Javascript类型错误：JSON数组对象未定义

我从服务器响应中得到了一个完美创建的JSON对象

例如：

{
    "users": [
        {
            "userId": 20410,
            "firstName": "Viral",
            "lastName": "Shah",
            "loginId": "viralp.shah@tcs.com",
            "userRole": 3
        },
        {
            "userId": 400881,
            "firstName": "Viral",
            "lastName": "Shah",
            "loginId": "viralpshah123@gmail.com",
            "userRole": 0
        },
        {
            "userId": 425622,
            "firstName": "Viral",
            "lastName": "Shah",
            "loginId": "viralpshah123@tcs-itontap.com",
            "userRole": 0
        }
    ]
}

我使用AJAX在JavaScript中使用它，如下所示：

var jsonobj2 = null;
var respObj = getSearchedWPUsers(firstname, lastname, loginid); //return json response
var len = respObj.length;
jsonobj2 = eval('(' + respObj + ')');


var tablehtml = "<table><tr><td><b>First Name</td><td><b>Last Name</td><td><b>Login Id</td><td><b>Editing Rights</td></tr><tr></tr>";

for (i = 0; i < len; i++) {
  tablehtml = tablehtml + "<tr>";
  //--------------
  tablehtml = tablehtml + "<td>";
  tablehtml = tablehtml + jsonobj2.users[i].firstName;
  tablehtml = tablehtml + "</td>";
  //--------------
  tablehtml = tablehtml + "<td>";
  tablehtml = tablehtml + jsonobj2.users[i].lastName;
  tablehtml = tablehtml + "</td>";
  //--------------
  tablehtml = tablehtml + "<td>";
  tablehtml = tablehtml + jsonobj2.users[i].loginId;
  tablehtml = tablehtml + "</td>";
  //--------------
  tablehtml = tablehtml + "<td><b>";
  var role = jsonobj2.users[i].userRole;
  if (role == 1 || role == 2 || role == 3) tablehtml = tablehtml + "<a href ='javascript:removeXML(" + jsonobj2.users[i].userId + ")'><u><font color='red'>Revoke access</font></a> ";
  else tablehtml = tablehtml + "<a href ='javascript:generateXML(" + jsonobj2.users[i].userId + ")'><u><font color='blue'>Assign access</font></a> ";

  tablehtml = tablehtml + "</td>";
  tablehtml = tablehtml + "</tr>";

}
tablehtml = tablehtml + "</table>";


document.getElementById("TableHolder").innerHTML = tablehtml;

//--------------------------

var jsonobj2=null；
var respObj=getSearchedWPUsers（firstname、lastname、loginid）//返回json响应
var len=响应长度；
jsonobj2=eval（'（'+respObj+'））；
var tablehtml=“First NameLast NameLogin IdEditing Rights”；
对于（i=0；i

它抛出如下错误

TypeError:jsonobj2.users[i]未定义

[在此错误上中断]

tablehtml=tablehtml+jsonobj2.users[i].firstName

尝试使用
JSON.parse（serverresponse）--什么也没发生

请帮助
尝试以下方法：

jsonobj2 = eval('(' + respObj + ')');
var len = jsonobj2.users.length;

据我所知，
respObj
是一个包含JSON响应的字符串，但它仍然需要解析。因此，如果执行
respObj.length
，则可以正确地获得字符串的长度，而不是用户数组的长度。另外，我建议您使用
JSON.parse
（只要浏览器可用）来解析此响应，而不是使用
eval
：

var respObj = getSearchedWPUsers(firstname, lastname, loginid);
var jsonobj2 = JSON.parse ? JSON.parse(respObj) : eval('(' + respObj + ')');
var len = respObj && respObj.users ? respObj.users.length : 0;

但是，正如Felix Kling和Aamir Adnan所说，ajax调用大多是异步的，因此您可以传递回调函数作为参数，它将按照您的意愿处理响应：

getSearchedWPUsers(firstname, lastname, loginid, function(respObj){
    var jsonobj2 = JSON.parse ? JSON.parse(respObj) : eval('(' + respObj + ')');
    var len = respObj && respObj.users ? respObj.users.length : 0;
    var tablehtml = "<table><tr><td><b>First Name</td><td><b>Last Name</td><td><b>Login Id</td><td><b>Editing Rights</td></tr><tr></tr>";

    for (i = 0; i < len; i++) {
        tablehtml = tablehtml + "<tr>";
        //--------------
        tablehtml = tablehtml + "<td>";
        tablehtml = tablehtml + jsonobj2.users[i].firstName;
        tablehtml = tablehtml + "</td>";
        //--------------
        tablehtml = tablehtml + "<td>";
        tablehtml = tablehtml + jsonobj2.users[i].lastName;
        tablehtml = tablehtml + "</td>";
        //--------------
        tablehtml = tablehtml + "<td>";
        tablehtml = tablehtml + jsonobj2.users[i].loginId;
        tablehtml = tablehtml + "</td>";
        //--------------
        tablehtml = tablehtml + "<td><b>";
        var role = jsonobj2.users[i].userRole;
        if (role == 1 || role == 2 || role == 3)
            tablehtml = tablehtml + "<a href ='javascript:removeXML(" + jsonobj2.users[i].userId + ")'><u><font color='red'>Revoke access</font></a> ";

        else tablehtml = tablehtml + "<a href ='javascript:generateXML(" + jsonobj2.users[i].userId + ")'><u><font color='blue'>Assign access</font></a> ";

        tablehtml = tablehtml + "</td>";
        tablehtml = tablehtml + "</tr>";

    }
    tablehtml = tablehtml + "</table>";

    document.getElementById("TableHolder").innerHTML = tablehtml;
});

getSearchedWPUsers（firstname、lastname、loginid、function（respObj）{
var jsonobj2=JSON.parse？JSON.parse（respObj）：eval（“（“+respObj+”）”）；
var len=respObj&&respObj.users？respObj.users.length:0；
var tablehtml=“First NameLast NameLogin IdEditing Rights”；
对于（i=0；i
你确定respObj是你认为的那样吗？从进行Ajax调用的函数返回值似乎很奇怪。请告诉我们它包含什么。您确定
respObj
包含什么吗？您可能需要执行
async:false
ajax，以便从ajax调用返回一些内容。因为ajax是异步的。你有没有检查firebug中respObj中的内容？检查firebug响应。getSearchedWPUsers（firstname、lastname、loginid）；方法是AJAX调用方法，它返回json对象。我已经在post respObj的上面粘贴了一个示例响应，但它确实返回了一个值。理解了这个错误的长度。谢谢你，马克