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我首先开发了一个简单的 jQuery 动画,然后我创建了一个 jQuery 函数来实现可重用性,它会做同样的事情。这是一个演示:http: //jsfiddle.net/SpMns/

我的代码正在运行,但并不可靠:当我单击按钮运行代码时,什么也没有发生;单击它两到三下将启动动画。为什么第一次就不行?

请看一下我的例行程序并告诉我需要纠正的区域:

jQuery.fn.busyToggle = function(ImgLoadSrc, marginBottom, opacity, speed, 
                                easing, callback) {
  var oDiv = $("<div id='BusyBox'><img src='" + ImgLoadSrc 
           + "'  alt='Loading...'/><div><em>Loading Wait...</em></div></div>");
  if ($("#BusyBox").exists() == false) {
    //alert('div not exist');
    oDiv.css("background", "-moz-linear-gradient(center top , #F1F2F2 0%, #F1F2F2 100%) repeat scroll 0 0 transparent");
    oDiv.css("border-top-left-radius", "5px");
    oDiv.css("border-top-right-radius", "5px");
    oDiv.css("bottom", "0px");
    oDiv.css("font-size", "0.8em");
    oDiv.css("font-style", "normal");
    oDiv.css("font-weight", "normal");
    oDiv.css("left", "50%");
    oDiv.css("margin-left", "-45px");
    oDiv.css("padding-top", "20px");
    oDiv.css("position", "fixed");
    oDiv.css("text-align", "center");
    oDiv.css("width", "90px");
    oDiv.css("height", "50px");
    oDiv.css("margin-bottom", "-70px");
    oDiv.css("background-repeat", "no-repeat");
    oDiv.css("background-position", "center center");
    oDiv.data('IsUp', 1)
    oDiv.appendTo('body');
  }

  // i work with jquery data function for achieving toggle behaviour
  if (oDiv.data('IsUp') == 1) {
    oDiv.data('IsUp', 0);
    return this.stop(true).animate({
      marginBottom: marginBottom,
      opacity: opacity
    }, {
      queue: false,
      duration: speed,
      complete: callback
    });
  }
  else {
    oDiv.data('IsUp', 1);
    return this.stop(true).animate({
      marginBottom: marginBottom,
      opacity: opacity
    }, {
      queue: false,
      duration: speed,
      complete: callback
    });
  }
};
$(document).ready(function() {
  $("#Process").click(function() {
    if (flag == 1) {
      $('#BusyBox').busyToggle('images/loader.gif', 0, 1, 500, 0, function() {
        alert('div visible')
      });
      flag = 0;
    }
    else {
      $('#BusyBox').busyToggle('images/loader.gif', -70, 0, 500, 0, function(){
        alert('div hide')
      });
      flag = 1;
    }
    return false;
  });
});

什么可能导致它在第一次运行时失败?

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1 回答 1

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问题是#busyToggle 中的这种用法

在第一次调用时,$('#BusyBox') 变成了一个空数组,这意味着这个(在#busyToggle 中)也是一个空数组。

A better solution to your probelm would be to call busyToggle this way:

$('#Process').busyToggle('images/loader.gif', 0, 1, 500, 0, function() {
    alert('div visible')
});

and use $('#BusyBox') instead of this in your function.

You can find all the code over there: http://jsfiddle.net/ubmCt/8/

P.S: I've also added following line to the ready function, otherwise it won't run in most browsers

var flag = 1;

于 2012-06-19T15:04:06.207 回答