jquery json 路径

Question

我有以下 json

{
    "id": "0001",
    "type": "donut",
    "name": "Cake",
    "ppu": 0.55,
    "batters": {
        "batter": [
                { "id": "1001", "type": "Regular" },
                { "id": "1002", "type": "Chocolate" },
                { "id": "1003", "type": "Blueberry" },
                { "id": "1004", "type": "Devil's Food" }
            ]
    },
    "topping": [
        { "id": "5001", "type": "None" },
        { "id": "5002", "type": "Glazed" },
        { "id": "5005", "type": "Sugar" },
        { "id": "5007", "type": "Powdered Sugar" },
        { "id": "5006", "type": "Chocolate with Sprinkles" },
        { "id": "5003", "type": "Chocolate" },
        { "id": "5004", "type": "Maple" }
    ]
}

我正在尝试将 xpath 作为变量传递。

$(document).ready(function(){
    var json_offset = 'topping.types'
    ...
    $.getJSON('json-data.php', function(data) {
        var json_pointer = data.json_offset;
        ...
    });
});

这不起作用。有人可以帮忙吗？

Answer 1

类似的东西应该可以工作(我实际上没有测试它):

Object.getPath = function(obj, path) {
    var parts = path.split('.');
    while (parts.length && obj = obj[parts.shift()]);
    return obj;
}
// Or in CoffeeScript:
// Object.getPath = (obj, path) -> obj=obj[part] for part in path.split '.'; obj

然后，像这样使用它:

Object.getPath(data, json_offset)

但是，除非路径是动态的并且您不能这样做，否则您应该简单地使用data.topping.types。另外，您将该路径称为“XPath”，但 XPath 是一个非常不同的东西，与您想要做的事情无关。