Question

我正在尝试遍历一个对象并获取名称的所有键值，我正在按以下方式进行操作：

var fakeData = {
     "manufacturer": "tesla",
     "cars": [
          {"title": "CALI", "name": "CALI", "type": "string" },
          {"title": "TEXAS", "name": "TEXAS", "type": "string" },
          {"title": "NY", "name": "NY", "type": "string" }
     ],
     "usedCars": [
          {"title": "FL", "name": "FL", "type": "string" }
     ],
}

returnName(fakeData) {
    for (key in fakeData.cars && fakeData.usedCars) {
     //return name property  of cars and usedCars arrays.
     //fakeData is the representation of the req.body data that might 
     //sometimes not necessarily have either cars or usedCars

    }

}

如果我这样做，它会返回 undefined。有什么方法可以将这些条件放入一个 for 循环中并获得所需的结果？

Answer 1

这将完成工作。

(fakeData.cars?fakeData.cars:[]).concat(fakeData.usedCars?fakeData.usedCars:[]).map(car => car.name)

输出：

["CALI", "TEXAS", "NY", "FL"]

说明：

首先，我们使用条件来检查 faekData.cars 是否确实存在。如果是，获取数组。如果不是，则在其位置返回一个空数组。

  (fakeData.cars)?fakeData.cars:[]

这转化为：（条件==真）？（如果为真则执行此操作）：（如果为假则执行此操作）

如果数组不存在，则不满足条件。因此，将执行“（如果为假，则执行此操作）”部分。

然后，我们对第二个数组做同样的事情。通过使用“concat”，我们将两个数组连接成一个。然后，我们使用“map”将整个对象转换为“name”属性的值。

Answer 2

希望这就是您要找的东西。

var fakeData = {
     "manufacturer": "tesla",
     "cars": [
          {"title": "CALI", "name": "CALI", "type": "string" },
          {"title": "TEXAS", "name": "TEXAS", "type": "string" },
          {"title": "NY", "name": "NY", "type": "string" }
     ],
     "usedCars": [
          {"title": "FL", "name": "FL", "type": "string" }
     ],
};

[].concat(
    fakeData.cars ? fakeData.cars : [],
    fakeData.usedCars ? fakeData.usedCars : [],
    fakeData.undefinedCars ? fakeData.undefinedCars : [],
).forEach(car => {
    console.log(car.name);
});

对于更多数组，只需将它们添加到以逗号分隔的 .concat() 函数中：

array1.concat(array2, array3, array4)

Answer 3

for (key in fakeData.cars && fakeData.usedCars) 的意思不是“把汽车和二手车的钥匙给我”，它的意思是“给我汽车评价的钥匙&&usedCars”（这是汽车，因为它是真实的）。

相反，只需使用 .map 并获取名称：

var fakeData = {
     "manufacturer": "tesla",
     "cars": [
          {"title": "CALI", "name": "CALI", "type": "string" },
          {"title": "TEXAS", "name": "TEXAS", "type": "string" },
          {"title": "NY", "name": "NY", "type": "string" }
     ],
     "usedCars": [
          {"title": "FL", "name": "FL", "type": "string" }
     ],
}

/*
This doesn't mean "get me the keys in cars and usedCars"
it means "get me the keys of the evaluation of cars && usedCars"
(which is cars since it's truthy)
returnName(fakeData) {
    for (key in fakeData.cars && fakeData.usedCars) {
     //return name property  of cars and usedCars arrays.
     //fakeData is the representation of the req.body data that might 
     //sometimes not necessarily have either cars or usedCars

    }

}
*/

// You can just map the arrays and spread them into a new array:
// If you have to deal with certain properties not being there,
// you can use optional chaining, the ? and ?? [] pieces
const names = [
  ...fakeData?.cars?.map(({name}) => name) ?? [],
  ...fakeData?.usedCars?.map(({name}) => name) ?? []
];

console.log(names); // ["CALI", "TEXAS", "NY", "FL"]

Answer 4

这应该打印所有汽车名称

var fakeData = {
     "manufacturer": "tesla",
     "cars": [
          {"title": "CALI", "name": "CALI", "type": "string" },
          {"title": "TEXAS", "name": "TEXAS", "type": "string" },
          {"title": "NY", "name": "NY", "type": "string" }
     ],
     "usedCars": [
          {"title": "FL", "name": "FL", "type": "string" }
     ],
}

var results = [];

function returnNames(){
    for(index in fakeData.cars){
        results.push(fakeData.cars[index].title);
    }
    for(index in fakeData.usedCars){
        results.push(fakeData.usedCars[index].title);
    }
    
    console.log("car names: " + results)
}

Answer 5

您可以将您想要的对象连接到一个数组中，并映射出您想要访问的任何值。这就是你要找的吗？

const fakeData = {
    manufacturer: 'tesla',
    cars: [
      { title: 'CALI', name: 'CALI', type: 'string' },
      { title: 'TEXAS', name: 'TEXAS', type: 'string' },
      { title: 'NY', name: 'NY', type: 'string' }
    ],
    usedCars: [{ title: 'FL', name: 'FL', type: 'string' }]
  };
  
  const combinedCarData = [...fakeData.cars, ...fakeData.usedCars];
  
  // Map whatever values you would like to access
  
  combinedCarData.map(car => {
    console.log(car.title);
    console.log(car.name);
    console.log(car.type);
  });

循环遍历一个对象并获取所有键的值

5 个答案: