Ajax 和 Java 网络服务奇怪错误

Ajax and Java web service strange error

我遇到了一个非常奇怪的错误。我成功 运行 我的代码并使用了网络服务。服务 return 200 代码。但是 jquery ajax 执行错误函数而不是成功。我 post 信息(控制台上没有):

Java脚本:

$(document).ready(
    function () {
        $.ajax({
            url: 'Test/Service/hello',
            dataType: 'json',
            contentType: 'application/json',
            data: {name: 'testing'},
            error: function () {
                $('#text').html('<p>An error has occurred</p>');
            },
            success: function (data) {

                $("#text").html(data.d);
            },
            type: 'POST'
        });
    });

Java:

package com.morethansimplycode.webservice;

import javax.jws.WebParam;
import javax.ws.rs.Consumes;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

@Path("/Service")
public class Service {

    public Service() {
    }

    @POST
    @Path("/hello")
    @Consumes({MediaType.APPLICATION_JSON})
    @Produces(MediaType.APPLICATION_JSON)
    public String hello(@WebParam(name = "name") String txt) {

        return "Hello " + txt + " !";
    }
}

Chrome 信息:

Remote Address:[::1]:8080
Request URL:http://localhost:8080/WebService/Test/Service/hello
Request Method:POST
Status Code:200 OK
Response Headers
view source
Content-Length:17
Content-Type:application/json
Date:Tue, 01 Dec 2015 21:24:26 GMT
Server:GlassFish Server Open Source Edition  4.1
X-Powered-By:Servlet/3.1 JSP/2.3 (GlassFish Server Open Source Edition  4.1  Java/Oracle Corporation/1.8)
Request Headers
view source
Accept:application/json, text/javascript, */*; q=0.01
Accept-Encoding:gzip, deflate
Accept-Language:es-ES,es;q=0.8,ca;q=0.6,en;q=0.4,pt;q=0.2,ru;q=0.2
Cache-Control:no-cache
Connection:keep-alive
Content-Length:9
Content-Type:application/json
Host:localhost:8080
Origin:http://localhost:8080
Pragma:no-cache
Referer:http://localhost:8080/WebService/
User-Agent:Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/46.0.2490.86 Safari/537.36
X-Requested-With:XMLHttpRequest
Request Payload
name=paco

预览:

Hello name=testing !

发生了什么事?

编辑 1:

适用于此调用:

$(document).ready(
        function () {
            $.ajax({
                url: 'Test/Service/hello',
                dataType: 'text',
                contentType: 'application/json',
                data: {name: 'paco'},
                error: function () {
                    $('#etiqueta').html('An error has occurred');
                },
                success: function (data) {

                    console.debug(data);
                    $("#etiqueta").html('<p>' + data + '</p>');
                },
                type: 'POST'
            });
        }
);

但我不知道如何将 Web 服务中的对象作为 Json 发送到 JS,以便将其与 data.d 等一起使用...

编辑 2:

正确工作如下:

JS:

$(document).ready(
        function () {
            $.ajax({
                url: 'Test/Service/hello',
                dataType: 'json',
                contentType: 'application/json',
                data: {name: 'paco'},
                error: function () {
                    $('#etiqueta').html('An error has occurred');
                },
                success: function (data) {

                    console.debug(data);
                    $("#etiqueta").html('<p>' + data.d + '</p>');
                },
                type: 'POST'
            });
        }
);

Java:

@Path("/Service")
public class Service {

    public Service() {
    }

    /**
     * This is a sample web service operation
     *
     * @param txt
     * @return
     */
    @POST
    @Path("/hello")
    @Consumes({MediaType.APPLICATION_JSON})
    @Produces(MediaType.APPLICATION_JSON)
    public String hello(@WebParam(name = "name") String txt) {

        return "{\"d\": \"Hello " + txt + " !\"}";
    }
}

而不是

public String hello(@WebParam(name = "name") String txt) {

        return "Hello " + txt + " !";
    }

尝试使用:

public String hello(@WebParam(name = "name") String txt) {

        return "{\"d\" : \"Hello " + txt + " !\"}";
    }

如果你return一些复杂的class,比如Product,它会被转换成json,但是如果你returnString,lib就不能转换它到 {key=value} 并认为您手动生成了 json。请参阅此 question 了解更多信息。