Convert JSON to a different format (nested json)

Question

This is really not a question. I was looking up this solution but couldn't find it anywhere. Thought of posting it here so it may save someone's time.

Here was my issue: I had a JSON coming from server which was not nested. Lets take example of Movies:

var movies = [
  {"name":"Ice Age 3", "language":"English", "year":"2012"},
  {"name":"Ice Age 3", "language":"French", "year":"2011"},
  {"name":"Ice Age 3", "language":"German", "year":"2013"}
];

Now effectively speaking, there is no need for server to return it like this, however this was my case and since I was using jquery-template I wanted to have it as a nested JSON object. Something like this:

var movies = [{
    "name": "Ice Age 3",
    "details": [
        {"language": "English", "year": "2012"}, 
        {"language": "French", "year": "2011"}, 
        {"language": "German", "year": "2013"}
    ]
}];

Here is the code which I wrote to convert it:

function convert(arr) {
    var convertedArray = [];

    $(arr).each(function () {
        var found = false;
        for (var i = 0; i < convertedArray.length; i++) {
            if (convertedArray[i].name === this.name) {
                found = true;
                convertedArray[i].details.push({
                    "language": this.language,
                    "year": this.year
                });
                break;
            }
        }
        if (!found) {
            convertedArray.push({
                "name": this.name,
                "details": [{
                    "language": this.language,
                    "year": this.year
                }]
            });
        }
    });

    return convertedArray;
}

I am sure there would be a better approach for this. But for me it worked.

It's okay to answer your own question, but please pose the question as a question, and post a separate, actual answer. meta.stackexchange.com/q/17463/133242 — Matt Ball
– Matt Ball, Commented Aug 7, 2012 at 19:39

Ivan Dyachenko · Accepted Answer · 2012-08-07 20:29:39Z

2

var found = false;

Using flag is not good practice.

This is better solution:

var movies = [
  {"name":"Ice Age 3", "language":"English", "year":"2012"},
  {"name":"Ice Age 3", "language":"French", "year":"2011"},
  {"name":"Ice Age 3", "language":"German", "year":"2013"},
  {"name":"Ice Age 4", "language":"German", "year":"2013"}
];

function convert(arr) {
    return $.map(unique(arr), function(name){
        return {
            name: name,
            details: $.grep(arr, function(item){
                return item.name == name
            })
        } 
    });
}

function unique(arr) {
    var result = [];
    $.map(arr, function(item){
        if ($.inArray(item.name, result))
            result.push(item.name);
    })
    return result;
}

console.log(convert(movies));

answered Aug 7, 2012 at 20:29

Ivan Dyachenko

\$\begingroup\$ Hi Ivan, Thanks for the wonderful code. However the following input displays the data incorrectly: var movies = [ {"name":"Ice Age 3", "language":"English", "year":"2012"}, {"name":"Ice Age 4", "language":"French", "year":"2011"}, {"name":"Ice Age 3", "language":"German", "year":"2013"}, {"name":"Ice Age 4", "language":"German", "year":"2013"}, {"name":"Ice Age 3", "language":"German", "year":"2014"} ]; \$\endgroup\$

Shantanu Wagh
– Shantanu Wagh

2012-08-08 20:34:24 +00:00
Commented Aug 8, 2012 at 20:34

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