我想在数组中的对象中返回一个属性，但是它导致代码中的其他问题

如何解决我想在数组中的对象中返回一个属性，但是它导致代码中的其他问题

因此，我正在使用此功能，该功能可以使汽车变道。只要我不退回车牌return vehicleObject_list[i]，它就会改变车道。但是，当我按照说明要求退回车牌时，汽车停止更改车道return vehicleObject_list[i].Number_Plate。详细信息如下：

function moveLanes(target_car)
{
    /*
    This function should do the following: 
     - move target_car from one lane to the other.
     - do the move in a single step without any extra animation.
     - use Lane_Position_a and Lane_Position_b to effect the change.
     - finally you should return target_car at the end of the function.
     hint: You will need to modify the x property of target_car.
    */
    if(checkCarInfront(target_car).x == Lane_Position_a ){
        target_car.x = Lane_Position_b;
    }   
    else {
        target_car.x = Lane_Position_a;
    }
    
}


function checkCarInfront( carObj )
{
    /*
    This function should do the following: 
     - determine if carObj is in the same lane and less than 200px behind any of the cars in vehicleObject_list.
     - do this by traversing vehicleObject_list and comparing each car's distance_Driven property to that of carObj.
     - if you find a car that matches these requirements then return the Number_Plate property for the car. Otherwise return false.
    */
    
    for (var i = 0; i < vehicleObject_list.length; i++)
    {
        if (carObj.x == vehicleObject_list[i].x && ((vehicleObject_list[i].distance_Driven - carObj.distance_Driven) < 200) && ((vehicleObject_list[i].distance_Driven - carObj.distance_Driven) > 0))
        {
            return vehicleObject_list[i].Number_Plate;
        }
    }
    return false;
    
    
}

var vehicleObject_list = [
{ x: 500,y: 0,distance_Driven: -200,Car_Type: 'greenCar',Number_Plate: 'MBH0WW',Gas_Amt: 2,exhaust: [  ]},{ x: 500,distance_Driven: 200,Car_Type: 'whiteCar',Number_Plate: 'RLDGCM',{ x: 300,distance_Driven: 600,Number_Plate: '9WGXXI',exhaust: [  ]} ]

Detective_CarObject = 
    {
        x: roadLeftEdge + roadWidth/4,y: 550,distance_Driven: 0,Gas_Amt: 3,Engineshudder_Value: 0,Car_Type: 'detective',Number_Plate: '5L3UTH',exhaust: []
    }

解决方法

您可以执行以下操作：

if(checkCarInfront(target_car) && target_car.x == Lane_Position_a ){
    target_car.x = Lane_Position_b;
} else {
    target_car.x = Lane_Position_a;
}

因为checkCarInfront(target_car)将在存在匹配项的情况下返回一个车牌字符串（该字符串将作为非空字符串求值为true）。您可以将条件链接到具有target_car属性的.x对象。

请确保手动测试数据。（例如，如果调用vehicleObject_list，则使用当前的Detective_CarObject和checkCarInfront(Detective_CarObject)值，这将是错误的（由于Distance_Driven）。不过，您可以临时设置{{1} } x到Detective_CarObject，并在列表中添加额外的测试车以检查您的逻辑：500）

作为旁注，我建议遵循JS命名约定。以下是几个示例：W3Schools，FreeCodeCamp，Google JavaScript Style Guide。过去我会说选择一个（例如，下划线/ snake_case（例如，{ x: 500,y: 0,Distance_Driven: 199,Car_Type: 'rainbowCar',Number_Plate: 'R41NB0W',Gas_Amt: 2,exhaust: [ ]}）或camelCase（例如，target_cat），但不能同时选择两者并保持一致，但如今看来大多数人坚持骆驼的情况。这样做将使以后的团队工作和共享/贡献开源库和项目变得更加容易。