例外情况如下: SEVERE:servlet [hello]中的Servlet.service()与path [/ Spring_MVC]的上下文引发异常[请求处理失败;嵌套异常是java.lang.IllegalArgumentException:没有找到具有根本原因的类型为:com.spring.mvc.model.User类的返回值的转换器 java.lang.IllegalArgumentException:找不到类型为:class com的返回值的转换器.spring.mvc.model.User未找到类型为返回值的转换器:class com.spring.mvc.model.User]
源代码都低于: UserController.java
@Controller
@RequestMapping("/user")
public class UserController {
private Map<String,User> users = new HashMap <String, User>();
public UserController(){
users.put("wang", new User("wang","pwang","wang a bo","123"));
users.put("chen", new User("chen","pChen","Chen Lin","1e23"));
}
@RequestMapping(value="{userName}",method=RequestMethod.GET, params="json")
@ResponseBody
public User show(@PathVariable String userName) {
return users.get(userName);
}
}
User.java
public class User {
@NotEmpty(message="User Name not null!!!")
private String userName;
@Size(min=6,max=12,message="Password need 6 to 12 Character!!!")
private String passWord;
private String nickName;
@Email(message="Email Fromat invalide!!")
private String email;
public User(){
}
public User(String userName, String passWord, String nickName, String email) {
super();
this.userName = userName;
this.passWord = passWord;
this.nickName = nickName;
this.email = email;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getPassWord() {
return passWord;
}
public void setPassWord(String passWord) {
this.passWord = passWord;
}
public String getNickName() {
return nickName;
}
public void setNickName(String nickName) {
this.nickName = nickName;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
}
弹簧4.3.5释放 同时杰克逊 - 清一色1.9.4 .jar
我用下面罐子代替同时杰克逊 - 全1.9.4.jar。此问题已得到解决。 –