Question

我有一个看起来像这样的JSON

{
secInfo: [
 {
  loginResult: 1,
  userId: "admin",
  userName: "admin",
  userpassword: "ramshyam"
 },
 {
  address: "",
  loginResult: 1,
  userEmail: "",
  userId: "anuraag",
  userName: "Anuraag",
  userpassword: "kk0903"
 }
]
}

当我使用下面的jquery代码在表中显示数据时，它会用“undefined”填充表格

<script>
$(document).ready(function(){
    //attach a jQuery live event to the button
    $('#submitlogin').live('click', function(){
        $.getJSON('/acc/services/json/security/loginresult.json', function(data) {
            //alert(data); //uncomment this for debug
            //alert (data.item1+" "+data.item2+" "+data.item3); //further debug
            // $('#showdata').html("<p>Result="+data.loginResult+" userid="+data.userId+" username="+data.userName+"</p>");
            $.each(data.secInfo, function(){

                $('#usertable').append(
                            function(this){
                                return "<tr>"+
                                            "<td>"+this.userId+"</td>"+
                                            "<td>"+this.userName+"</td>"+
                                            "<td>"+this.userpassword+"</td>"+
                                            "<td>1</td>"+
                                        "<tr>"; 
                            }
                            );

                })
        });
    });
});
</script>

我是jquery的新手，并从http://forum.jquery.com/topic/jquery-create-table-from-json-data这篇帖子中得到了一些想法。

请帮帮我

Answer 1

问题可能是函数回调中的“this”，即保留关键字。另外，尝试将JSON条目作为数组而不是类属性进行访问。

Answer 2

这个＆＃39;的范围我猜想已经改变了。试试这个：

$(document).ready(function(){
    //attach a jQuery live event to the button
    $('#submitlogin').live('click', function(){
        $.getJSON('/acc/services/json/security/loginresult.json', function(data) {
            //alert(data); //uncomment this for debug
            //alert (data.item1+" "+data.item2+" "+data.item3); //further debug
            // $('#showdata').html("<p>Result="+data.loginResult+" userid="+data.userId+" username="+data.userName+"</p>");
            $.each(data.secInfo, function(dataItem){
                $('#usertable').append(
                    function() {
                        return "<tr>"+
                                    "<td>"+dataItem.userId+"</td>"+
                                    "<td>"+dataItem.userName+"</td>"+
                                    "<td>"+dataItem.userpassword+"</td>"+
                                    "<td>1</td>"+
                                "<tr>"; 
                    }
                );
            })
        });
    });
});

Answer 3

这肯定有效

<script>
$(document).ready(function(){
    //attach a jQuery live event to the button
    $('#submitlogin').live('click', function(){
        $.getJSON('/acc/services/json/security/loginresult.json', function(data) {

     var genericlist =  eval('(' + data+ ')');

var table = document.getElementById("usertable");
var tabledata = "";

for(i=0;i<genericlist.secInfo.length;i++){

tabledata += "<tr>";
tabledata += "<td>" + genericlist.secInfo[i].useID += "</td>";
tabledata += "<td>" + genericlist.secInfo[i].useName += "</td>";
tabledata += "<td>" + genericlist.secInfo[i].usePassword += "</td>";
tabledata += "</tr>";

}

table.innerHTML = tabledata;

使用Jquery获取JSON数据以构建显示未定义的表

3 个答案: