从JSON对象中删除元素

• Z时代
• 2024-01-10
• 分类：问答

我有一个看起来像这样的json数组：

  {
    "id": 1,
    "children": [
    {
        "id": 2,
        "children": {
            "id": 3,
            "children": {
                "id": 4,
                "children": ""
            }
        }
    },
    {
        "id": 2,
        "children": {
            "id": 3,
            "children": {
                "id": 4,
                "children": ""
            }
        }
    },
    {
        "id": 2,
        "children": {
            "id": 3,
            "children": {
                "id": 4,
                "children": ""
            }
        }
    },
    {
        "id": 2,
        "children": {
            "id": 3,
            "children": {
                "id": 4,
                "children": ""
            }
        }
    },
    {
        "id": 2,
        "children": {
            "id": 3,
            "children": {
                "id": 4,
                "children": ""
            }
        }
    },
    {
        "id": 2,
        "children": {
            "id": 3,
            "children": {
                "id": 4,
                "children": ""
            }
        }
    },
    {
        "id": 2,
        "children": {
            "id": 3,
            "children": {
                "id": 4,
                "children": ""
            }
        }
    }]
}

我想有一个函数来删除“孩子”为空的元素。我该怎么做？我不是要答案，只是建议

回答：

要遍历对象的键，请使用for .. in循环：

for (var key in json_obj) {
    if (json_obj.hasOwnProperty(key)) {
        // do something with `key'
    }
}

要测试空元素的所有元素，可以使用递归方法：遍历所有元素，然后也递归地测试它们的孩子。

可以使用delete关键字删除对象的属性：

var someObj = {
    "one": 123,
    "two": 345
};
var key = "one";
delete someObj[key];
console.log(someObj); // prints { "two": 345 }

说明文件：

• https://developer.mozilla.org/zh-CN/docs/JavaScript/Guide/Working_with_Objects
• https://developer.mozilla.org/zh-CN/docs/JavaScript/Reference/Statements/for...in
• https://developer.mozilla.org/zh-CN/docs/JavaScript/Reference/Operators/delete

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