在双跳游戏中，郊狼的计时不起作用。(C#，统一)

Question

我试着用双跳和土狼的时间来做一个跳台，但是土狼的时间似乎不适用于双跳。每当狼跳被认为是触发，角色使用他们的双跳代替。

if (Input.GetKeyDown(KeyCode.UpArrow) && extraJumps > 0 
|| Input.GetKeyDown(KeyCode.Space) && extraJumps > 0
|| Input.GetKeyDown(KeyCode.W) && extraJumps > 0)
{
    rigidbody2D.velocity = Vector2.up * jumpForce;
    extraJumps--;
    audioSource.Play();
}
else if (Input.GetKeyDown(KeyCode.UpArrow) && extraJumps == 0 && coyoteCounter > 0f 
|| Input.GetKeyDown(KeyCode.Space) && extraJumps == 0 && coyoteCounter > 0f 
|| Input.GetKeyDown(KeyCode.W) && extraJumps == 0 && coyoteCounter > 0f)
{
    rigidbody2D.velocity = Vector2.up * jumpForce;
    audioSource.Play();
}

Answer 1

在检查郊狼的时间之前，你正在检查双跳。当您检查第一个代码块时，它总是在检查郊狼时间之前使用双跳。这应该可以解决您的问题：

if (Input.GetKeyDown(KeyCode.UpArrow) || Input.GetKeyDown(KeyCode.Space) || Input.GetKeyDown(KeyCode.W))
    {
        // Check coyote timer before we activate a double jump
        if (coyoteCounter > 0f)
        {
            rigidbody2D.velocity = Vector2.up * jumpForce;
            audioSource.Play();
        }
        // Now we have run out of coyoteCounter, time to use double jumps
        else if (extraJumps > 0)
        {
            rigidbody2D.velocity = Vector2.up * jumpForce;
            extraJumps--;
            audioSource.Play();
        }
    }

您还可能希望确保播放机只能触发一次coyoteCounter ;)