javascript - 将 var 从一个扩展函数传递到另一个

Question

我在传递$.PhotoUpdater.doUpdate(url)执行该doUpdate功能的 url 时遇到问题。

Firefox 返回此错误：

useless setTimeout call (missing quotes around argument?)
[Break on this error] timer = setTimeout($.PhotoUpdater.doUpdate(url), 5000) 

我的代码：

$.extend({
  PhotoUpdater: {

    startUpdate: function(organization, gallery){
      url = "/organizations/" + organization + "/media/galleries/" + gallery + "/edit_photo_captions"
      timer = setTimeout($.PhotoUpdater.doUpdate(url), 5000)
    },
    stopUpdate: function(){
      clearTimeout(timer);
      timer = 0;
    },
    doUpdate: function(url){
      $.ajax({type: "GET", url: url, dataType: "script"});
    }
  }
});

我怎么称呼它：

$.PhotoUpdater.startUpdate("#{@organization.id}", "#{@gallery.id}");

Answer 1

您需要将函数传递给window.setTimeout，而不是调用函数的结果：

startUpdate: function(organization, gallery){
    url = "/organizations/" + organization + "/media/galleries/" + gallery + "/edit_photo_captions";
    timer = window.setTimeout(function() {
        $.PhotoUpdater.doUpdate(url)
    }, 5000);
},