Get Json arrays values in a javascript function

Question

I have defined a json array in a file as

{
    mainServiceBar= [
        { "mainServiceName":"ECONOMIC SERVICES" , "mainServiceValue":"23.2" },
        { "mainServiceName":"SOCIAL SERVICES" , "mainServiceValue":"34.5" },
        { "mainServiceName":"DEFENSE" , "mainServiceValue":"4.5" },
        { "mainServiceName":"GENERAL PUBLIC SERVICES" , "mainServiceValue":"19" },
        { "mainServiceName":"DEBT BURDEN" , "mainServiceValue":"18.8" },
    ];


    subServiceBar1: [
        { "subServiceName":"agriculture" , "subServiceValue":"12.5" },
        { "subServiceName":"trade" , "subServiceValue":"12.5" },
        { "subServiceName":"tourism" , "subServiceValue":"12.5" },
        { "subServiceName":"power and energy" , "subServiceValue":"12.5" },
        { "subServiceName":"water resource" , "subServiceValue":"12.5" }
        { "subServiceName":"communication roads" , "subServiceValue":"12.5" }
        { "subServiceName":"other economic services" , "subServiceValue":"12.5" },
        { "subServiceName":"subsidy" , "subServiceValue":"12.5" },
    ];
}

now i want these json arrays values in a javascript function , I have done few things but it I didn't get the values , my code is as follows :

function getJsonArray() {
    return $.getJSON("myjson.js"); 
}

function socialservice(thisv) {
    json = getJsonArray();

    for (var key in json) {
        if (json.hasOwnProperty(key)) {
            var item = json[key];

            alert(item.mainServiceName);
        }
    }

    return false;
}

**EDITED I have also tried this :

$.getJSON("myjson.json", function(data) {
    alert("val");

    var items = [];

    $.each(data, function(key, val) {
        alert(val);
    });
});

but it do\esn't alert anything means it do not get json data

It alerts undefined , Please suggest me some way to get the values , Thanks

It is true that the json is incorrect but the function to read the json is also incorrect, see my answer. — Yussuf S
– Yussuf S, Commented Sep 20, 2013 at 7:49

Yussuf S · Accepted Answer · 2013-09-20 08:22:07Z

1

please change your json file to below (it was wrongly formatted)

{
    mainServiceBar: [
        { "mainServiceName":"ECONOMIC SERVICES" , "mainServiceValue":"23.2" },
        { "mainServiceName":"SOCIAL SERVICES" , "mainServiceValue":"34.5" },
        { "mainServiceName":"DEFENSE" , "mainServiceValue":"4.5" },
        { "mainServiceName":"GENERAL PUBLIC SERVICES" , "mainServiceValue":"19" },
        { "mainServiceName":"DEBT BURDEN" , "mainServiceValue":"18.8" }
    ],
    subServiceBar1: [
        { "subServiceName":"agriculture" , "subServiceValue":"12.5" },
        { "subServiceName":"trade" , "subServiceValue":"12.5" },
        { "subServiceName":"tourism" , "subServiceValue":"12.5" },
        { "subServiceName":"power and energy" , "subServiceValue":"12.5" },
        { "subServiceName":"water resource" , "subServiceValue":"12.5" },
        { "subServiceName":"communication roads" , "subServiceValue":"12.5" },
        { "subServiceName":"other economic services" , "subServiceValue":"12.5" },
        { "subServiceName":"subsidy" , "subServiceValue":"12.5" }
    ]
}

also change the socialservice function code to below it was wrongly retrieving the values :

function socialservice(thisv) {
    $.getJSON("myjson.js", function(json) {
        for (key in json) {
          $.each(json[key], function(k, arrayItem) {
            alert(arrayItem.mainServiceName);
          });
        }
    });
    return false;
}

call the function with socialservice(); anywhere in your code.

edited Sep 20, 2013 at 8:22

answered Sep 20, 2013 at 7:36

Yussuf S

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5 Comments

Inforian Over a year ago

Still function didn't alert anything (didn't go inside this $.getJSON( "myjson.js", function( json ) { })

Yussuf S Over a year ago

this might sound silly but did you call the function ? try the new code.

Inforian Over a year ago

Yes , I have . Infact alert outside this $.getJSON() is working but not inside

Yussuf S Over a year ago

the only reason I can think of is that the json (js) file is not in the right place.

Inforian Over a year ago

may be , but its in the same directory , let me define full path to file

Luca Rainone · Accepted Answer · 2013-09-20 07:36:37Z

1

Accordin with getJson doc you should write a callback in order to manipulate the data.

$.getJSON( "myjson.js", function( json ) {
    for (var key in json) {
          // code
    }
});

NB: Your JSON code posted is invalid. There are a lot of errors. Try to replace with this:

{
    "mainServiceBar": [
        { "mainServiceName":"ECONOMIC SERVICES" , "mainServiceValue":"23.2" },
        { "mainServiceName":"SOCIAL SERVICES" , "mainServiceValue":"34.5" },
        { "mainServiceName":"DEFENSE" , "mainServiceValue":"4.5" },
        { "mainServiceName":"GENERAL PUBLIC SERVICES" , "mainServiceValue":"19" },
        { "mainServiceName":"DEBT BURDEN" , "mainServiceValue":"18.8" }
    ],
    "subServiceBar1": [
        { "subServiceName":"agriculture" , "subServiceValue":"12.5" },
        { "subServiceName":"trade" , "subServiceValue":"12.5" },
        { "subServiceName":"tourism" , "subServiceValue":"12.5" },
        { "subServiceName":"power and energy" , "subServiceValue":"12.5" },
        { "subServiceName":"water resource" , "subServiceValue":"12.5" },
        { "subServiceName":"communication roads" , "subServiceValue":"12.5" },
        { "subServiceName":"other economic services" , "subServiceValue":"12.5" },
        { "subServiceName":"subsidy" , "subServiceValue":"12.5" }
    ]
}

edited Sep 20, 2013 at 7:36

answered Sep 20, 2013 at 7:21

Luca Rainone

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1 Comment

Inforian Over a year ago

I have tried that to but , it didn't go inside this , I have edited my question please check

LondonAppDev · Accepted Answer · 2013-09-20 07:24:35Z

0

Change it to this:

function getJsonArray(){
    $.ajax({
        url : 'myjson.js',
        type: 'GET',
        success : socialservice
    })
}


function socialservice(data){

json= data;
for (var key in json) {
    if (json.hasOwnProperty(key)) {
    var item = json[key];
    alert(item.mainServiceName); }}
return false;
  }

answered Sep 20, 2013 at 7:24

LondonAppDev

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