json parsing in JavaScript issue

Question

I have the following json:

{
"result":[
   {
     "a": 500000,
     "b": null,
     "c": 0,
     "d": 0,
     "e": 1855
  },
  {
     "a": 500001,
     "b": null,
     "c": 0,
     "d": 0,
     "e": 3770
  }
],
"host": "43252ed565f9",
"time": "Wed Mar 23 09:57:43 UTC 2016",
"status": "Ok"
}

I'm trying to parse it, but got a strange value of "[", instead each one of the keys.

    function f() {
              var request1 = $.ajax({
                  url: '/a', 
                  type: 'GET',
              });
              var request2 = $.ajax({
                   url : '/b',
                   type: "GET"
              });
              $.when(request1, request2).done(function(result1, result2){
                        //result1 = [Object, "success", Object]
                        json =  JSON.stringify(result1)

              for( key in json){
                    myJson = json[key] //The value is: "["

               a = myJson['a']
               b = myJson['b']
               c = myJson['c']
               d = myJson['d']
               e = myJson['e']

              }
})
          }

What could be the issue?

Is this because of the JSON.stringify? How can i parse it as json? — Omri
– Omri, Commented Mar 26, 2016 at 22:03
Possible duplicate of SyntaxError: JSON.parse: unexpected character — Daedalus
– Daedalus, Commented Mar 26, 2016 at 22:24

Ultron · Accepted Answer · 2016-09-20 20:43:44Z

3

No need to parse by JSON.parse() because the answer is already an object itself you can do this.

function f() {
              var request1 = $.ajax({
                  url: '/a', 
                  type: 'GET',
              });
              var request2 = $.ajax({
                   url : '/b',
                   type: "GET"
              });
              $.when(request1, request2).done(function(result1, result2){
                        //result1 = [Object, "success", Object]


              for( key in result1){
                    a = result1[key] //The value is: "["

              }
})
          }

In fact ill show a snippet:

var Json = {
  "result": [{
    "a": 500000,
    "b": null,
    "c": 0,
    "d": 0,
    "e": 1855
  }, {
    "a": 500001,
    "b": null,
    "c": 0,
    "d": 0,
    "e": 3770
  }],
  "host": "43252ed565f9",
  "time": "Wed Mar 23 09:57:43 UTC 2016",
  "status": "Ok"
}

for (key in Json) {
  a = Json[key] //The value is: "["

}
document.getElementById("result").innerHTML = a

<p id="result"></p>

Reason Behind This

The reason behind this is because the JSON is already an object. Therfore there is no need to JSON.parse() it or JSON.stringify() it. Simple as that! Hope this helped!

FOR LOOP

In the for loop im looping over the objects.. And constantly rewritting the variable a. So at the end the variable a would equal "Ok", because thats the last "Object" in the JSON. Now lets say you want to keep every "Objects", then you will need a "count":

count=0; //GLOBAL VARIABLE.
  for ( key in result1){
    actualJSON[count]
    count++;
  }

Now the actualJSON Has the all of the "objects" of that JSON.

edited Sep 20, 2016 at 20:43

Ultron

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answered Mar 26, 2016 at 22:09

amanuel2

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6 Comments

user2864740 Over a year ago

Only 'the person' would know; and there is no requirement to explain downvotes.. annoying as it may be.

conventi Over a year ago

But result1[1] should be statusText as documented at api.jquery.com/jquery.when. Why you write a = result1[key]?

Omri Over a year ago

@AmanuelBogale I din't understand your for loop. Can you explain and add more details regarding how to extract each key and key into a parameter?

conventi Over a year ago

I'm referring to the last example in that page where is write: Each argument is an array with the following structure: [ data, statusText, jqXHR ]. So he must take the first item of the array? I do not want to be right and i'm sorry if i misunderstood your code.

Omri Over a year ago

@AmanuelBogale I'm not sure if i understood the loop section. I got the following error: Uncaught ReferenceError: actualJSON is not defined.

|

conventi · Accepted Answer · 2016-03-26 22:27:33Z

1

Result1 is [Object, "success", Object] and not the JSON as described in final example here https://api.jquery.com/jquery.when/.

result1[0] should be the response of the ajax request and so it should be the javascript object that map the JSON of the response.

edited Mar 26, 2016 at 22:27

answered Mar 26, 2016 at 22:06

conventi

4403 silver badges8 bronze badges

Comments

isvforall · Accepted Answer · 2016-03-26 22:48:43Z

1

result1 and result2 is already the objects, you don't need to parse or stringify it.

Just iterate it:

for(var key in result1){
    var a = result1[key];
    // do something with 'a'
}

edited Mar 26, 2016 at 22:48

answered Mar 26, 2016 at 22:09

isvforall

8,9166 gold badges39 silver badges52 bronze badges

1 Comment

Rajshekar Reddy Over a year ago

replace json with result

JoniJnm · Accepted Answer · 2016-03-26 22:10:52Z

0

JSON.stringify returns a string.

Use:

var json = typeof(result1[0]) === 'string' ? JSON.parse(result1[0]) : result1[0];

The [0] is because $.ajax() returns the data in the first argument in done function.

answered Mar 26, 2016 at 22:10

JoniJnm

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Comments

Rajshekar Reddy · Accepted Answer · 2016-03-26 22:29:51Z

-1

After reading all the comments I understood that your result1 is an object. Hence no need to parse it, you can directly access the values. Here is a example snippet.

var Json = {
  "result": [{
    "a": 500000,
    "b": null,
    "c": 0,
    "d": 0,
    "e": 1855
  }, {
    "a": 500001,
    "b": null,
    "c": 0,
    "d": 0,
    "e": 3770
  }],
  "host": "43252ed565f9",
  "time": "Wed Mar 23 09:57:43 UTC 2016",
  "status": "Ok"
}

for (key in Json) {
  a = Json[key] 
  document.getElementById("result").innerHTML += a + '<br/>';
}

<p id="result"></p>

edited Mar 26, 2016 at 22:29

answered Mar 26, 2016 at 22:04

Rajshekar Reddy

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6 Comments

user2864740 Over a year ago

IIRC, the returned object is already deserialized. Using parse probably results in errors about "o".

Rajshekar Reddy Over a year ago

@user2864740 yes I read that the OP is getting the error in the other answers comment

amanuel2 Over a year ago

@Daedalus lets remove the noise here. As well as Reddy. Lets polish up this question and anwser a bit!

Rajshekar Reddy Over a year ago

@Daedalus changd my answer.

amanuel2 Over a year ago

Im saying that snippet is confusing. It gives a anwser i didnt expect.

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