jquery函数返回数组中的json值

I would make a jquery function that import a json value and return an array when we call it. I coded the following function but the returned array is empty.

我将创建一个jquery函数，该函数导入json值并在调用它时返回数组。我编写了下面的函数，但是返回的数组是空的。

The json file is very basic like this example :

json文件非常基本，如下例所示:

{
"fruit": [ "babana", "pineapple", "apple" ],
"vegetable" :  [ "salad", "carrot", "tomato" ]
}

I call my json with the following function

我使用以下函数调用json

$( function() {
var fruit = new Array();
var vegetable = new Array();
var food = new Array();

function loadJson(result) {
    $.getJSON('myfile.json', function (data) {
        fruit = data.fruit;
        vegetable = data.vegetable;
    })
    .error(function() {
        console.log('error: JSON not loaded');
    })
    .done(function() {
        //console.log( "JSON loaded!" );
        var food = fruit.concat(vegetable);
        return result;
    });
}

After I call this function in another function with

我把这个函数命名为另一个函数。

loadJson(food);   loadJson(fruit);

Can you help me?

你能帮我吗?

1 个解决方案

#1

I think this is what you're looking for, I've fired mine on document ready just so I could test the function quickly, but you should give you the result you're after.

我想这就是你要找的，我已经把我的文档准备好了，这样我可以快速地测试函数，但是你应该给你你想要的结果。

Answer Updated:

回答更新:

function loadJson( type ) {

var final_result = null;
    $.ajax({
        type: 'GET',
        url: "test.json",
        async: false,
        dataType: "json",
        success: function (data) {
            var fruit = data['fruit'],
                vegetable = data['vegetable'],
                food = fruit.concat(vegetable),
                result = food;

            switch( type ) {
                case 'fruit':
                    result = fruit;
                    break;
                case 'vegetable':
                    result = vegetable;
                    break;
                case 'food':
                    result = food;
                    break;
            }
            final_result = result;
        }
    });
    return final_result;
}


$( document ).ready(function() {
    loadJson( 'food' );
    loadJson( 'fruit' );
    loadJson( 'vegetable' );

    var test = loadJson( 'fruit' );
    console.log( test );
 });

#1