Question

所以问题是...它不起作用。而且更正确地...我该如何更好地执行这些嵌套的if？

document.body.onkeypress = function(e) {

                e = e || window.event;

                if (e.keyCode == '38') { //arrow up

                }
                if (e.keyCode == '40') { //arrow down
                    if (top == 1) {
                        window.setTimeout(show2, 100);
                        alert('2 layer works');
                    }
                    if (top == 2) {
                        window.setTimeout(show3, 100);
                    }
                    if (top == 3) {
                        window.setTimeout(show4, 100);
                    }
                }

            }

我已经尝试了一切。请帮助我...

Answer 1

从此示例开始。它显示了如何正确地将keydown事件连接到文档主体。从这里，您应该可以对其进行修改，以添加示例中具有的其他逻辑。请务必确保正确设置了top之类的外部变量。

function handleKeyDown(e) {
  console.log('got keyDown event.  e.keyCode =', e.keyCode)
}

document.body.addEventListener("keydown", handleKeyDown, false);

<p>Press a key</p>

Answer 2

您在哪里更改top的值...？同样，第一个if语句（如果它像您在此处显示的那样空白）将引发错误，或者至少不会执行任何操作。你为什么说e = e || window.event ..？同样，这可能只是我，但不要调用此类函数。更好（至少在我看来也是）：

document.body.addEventListener("keypress", e => {
// Get rid of that e = e || window.event, I have no clue why you'd do that.
if (e.keyCode == 38) {
 // Actually give it some parameters to do here, leaving it empty will either 
 // throw an error or in "best" case won't do anything.
}
if (e.keyCode == 40) {
               // Again, you said you did var top = 1. It will work here.
                if (top == 1) {
                    window.setTimeout(show2, 100);
                    alert('2 layer works');
                }
               // But you dont INCREMENT the top variable anywhere, so this wont work.
                if (top == 2) {
                    window.setTimeout(show3, 100);
                }
                // But you dont INCREMENT the top variable anywhere, so this wont work.
                if (top == 3) {
                    window.setTimeout(show4, 100);
                }
  }

})

这个问题的格式确实非常错误，您没有提供太多信息。向我们显示您收到的错误，请始终使用console.log，这是您最好的朋友。

如果条件正确，则在按键上执行功能

2 个答案: