Array In Array in JSON

Question

My javascript code;

function callDist(sel)
{
   $.getJSON('Dist.json', function(data){
    var $container = $('#parameters').empty();

    $.each(data.distributions, function(i, distribution){
         $.each(distribution, function(key,value) 
         {
          if(distribution.type==sel.value) {
            alert(distribution.name);
            $.each(distribution.parameters, function(key, value) {
            $container.append(key + ':' + value + '<br />');
            $container.append('<hr />');
          });
         }
    });
 }
 );
 });
 }

My Json code;

{
"distributions":[
  { 
     "name":"Uniform",
     "type":1,
     "parameters":[{ "minValue":2 , "maxValue":4 }]
  },
  {  "name":"Normal",
     "type":2,
     "parameters":[{ "mean":5 , "standartDeviation":3 }]
  },
  {
     "name":"Exponential",
     "type":3,
     "parameters":[{"lamda":2}]
  },
  {
     "name":"Geometric",
     "type":4,
     "parameters":[{"probability":0.2}]
  }
 ]
}

Html code;

<select name="selectDistribution" class="span12"  id="Options"       

 onchange="callDist(this);" >
      <option value="0">Choose one distribution</option>
      <option value="1">Uniform</option>
      <option value="2">Normal</option>
      <option value="3">Exponential</option>
      <option value="4" >Geometric</option>
</select>

In Json there is an array that created by some distributions.when the user selects some of them from select option I am triggering callDist(sel) function. When the passed value sel is equal to distribution.type , I want to write the parameters of that distribution to a div.How can I do that?Thanks.

Never call a variable JSON, as that will override the native JSON object. And what have you tried to accomplish your task? — MaxArt
– MaxArt, Commented Mar 21, 2013 at 15:34

000 · Accepted Answer · 2013-03-21 16:04:23Z

2

function callDist(sel) {
    $.getJSON('Dist.json', function(json){
        var $container = $('#parameters').empty();


        $.each(json.distributions, function(i, distribution) {
            if(distribution.type==sel.value) {
                $.each(distribution.parameters, function(j, parameters) {
                    $.each(parameters, function(parameter_key, parameter_value) {
                        $container.append(parameter_key + ':' + parameter_value + '<br />');
                        $container.append('<hr />');
                    });
                });
             }
        });
    });
}

edited Mar 21, 2013 at 16:04

answered Mar 21, 2013 at 15:39

000

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6 Comments

ntf Over a year ago

It seems true but this time, there is no output.

ntf Over a year ago

I run my project from wamp and localhost.And there is no error.Only div tag for paramters is empty now.

000 Over a year ago

I have it running in a jsfiddle.

ntf Over a year ago

I see the result is correct, but why I cannot get the same result in my project:(

000 Over a year ago

Have you copy/pasted my solution?

|

ckersch · Accepted Answer · 2013-03-21 15:51:51Z

With distributions set up as an array, you'll have to loop through distributions and check to see if the type value of each element matches sel, and then append the parameters of that object to the div in question, unless position always translates to a certain type. If that's the case, something like

var distribution = data[(sel-1)]

can be used to set the distribution.

What might make more sense, since there may be cases where position in the array does not correspond to type, is to restructure your JSON so that you have an object instead of an array, like the following:

{"distributions":{
    "Uniform":{
        "name":"Uniform",
        "type":1,
         "parameters":[{ "minValue":2 , "maxValue":4 }]
    },
    "Normal":{
         "name":"Normal",
         "type":2,
         "parameters":[{ "mean":5 , "standartDeviation":3 }]
    }
  }
}

In this case, I've used the name field as the attribute names of the distributions object, and can get my parameters for any given distrubution by typing data[selName].parameters, in which selName is the name corresponding to the selection, but you could just as easily use the type as the attribute name.

Both of these methods would remove the need to loop through the data.

jagzviruz · Accepted Answer · 2013-03-21 15:57:20Z

0

I made a fiddle around it here . I am able to see the alert with the correct value. I used

$('#Options').change()

to trigger the function and a few other modifications to work on fiddle.

answered Mar 21, 2013 at 15:57

jagzviruz

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1 Comment

ntf Over a year ago

I want to get the parameters of the wanted distribution.And I want to write the parameters to parameters div

jagzviruz · Accepted Answer · 2013-03-21 15:40:32Z

-1

Remove the 2nd comma in

$.each(data.distributions, function(i, distribution,) {

answered Mar 21, 2013 at 15:40

jagzviruz

1,52313 silver badges28 bronze badges

2 Comments

ntf Over a year ago

I changed it but this time I got 0:[object Object] at three times.

ntf Over a year ago

but I can get only names of the distributions with your solution.

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