Commented Jul 10, 2015 at 20:56
The data in your Edit is invalid syntax. When you have keys and values, they have to be in an object, not an array. It should be data = { locations: ..., notes: "Sample note" } — Barmar
– Barmar, Commented Jul 10, 2015 at 21:20
And if you're going to pass it to JSON.parse, the value of data should be a string, not an array. — Barmar
– Barmar, Commented Jul 10, 2015 at 21:21
The JSON is invalid. Strings in JSON have to be in double quotes, not single quotes. — Barmar
– Barmar, Commented Jul 10, 2015 at 21:50

Barmar · Accepted Answer · 2015-07-10 21:56:25Z

1

It looks like you're trying to put a JSON object in data.locations, but you have the syntax wrong. Strings in JSON must be enclosed in double quotes, not single quotes.

public JsonResult Foo()
{
    var data = new JObject();
    data["locations"] = "[{\"value\": 609,\"label\": \"Wyandotte, MI\"},{\"value\": 3141,\"label\": \"Wilmington, NC\"}]";
    data["supervisor"] = "John Doe";
    data["notes"] = "Sample note";

    return new JsonResult()
    {
        Data = JsonConvert.SerializeObject(data),
        JsonRequestBehavior = JsonRequestBehavior.AllowGet
    };
}

Then in the Javascript, when you want to iterate over it, you need to call JSON.parse(data.locations):

var locations = JSON.parse(data.locations);
$.each(locations, function() {
    alert('value: ' + this.value + ' label: ' + this.label);
});

answered Jul 10, 2015 at 21:56

Barmar

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2 Comments

Matthew Over a year ago

The second part was the problem. I was calling JSON.parse on the data and then calling $(data.divisions).each(...) but I was not calling $(JSON.parse(data.divisions)).each()!!! Thank you thank you thank you! :-)

Barmar Over a year ago

Why are you putting JSON into data["locations"] in the first place? Why not just put a C# array of objects there?

iCollect.it Ltd · Accepted Answer · 2015-07-10 20:56:41Z

1

That should be the static version of each, passing the array (faster than wrapping the array in a jQuery object, just to iterate over it):

$.each(data, function () {
     alert('value: ' + this.value + ' label: ' + this.label);
});

Note: Your current code does not give that error and appears to work fine:

http://jsfiddle.net/yk4wb7dv/

Have you included JQuery etc?

edited Jul 10, 2015 at 20:56

answered Jul 10, 2015 at 20:53

iCollect.it Ltd

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6 Comments

Pointy Over a year ago

Probably better, but the code as posted won't cause the error reported.

iCollect.it Ltd Over a year ago

@Pointy: $(data) probably would cause that error if data is his object as I suspect.

Pointy Over a year ago

Well the OP says that data is an array. If that's the case, then jQuery will just wrap itself around the array.

Fiid Over a year ago

I've always done function(index, item) { ... } and then referenced item in the block. "this" handling in JS scares me. :)

iCollect.it Ltd Over a year ago

@Fiid: You can do that too. No need to be afraid of this though :)

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