Question

Ajax怎么获取Json数据

$("a").click(function(){
$.ajax({
type:"get",
url:"test1.json",
dataType:"json",
success:function(response,status,xhr){

$.each(response,function(index,value){

$firstName = response[index].firstName;
$lastName = response[index].lastName;

$("ul").append("<li>"+index+"----firstName:"+$firstName+"///lastName:"+$lastName+"</li>");
});
}
});
});

json:

{
"programmers": [
{ "firstName": "Brett", "lastName":"McLaughlin", "email": "brett@newInstance.com" },
{ "firstName": "Jason", "lastName":"Hunter", "email": "jason@servlets.com" },
{ "firstName": "Elliotte", "lastName":"Harold", "email": "elharo@macfaq.com" }
],
"authors": [
{ "firstName": "Isaac", "lastName": "Asimov", "genre": "science fiction" },
{ "firstName": "Tad", "lastName": "Williams", "genre": "fantasy" },
{ "firstName": "Frank", "lastName": "Peretti", "genre": "christian fiction" }
],
"musicians": [
{ "firstName": "Eric", "lastName": "Clapton", "instrument": "guitar" },
{ "firstName": "Sergei", "lastName": "Rachmaninoff", "instrument": "piano" }
]
}

Answer 1

$("a").click(function() {
    $.ajax({
        type: "get",
        url: "test1.json",
        dataType: "json",
        success: function(response, status, xhr) {
          for (var p in response) {
            var li = $("<li>" + p + "</li>"),
              arr = response[p];
            var ul = $("<ul>");
            li.append(ul);
            for (var i = 0; i < arr.length; i++) {
              var lii = $("<li>");
              var obj = arr[i];
              for (var k in obj) {
                lii.append("<div>" + k + ": " + obj[k] + "</div>");
              }
              ul.append(lii);
            }
            $("ul").append(li);
          }
        });
    });
});

追问

为什么我在网上查看会用到eval()这个的，这个是什么时候用到

追答

你的response是字符串类型还是对象，如果是字符串的json，那就eval转为object就可以了
如果是object，就不需要了

var str = "{}";
var obj = eval("("+str+")");

好了，更多的细节不多说了，多查查网就知道了，采纳即可