无法在Postman中执行管理员登录

7uzetpgm  于 2022-11-07  发布在  Postman
关注(0)|答案(1)|浏览(320)

我试图在postman中为admin创建一个登录函数,但无法获得正确凭据所需的输出。
这是登录控制器

@RequestMapping(value="/", method = RequestMethod.GET)
    public String loginPage() {
        return "Welcome to ABC Bank login page";
    }

    @PostMapping(value="/login", consumes = "application/json", produces ="application/json")
    public String display(@RequestBody String userId, String password,Model m){
        //System.out.println("Please Login");
    System.out.println("userId" + userId);
    if(userId.equals("admin") && password.equals("root")) {
        return "Welcome to ABC Bank";
    }else {
        return "Login UnSuccessful ";
    }

    }

这是登录模型

@Entity
@Table(name = "login_page")
public class Login {

    @Id
    //@GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "user_Id", nullable = false)
    private String userId;

    @Column(name = "password", nullable = false)
  // @Length(min = 5, message = "*Your password must have at least 5 characters")
  // @NotEmpty(message = "*Please provide your password")
   // @JsonIgnore
    private String password;

    public String getUserId() {
        return userId;
    }

    public void setUserId(String userId) {
        this.userId = userId;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

}

因此,当管理员使用正确的凭据登录时,我需要执行Welcome to ABC bank。如何执行?
enter image description here

fnx2tebb

fnx2tebb1#

而不是

public String display(@RequestBody String userId, String password,Model m){}

我会考虑做一些

public String display(@RequestBody LoginRequest request){}

这可以通过标头和令牌的Map来进一步增强安全性。

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