javascript - 返回未定义的 jQuery JSON

Question

我的 php json 中有这种形式的数据:

string(170) "[{"id":"3","Name":"Kontrahent#322","NIP":"753","Adress":"Wiosenna29","PostCode":"20-201","City":"Olkusz","Phone":"12312312","Email":"jakub@edu.pl","Value":"0"}]"

和我的功能:

function showUser(str) {
    if (str == "") {
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else { 
         var val = $('#test').val()       
         var id = $('#clientsname option').filter(function() {
            return this.value == val;
        }).data('id');
        if (window.XMLHttpRequest) {
            xmlhttp = new XMLHttpRequest();
        } else {
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
                var data = xmlhttp.responseText;
                alert(data[0].Name);

            }
        }
        xmlhttp.open("GET","getclients/"+id);
        xmlhttp.send();
    }
}  

警报(数据[0].名称)；或警报(数据。名称)；返回未定义。 控制台日志(数据)； 返回:

string(141) "[{"id":"1","Name":"Kontrahent #1","NIP":"735256985","Adress":"","PostCode":"","City":"","Phone":"777555888","Email":"e@mail.pl","Value":"0"}]"

我不知道我的脚本有什么问题。任何人都可以帮助我吗？

Answer 1

您需要使用 JSON.parse 将响应解析为 JSON方法，因为 xmlhttp.responseText 只是一个字符串:

xmlhttp.onreadystatechange = function() {
    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
        document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
        var data = JSON.parse(xmlhttp.responseText);
        alert(data[0].Name);
    }
}

演示: http://plnkr.co/edit/LygRQEu89LnQXW6TWDMa?p=preview