Question

我需要使用外部函数来成功回调，我不知道如何将json对象传递给我的函数。

$.ajax({
url:"get_box.php",
type:"POST",
data:data,
dataType:"json",
success: myFunction(data);  
    });

我的功能看起来像这样：

function myFunction(result2){
...
}

错误是：undefined result2 ...

Answer 1

试试这种方式，

 success: function(data){
        myFunction(data);
    });

或......

success: myFunction 
    });

Answer 2

如何实现成功和失败回调方法（jquery documentation）。您也可以链接这些而不是在初始的ajax设置对象中提供它们，如下所示：

Here is a fiddle

jQuery.ajax({
    // basic settings
}).done(function(response) {
    // do something when the request is resolved
    myFunction(response);
}).fail(function(jqXHR, textStatus) {
    // when it fails you might want to set a default value or whatever?
}).always(function() {
    // maybe there is something you always want to do?
});

Answer 3

<script>
function fun(){
    $.ajax({
        url : "http://cdacmumbai.in/Server.jsp?out=json&callback=?",
        dataType: "json",
        contentType: "application/json;charset=utf-8",
        type: "GET",
        success: function ( output ) {
            var data = eval( output );
            document.getElementById("datetime").innerHTML = "Server Date&Time: "+data.servertime;
            document.getElementById("hostname").innerHTML = "Server Hostname: "+data.hostname;
            document.getElementById("serverip").innerHTML = "Server IP Address: "+data.serverip;
            }
        });
      }
</script>

Jquery ajax外部回调函数

3 个答案: