Java - 如何使用邮递员插入数据

时间:2018-03-01 06:00:36

标签: java spring-mvc postman

我的代码运行正常,但在邮递员方面显示错误

  

"此请求标识的资源只能根据请求生成具有不可接受特征的响应"接受"报头&#34。

发布的数据来自restControllerserviceImplStudentimpl

我的restController

        @RequestMapping(value="save",method=RequestMethod.POST,produces="application/json")

        GetStudentSaveResponse saveResponse(@RequestBody GetStudentSaveRequest saveRequest)
        {
            System.out.println(saveRequest.getName());
            System.out.println(saveRequest.getAddress());
            return serviceIntf.saveResponse(saveRequest);
        }
         RestImpl is
         @Override
        public GetStudentSaveResponse saveResponse(GetStudentSaveRequest saveRequest) {
                GetStudentSaveResponse saveResponse = new GetStudentSaveResponse();
                studentIntf.SaveStudent(saveRequest.getName(), saveRequest.getAddress());           
                System.out.println("SERVICE"+saveRequest.getName());
                return saveResponse;
            }

实施类是:

public void SaveStudent(String name, String address) {
    try
    {
        java.sql.Date date = new java.sql.Date(new java.util.Date().getTime());
        con=jdbctemplate.getDataSource().getConnection();

        CallableStatement call = con.prepareCall("{Student_pro(?,?,?,?,?)}");
        call.setString(1, "insertStudent");
        call.setInt(2, 0);
        call.setString(3, name);
        call.setString(4, address);
        call.setDate(5, date);

        System.out.println(name);
        System.out.println(address);

    }catch(Exception e){e.printStackTrace();}
}

1 个答案:

答案 0 :(得分:0)

确保使用以下

注释Controller类
@Controller
@EnableWebMvc 

您正在获取HTTP 406,表示媒体类型不正确。将内容类型设置为application / json,如下所示

enter image description here

<强>更新

问题是,如果要从POST方法返回Object,请使用ResponseEntity<Object>而不是直接使用Object。因此,在这种情况下,您将返回类型更改为ResponseEntity<GetStudentSaveResponse>