循环一个 object 并获取所有键的值

Question

我正在尝试遍历 object 并获取名称的所有键值，我正在通过以下方式进行操作：

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，它会返回未定义。 有没有办法在一个 for 循环中适应这些条件并获得所需的结果？

Answer 1

这将完成这项工作。

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

Output：

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

解释：

首先，我们使用条件来检查 faekData.cars 是否真的存在。 如果是，请获取数组。 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，请执行此操作）：（如果为假，请执行此操作）

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

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 arrays 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

希望这就是你要找的。

 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); });

更多 arrays 只需将它们添加到.concat() function 中，用逗号分隔：

array1.concat(array2, array3, array4)

Answer 3

for (key in fakeData.cars && fakeData.usedCars)并不意味着“给我汽车和二手车的钥匙”，它的意思是“给我汽车和二手车评估的钥匙”（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

这应该打印所有的汽车名称

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，然后将 map 输出您想要访问的任何值。 那是你要找的吗？

 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); });