使用不同数据选项之间的循环迭代对象 [英] Iterate object with loop between different data options

问题描述

假设我在人类对象中有几个嵌套对象（human1，human2，human3）。

Suppose I have several nested objects(human1, human2, human3) in "human" object.

human: {
    "human1": {
        "name" : "John",
        "sex" : "male",
        "age" : 18
    }

    "human2": {
        "name" : "Peter",
        "sex" : "male",
        "age" : 16
    }

    "human3": {
        "name" : "May",
        "sex" : "female",
        "age" : 19
    }
}

我在下面有另一个名为 currentPlayer 的对象，我希望它是一个容器，以便从human1，human2访问数据，或human3用于不同用途。

And I have another object called currentPlayer below, which I want it to be a vessel, in order to access the data from "human1", "human2", or "human3" for different use.

currentPlayer: {
    "name" : "default",
    "sex" : "default",
    "age" : 0
}

示例：今天我希望currentPlayer成为John，它就是

Example: today I want currentPlayer to be John, and it goes

currentPlayer: {
    "name" : "John",
    "sex" : "male",
    "age" : 18
}

然后我希望currentPlayer成为Peter，它就是：

And then I want currentPlayer to be Peter, and it goes:

currentPlayer: {
    "name" : "Peter",
    "sex" : "male",
    "age" : 16
}

如何使用循环迭代currentPlayer的属性值，而不仅仅是逐个键入？谢谢...

How do I iterate property values of currentPlayer like this with loop, not just key in one by one? Thanks...

推荐答案

Bellow代码将遍历人类对象的所有属性

Bellow code will iterate through all Properties of human object

listofhuman = Object.getOwnPropertyNames(human);
var currentPlayer;
for (var objHumanName in listofhuman) {
     if (listofhuman[objHumanName].Name === "Jonh") {
         currentPlayer = Object.create(listofhuman[objHumanName]);
         break;
     }
}

在这个循环结束时你会得到人你想要

at the end of this loop you will get human which you wonted

如果你做 Object.getOwnPropertyNames（currentPlayer）这将返回字符串数组，这是实际的键在对象currentPlayer中，您可以通过 currentPlayer [arryofProp [0]]

if you do Object.getOwnPropertyNames(currentPlayer) this will return array of string which are the actual keys in object currentPlayer, and you can access those values by currentPlayer[arryofProp[0]]

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