我有一个带有以下控制器的Spring应用程序：
@RestController
@RequestMapping("/app")
public class RegisterRestController {
@Autowired
UserRepository userRepository;
@Autowired
PasswordEncoder passwordEncoder;
@Autowired
UserService userService;
@RequestMapping( value="/loginuser",method =RequestMethod.POST,produces="application/json")
public String loginUser(@RequestBody String requestBody) {
System.out.println("inside");
JSONObject responseJsonObject = new JSONObject();
String phonenumber;
String password;
try{
JSONObject object = new JSONObject(requestBody);
phonenumber = object.getString("phonenumber");
password = object.getString("password");
User user = userService.findByNumber(phonenumber);
String sha256Password = passwordEncoder.encode(password);
if(sha256Password.equals(user.getPassword())){
responseJsonObject.put("response","Login Successful");
}
else {
responseJsonObject.put("repsonse","Login failed");
}
}
catch (Exception e){
e.printStackTrace();
try {
responseJsonObject.put("response","Invalid Credentials");
} catch (JSONException e1) {
e1.printStackTrace();
}
}
return responseJsonObject.toString();
}
但是,当我从Postman发送包含以下内容的POST请求时：
{
"phonenumber":"9123456789","password":"password"
}
我收到以下回复：
{
"timestamp": 1456043810789,"status": 400,"error": "Bad Request","exception": "org.springframework.http.converter.HttpMessageNotReadableException","message": "Could not read JSON: Can not deserialize instance of java.lang.String out of START_OBJECT tokenn at [Source: java.io.PushbackInputStream@eaa3acb; line: 1,column: 1]; nested exception is com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of java.lang.String out of START_OBJECT tokenn at [Source: java.io.PushbackInputStream@eaa3acb; line: 1,column: 1]","path": "/app/loginuser"
}
另外,我也在试验Spring Security.服务器没有显示任何错误,控制器似乎没有收到请求,因为没有打印“内部”.我试图了解Spring,但是我找不到出现这种错误的原因.我将不胜感激任何帮助.提前致谢
您的代码中存在两个问题：
>您尝试将JSON转换为控制器内的对象.
这已经由Spring完成了.它接收请求的主体并尝试将其转换为控制器方法中相应参数的Java类.
>您的控制器方法需要一个字符串：
@RequestBody String requestBody
并且您发送的对象具有两个属性：
{
"phonenumber": "9123456789","password": "password"
}
解：
为您需要登录的值创建一个类：
public class Login {
public String phonenumber;
public String password;
// you need a zero argument constructor
// maybe you have to add getter and setters
}
更改您的控制器方法,以便它期望这种类型的对象
@RequestBody Login requestBody
