精氨酸!!我的工作完美无瑕,现在我又回到了敲击键盘的位置。
我希望在数组中访问已定义的列,但是我得到了未定义但是如果我使用警告显示结果,详见下面的剪辑代码,我会看到以下内容:
[{"firstname":" Mr","0":" Mr","lastname":" Two","1":" Two","user_public_info_id":"3","2":"3","st_usage_id":null,"3":null},{"firstname":" Mr","0":" Mr","lastname":" Three","1":" Three","user_public_info_id":"5","2":"5","st_usage_id":null,"3":null}]
***
g
***
e
***
undefined
这是Ajax代码:
$.ajax({
type: "POST",
url: "models/ajaxHandler.php",
data: "handler=getStActivitySharingList&stu_id="+stu_id,
datatype: "json",
success: function(result){
var count = 0;
if (result !== null)
{
//display results
alert(result + " <br />*** <br />" + result[0] +" <br />*** <br />" + result[1] + " <br />*** <br />" + result[0]["firstname"]);
//clears choice list
clearOptions();
//result = $.parseJSON(result); //if this is used cannot display result again
alert (result);
$.each(result, function (i, elem) {
alert("elem"+elem.st_usage_id ); //displays as undefined and won't break
if (elem.st_usage_id === null)
{
count++;
alert(elem.firstname + " "+ elem.lastname + " " + elem.user_public_info_id);
appendOption(elem);
}
});
}
alert(count);
if (count === 0){
noResultsAvailableOption();
}else{
resultsAvailableOption();
}
ShowDialog(false);
e.preventDefault();
},
error: function(){
alert("ajax failure: could not retrieve a list of contacts");
}
});
答案 0 :(得分:0)
更好地回答这个问题是要实现更好的调试程序。
以下是我用于调试此问题的代码。 xmlHttpRequest的分解清楚地向我显示问题与数据有关,并且在尝试将数据编码为json时遇到了非法字符异常。
解决任何问题的一个好方法是首先实现正确的调试程序,其他一切都将自行解决。
error: function(xmlHttpRequest, status, error){
alert("ajax failure: could not populate list of countires | " + status + " | error:" + error);
var xmlHttpRequestStr= "";
for(x in xmlHttpRequest)
xmlHttpRequestStr = xmlHttpRequestStr + xmlHttpRequest[x];
alert(xmlHttpRequest);
}
答案 1 :(得分:0)
我不知道你是如何从PHP返回的,但是在jquery中尝试:
sucess: function (result)
{
console.log(JSON.parse(result)); // parse string to json
}
请参阅json.org