Question

我是初学者，并且不知道如何从JSON获取数据。我有以下JSON结构，

{
"title": "Delicate style of embroidery lawn dress",
"field": [
    {
        "row1": {
            "key": "free_homedelivery",
            "value": "yes"
        }
    },
    {
        "row2": {
            "key": "_price",
            "value": "60"
        }
    },
    {
        "row3": {
            "key": "_regular_price",
            "value": "60"
        }
    },
    {
        "row4": {
            "key": "_stock_status",
            "value": "instock"
        }
    }
         ]
}

我试过了，但控制台说undefined

for(var i =0 ;i < s.length; i++)
{
console.log('key ' + s.field['row'+i].key);
}

这似乎没有获取值......

Here is the Fiddle

Answer 1

问题

什么是s？ - 假设s是您粘贴的json
迭代在s.field上，但s.length已使用
将field称为数组，但尝试访问数组内对象的属性作为数组的属性。
数组迭代从0开始，row名称以row1

var numFields = s.field.length; // Good practice to keep length calculated
for(var i = 0; i < numFields; i++){
    console.log('key ' + s.field[i]['row'+(i+1)].key);
}

脚注 - JSON太复杂了，它可以像下面这样简单，

var s = {
    "title": "Delicate style of embroidery lawn dress",
    "field": [
        {
            "key": "free_homedelivery",
            "value": "yes"
        },
        {
            "key": "_price",
            "value": "60"
        }
    ]
};
// Updated loop
var numFields = s.field.length; // Good practice to keep length calculated
for(var i = 0; i < numFields; i++){
    console.log('key ' + s.field[i].key);
}

希望这有帮助！

小提琴 - http://jsfiddle.net/n4J5U/

Answer 2

你在json中的对象有点不同（属性名称是row1，row2等），所以你必须做这样的事情：

var json = {
    "title": "Delicate style of embroidery lawn dress",
        "field": [{
        "row1": {
            "key": "free_homedelivery",
                "value": "yes"
        }
    }, {
        "row2": {
            "key": "_price",
                "value": "60"
        }
    }, {
        "row3": {
            "key": "_regular_price",
                "value": "60"
        }
    }, {
        "row4": {
            "key": "_stock_status",
                "value": "instock"
        }
    }]
}

for (i = 0; i < json.field.length ; i++){
    var item = json.field[i];    
    var key = item["row" + (i + 1)].key;
    alert(key);
}

Fiddle

遍历表行只选择特定的列名和值

2 个答案: