Question

我有此代码：

if(car.position == a1.position)
  {
    car.speed = 0;
    a1 = 1;
   if(cooldow == 0){
        car.taget.position = targetHere.position
        car.speed = 5;
        a1 = 0;
        Destroy(obj, 5)
       }
  }

第一个条件[if（car.position == a1.position）]我将重复10次以上，这意味着我将检查（car.position）是否与 a2.position < / strong>，然后是 3.position ，然后是 a4.position ，然后是 a5.position ，然后是 a6.position 然后是< strong> b1.position ，然后 b2.position ，然后 b3.position ，然后 b4.position ，然后 b5.position < / strong>然后 b6.position ....十二次...，我找不到将其放入（ For ）循环。

任何想法如何做到！！！ TY

Answer 1

我假设a1是一个变换，因为您正在访问它的位置变量。我还假设您是在检查器中分别分配这些游戏对象。

我会执行以下操作：

// Make sure to put all your a objects in this list in the inspector
public List<Transform> myABCObjects; 

void CheckPositions()
{
    // This foreach loop, will loop through your list of transforms, so your a b c 
    // objects would be added to this list via the inspector, or through whatver
    // code you are using to set them.
    foreach(Transform a in myABCObjects)
    {
        // Not to sure if you can do an == on a vector3, but I wouldn't want to
        // considering the data members are float, and can cause floating point error
        // Because of this I am getting the distance to the location, and using
        // Mathf.Approximately to check if the distance is close to 0f.
        if(Mathf.Approximately(car.position.distance(a.position), 0f))
        {
            // Your logic here if location is the same...
            // so if you are trying to remove say a1
            // add this object to a list, to be removed after the loop
            // then continue on.
        }
    }

    // if you are removing an object, check the list size, then loop through it to remove the objects from your original list...
}

使用for循环而不是foreach的示例：

// Make sure to put all your a objects in this list in the inspector
public List<Transform> myABCObjects; 

void CheckPositions()
{
    // This for loop, will loop through indices 0 - myABCObjects.count - 1, so your a b c 
    // objects would be added to this list via the inspector, or through whatver
    // code you are using to set them.
    for(int i = 0; i < myABCObjects.count; ++i)
    {
        // Not to sure if you can do an == on a vector3, but I wouldn't want to
        // considering the data members are float, and can cause floating point error
        // Because of this I am getting the distance to the location, and using
        // Mathf.Approximately to check if the distance is close to 0f.
        if(Mathf.Approximately(car.position.distance(myABCObjects[i].position), 0f))
        {
             // Your logic here if location is the same...

        }
    }

}

Answer 2

如果[a1..a6]和[b1..b6]具有相同的类型，则尝试这样做：

var list = new List<TypeOfa1> {a1, a2, a3, a4, a5, a6, b1, b2, b3, b4, b5, b6};             

foreach (var item in list)
{
    if (car.position == item.position)
    {
        car.speed = 0;
        item = 1;
        if (cooldow == 0)
        {
            car.taget.position = targetHere.position;
            car.speed = 5;
            item = 0;
            Destroy(obj, 5);
        }
    }
 }

Answer 3

您可以创建变量的集合，然后通过它们运行foreach。

List<CarType> collection = new List<CarType> {a1, b1, a2, b2 ...};
foreach (CarType c in collection)
{
    // do stuff
}

unity ..用于检查12位置的逻辑回路

3 个答案: