如何使用javascript計時來控制鼠標停止和鼠標移動事件

Question

所以我在aspx頁面上有一個控件（一張地圖）。 我想寫一些javascript來onload設置如下：

1. 當鼠標停在控制=某些代碼時

2. 當鼠標移動=某些代碼時（但僅當移動時間超過250毫秒時）

這可以觸發代碼停止然后移動...

function setupmousemovement() {
var map1 = document.getElementById('Map_Panel');
var map = document.getElementById('Map1');
map1.onmousemove = (function() {
    var onmousestop = function() {
            //code to do on stop
    }, thread;

    return function() {
        //code to do on mouse move
        clearTimeout(thread);
        thread = setTimeout(onmousestop, 25);
    };
    })();
};

但我無法弄清楚如何在移動代碼中引入延遲。 我以為我有這個......

function setupmousemovement() {
var map1 = document.getElementById('Map_Panel');
var map = document.getElementById('Map1');
map1.onmousemove = (function() {
    var onmousestop = function() {
            //code to do on stop
            clearTimeout(thread2);
    }, thread;

    return function() {
        thread2 = setTimeout("code to do on mouse move", 250);
        clearTimeout(thread);
        thread = setTimeout(onmousestop, 25);
    };
    })();
};

但它並沒有像我想象的那樣表現。 移動“thread2”永遠不會被停止清除。 我錯過了什么？

Answer 1

這是一個棘手的問題。 一點點的修補導致了這個：

function setupmousemovement() {

  var map1 = document.getElementById('Map_Panel');
  map1.onmousemove = (function() {
    var timer,
        timer250,
        onmousestop = function() {

          // code to do on stop

          clearTimeout( timer250 ); // I'm assuming we don't want this to happen if mouse stopped
          timer = null;  // this needs to be falsy next mousemove start
        };
    return function() {
      if (!timer) {

        // code to do on start

        timer250 = setTimeout(function () { // you can replace this with whatever

          // code to do when 250 millis have passed

        }, 250 );
      }
      // we are still moving, or this is our first time here...
      clearTimeout( timer );  // remove active end timer
      timer = setTimeout( onmousestop, 25 );  // delay the stopping action another 25 millis
    };

  })();

};

你的代碼不起作用的原因是當鼠標移動時你會反復觸發mousemove並且你每次都會開始新的超時。