为什么脚本只能重新加载页面?

时间:2016-01-20 15:13:07

标签: javascript jquery

我需要使用jquery在表中加载elem。这是代码:

在file.html中,表的代码是:

<table id="tabella_ban_email" class="table table-condensed table-responsive table-hover"></table>

并在script.js中代码为:

$(document).ready(function() {
  var email_table=$('#tabella_ban_email');
  if(email_table!=null && email_table!=undefined){
    $.ajax({
      url:'....',
      success: function(data) {
        data.split(',').forEach(function(email){
          email_table.append('<tr><th>'+email+'</th><th><input type="button" name="'+email+'" value="X">');
        });
      }
    });
  }
});

该程序仅在我重新加载程序无法正常工作的页面时刷新页面。

1 个答案:

答案 0 :(得分:0)

$(document).ready(function() {
 // executes when HTML-Document is loaded and DOM is ready
 alert("document is ready");
});


$(window).load(function() {
 // Executes when page is fully loaded
 alert("window is loaded");
});

Use $(window).load ... it might work in your scenario