停止在 backbone 中应用于同一事件对象的事件传播

Question

我使用 2 个不同的回调将相同的自定义事件应用于 backbone 中的事件对象。

var ourObject = {};

_.extend(ourObject, Backbone.Events);
function dancing (msg) { console.log("We are dancing. " + msg); }
function jumping (msg) { console.log("We are jumping. " + msg); }
ourObject.on("move", dancing);
ourObject.on("move", jumping);

当我使用 ourObject.trigger("move", "Yeah!"); 触发移动事件时，它会触发两个回调。

如何防止第一次回调触发第二次回调？

Answer 1

您可以将包含标志的对象作为事件对象作为第二个参数传递：

function dancing(msg, event) {
  console.log(event); // flag: false
  event.flag = true;
  console.log("We are dancing. " + msg);
}

function jumping(msg, event) {
  console.log(event); // flag: true
  if(!event.flag) return;
  console.log("We are jumping. " + msg);
}

ourObject.trigger("move", 'test', {
  flag: false
});

或者您可以简单地执行以下操作：

function dancing (msg) {
  this.dancingTriggered = true;
  console.log("We are dancing. " + msg); 
}
function jumping (msg) {
 if(this.dancingTriggered){
   this.dancingTriggered = false;
   return;
 }
 console.log("We are jumping. " + msg); 
}

或者只使用在这些回调范围内可用的简单变量

停止在 backbone 中应用于同一事件对象的事件传播

Stop event propagation applying on same event object in backbone

javascript

jquery

events

backbone.js

eventtrigger