JSON Data cannot be stored in variable

Question

Im trying to load the content of a JSON File into an variable. Before, the variable looked something like this:

var Data = {
    teams : [
      ["Team 1", "Team 2"], 
      ["Team 3", "Team 4"]  
    ],
    results : [
      [[1,2], [3,4]],       
      [[4,6], [2,1]]        
    ]}

Now I have a JSON File looking something like this:

{"teams":[["Team 1","Team 2"],["Team 3","Team 4"],"results":[[[[1,2],[3,4]],[[4,6],[2,1]]]}

Now I want that the the content of the JSON File is stored in the Data Variable before. I tried it with Ajax which looks like this:

$.ajax({
          type: "GET",
          dataType : 'json',
          async: true,
          url: 'data.json',
          success: function(data) {
          console.log(data)
          var Data = data
          },
      });

Console.log works perfect, but the Data is not saved in the variable and I'm getting the error: Uncaught ReferenceError: Data is not defined.

I also tried it with var Data = JSON.parse(data), but this doesn't seem to work either.

And now I'm stuck without any clue.

Thanks for help in advance.

ajax commands are asynchronous, you first have to learn about promises to fix this question — Mister Jojo
– Mister Jojo, Commented Jun 8, 2021 at 23:45
Not only might there be a asynchronous issue, but var Data will only be available within the scope of the success callback. You'll need a way to store your state where it's available outside that function. — jnpdx
– jnpdx, Commented Jun 9, 2021 at 0:02
Uncaught ReferenceError: Data is not defined means that the JS Interpreter found a reference in the name of Data that is not defined in the Scoop within which it's being used. SHORT STORY: At some other place, you use Data outside of async ajax call. You should put that code into success function, or put it into a function with data argument and call that function inside sucess function while passing Data as parameter — KeitelDOG
– KeitelDOG, Commented Jun 9, 2021 at 0:02

erics2783 · Accepted Answer · 2021-06-11 13:45:12Z

3

I'm not sure what your code looks like after the ajax call, but I'm guessing the the code where you are using Data is after the ajax call. ajax is asynchrounous. That means that your code doesn't wait for it to finish before moving on. Any code that needs to wait until after it's done fetching the data, you can put in the .success function. Also, it's worth noting that success only gets called when the ajax request is successful. If you want to handle errors, you can use .error Something like this should work:

$.ajax({
    type: "GET",
    dataType : 'json',
    async: true,
    url: 'data.json',
    success: function(data) {
      console.log(data)
      var Data = data;
      // Anything that needs to use Data would go inside here
    },
    error: function(err) {
      // handle errors
      console.error(err);
    }
});

// Any code here will not wait until `Data` is defined
// console.log(Data) // error

edited Jun 11, 2021 at 13:45

answered Jun 8, 2021 at 23:45

erics2783

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6 Comments

BvstedCrew Over a year ago

gives me the following error Uncaught Error: Trying to wrap Option into an Option

erics2783 Over a year ago

@BvstedCrew I updated my answer. I didn't think about ajax being async. The previous answer wouldn't work.

BvstedCrew Over a year ago

with a simple change from async: false to async: true everything seems to work fine. Thanks!

KeitelDOG Over a year ago

The 1st code was incorrect. Now the edit is good. You can also define the code outside in a function and call that function inside success callback to be more readable if the code is very long.

Matt Browne Over a year ago

That catch wouldn't do anything. You can use either the error option, or you can use .fail() like this: $.ajax({...}).fail(err => { ... }). There is no catch option or callback for jQuery (you're probably thinking of native promises).

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