[英]JSONP callback function not working correctly
I have jsonp callback functionalty. 我有jsonp回调功能。 Response coming from server is undefined.I do not know where the problem is.
来自服务器的响应是不确定的。我不知道问题出在哪里。 I have made RND for jsonp.
我为jsonp制作了RND。 I am posting code
我正在发布代码
$.ajax({
url : 'http://192.168.16.111:8081/MiddleWareUsman/androidServlet',
type: "GET",
dataType: "jsonp",
crossDomain: true,
async: false,
data : {"fname": "chaaaaapiio","lname": "gya"},
jsonpCallback: function(data, status){
alert('callback');
alert(data);
},
success: function(data, status){
alert('sucess');
},
error : function(xhr, ajaxOptions, thrownError) {
alert(thrownError);
}
});
And Servlet code is Servlet代码是
String a=request.getParameter("fname");
String b=request.getParameter("lname");
String cb=request.getParameter("callback");
response.getWriter().write(cb+"("+a+" "+b+")");
First, jsonpCallback
is used when you want to override the default function name. 首先,当您想覆盖默认函数名称时,使用
jsonpCallback
。 If you assign a function to it, then the return value of that function should be the name. 如果为它分配一个函数,则该函数的返回值应为名称。 Giving it a function that returns
undefined
is just going to break things. 给它一个返回
undefined
的函数只会破坏事情。
Remove the jsonpCallback
property from your object . 从您的对象中删除
jsonpCallback
属性 。 Handle things in success
. success
处理事情。
Second, that servlet code is going to generate: 其次,该servlet代码将生成:
jQueryCallback23235(chaaaaapiio gya)
This isn't valid JavaScript. 这不是有效的JavaScript。 You need to have a real JavaScript data structure as your function arguments.
您需要具有真实的JavaScript数据结构作为函数参数。
Typically, a JSONP response would consist of an object literal: 通常,JSONP响应将由对象文字组成:
jQueryCallback23235({ "something": "chaaaaapiio", "something": "gya")
Find a Java library for generating JSON and use that to produce the contents of the parens, don't try to write JSON by hand. 找到一个用于生成JSON的Java库,并使用该库生成paren的内容,不要尝试手工编写JSON。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.