Parsing JSON with JQuery

Question

I have some JSON that looks like the following:

[{
    "Age": 35,
    "FirstName": "Peyton",
    "LastName": "Manning"
  },
  {
    "Age": 31,
    "FirstName": "Drew",
    "LastName": "Brees"
  },
  {
    "Age": 58,
    "FirstName": "Brett",
    "LastName": "Favre"
  }
]

This JSON is passed into a function I have created called parseJSON. This function looks like this:

function parseJSON(result)
{
  $.each(result.items,function(i,item) {
    alert("First Name: " + item.FirstName);
    alert("Age: " + item.Age);
  });
}

When I run this code though, I get a JavaScript error that says "G is undefined". How do I parse my JSON with JQuery such that I can read the FirstName and Age properties?

Thank you!

ingeniarius · Accepted Answer · 2010-01-19 18:49:25Z

2

Most simple parse JSON by native JS:

function parseJSON(result)
{
  for(var i in result) {
    var item = result[i];
    alert("First Name: " + item.FirstName);
    alert("Age: " + item.Age);
  } 
}

answered Jan 19, 2010 at 18:49

ingeniarius

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1 Comment

Kristoffer Sall-Storgaard Over a year ago

I agreed, bypass jQuery entirely

Singularity · Accepted Answer · 2010-01-19 15:31:31Z

Here is a working function that uses your sample case.

var result = [{
    "Age": 35,
    "FirstName": "Peyton",
    "LastName": "Manning"
}, {
    "Age": 31,
    "FirstName": "Drew",
    "LastName": "Brees"
}, {
    "Age": 58,
    "FirstName": "Brett",
    "LastName": "Favre"
}];

function parseJSON(result){
    $.each(result, function(index, object){
        $.each(object, function(i, o){
            alert(i + " = " + o);
        })
    })
}

parseJSON(result);

Christian Waidner · Accepted Answer · 2010-01-19 15:00:36Z

0

Check your complete code, not only the sample shown here for "G". I suppose, there is a syntax error somewhere. As to the use of "result" - it could be a global variable, but tvanfosson is right I think. Doesn't look good here.

answered Jan 19, 2010 at 15:00

Christian Waidner

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1 Comment

tvanfosson Over a year ago

I think G is probably an internal jQuery function after minification.

geowa4 · Accepted Answer · 2010-01-19 19:41:25Z

0

From json.org: json_sans_eval. This parses a JSON string without the hazards of using eval.

answered Jan 19, 2010 at 19:41

geowa4

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Comments

tvanfosson · Accepted Answer · 2010-01-28 15:21:22Z

0

You are using a variable called result that doesn't exist in your function. Perhaps, it should be replaced with json.

function parseJSON(json) 
{ 
  $.each(json,function(i,item) { 
    alert("First Name: " + item.FirstName); 
    alert("Age: " + item.Age); 
  }); 
}

EDIT: based on your update, I think it's because you are referencing a property on result that doesn't exist, namely items. The null value being passed into the each function is causing jQuery to choke on an internal function.

edited Jan 28, 2010 at 15:21

answered Jan 19, 2010 at 14:54

tvanfosson

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6 Comments

tvanfosson Over a year ago

Nope. The object doesn't have an items property. It's simply an array of objects. Each object has an Age, FirstName, etc. property, so I think it is correct as I have it. -- this was in response to a comment that it should be json.items instead of simply json.

tvanfosson Over a year ago

I should also add that I'm assuming that it isn't a string, i.e., the json string has already been "objectified."

naugtur Over a year ago

How should changing a variable name fix any problems?

tvanfosson Over a year ago

@naugtur -- check the edit history on the question. Originally, the function didn't have result in it. You're downvoting me because my answer reflected the OP's omission.

naugtur Over a year ago

That's a big "excuse me" then. I'll make it up for You.

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