Question

我编写的脚本如下：

<script type="text/javascript">
    $(document).ready(function(){
        $("#Click").click(function(){
            var data1 = new Object();
            data1.name = $('#databases').val();
            $.ajax({
          url : "form_submit",
          type : 'POST',
          data : data1,
          dataType : "text",              
          success : function(data){
            data = JSON.parse(data);
            var output="<table id=tableStyle>";
            output+="<tr>" + "<th>" + "REPORTSUITE_ID" + "</th>" + "<th>" + "REPORTSUITE_NAME" + "</th>" + "<th>" + "STAGING_DATABASE" + "</th>" + "<th>" + "DWH_DATABASE" + "</th>" + "</tr>";
            for (var i in data)
            {
                output+="<tr>" + "<td>" + data[i].REPORTSUITE_ID + "</td>" + "<td>" + "<button type = \"buttion\" id =\"tableList\">" + data[i].REPORTSUITE_NAME + "</button>" + "</td>" + "<td>" + data[i].STAGING_DATABASE + "</td>" + "<td>" + data[i].DWH_DATABASE + "</td>" + "</tr>";
            }
            output += "</table>";
            document.getElementById("placeholder").innerHTML=output;
          }
          });
        });
        });
    $("#tableList").click(function(){
            alert("this is data");
            });
</script>

在点击id = tableList的按钮时，我想要显示带有“this is data”消息的警报，但它不会显示任何内容。我也试着粘贴与不同标签之间的id tableList按钮相关的代码，但它仍然无法正常工作。请帮忙

Answer 1

您使用了上述问题的事件授权，如下所示：

<script>
    $(document).on("ready",function(){
        $("#placeholder").on("click","button",function(event){
            alert("id is " + this.id + "for database " + $('#databases').val())
        })
    });
</script>

正确传递了该值。谢谢你的指针。

单击按钮调用函数

1 个答案: