無法添加表行，因為“當 IDENTITY_INSERT 設置為 OFF 時，無法為表‘用戶’中的標識列插入顯式值”錯誤

Question

我正在嘗試使用 Spring boot 和 JPA 將用戶保存到數據庫中，但由於此錯誤，它對我不起作用。

com.microsoft.sqlserver.jdbc.SQLServerException: 
Cannot insert explicit value for identity column in table 'Users'
when IDENTITY_INSERT is set to OFF.

錯誤很明顯，所以我嘗試在我的數據庫中更改身份插入成功，但錯誤仍然發生。 在那個問題中很奇怪的第二件事是我沒有嘗試將特定值作為 id 放入我的數據庫。 這是負責保存數據的確切行：

userService.save(new User("wikimmax","test","testmail@op.pl","test", Arrays.asList(new Role("ROLE_USER"))));

還有我的 User 類

import javax.persistence.*;
import java.util.Collection;

@Entity
@Table(name = "Users",uniqueConstraints = @UniqueConstraint(columnNames = "email"))
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id")
    private Long id;

    private String firstName;
    private String lastName;
    private String email;
    private String password;

    @ManyToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinTable(
            name = "users_roles",
            joinColumns = @JoinColumn(
                    name = "user_id", referencedColumnName = "id"),
            inverseJoinColumns = @JoinColumn(
                    name = "role_id", referencedColumnName = "id"))
    private Collection<Role> roles;

    public User() {
    }

    public User(String firstName, String lastName, String email, String password) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.email = email;
        this.password = password;
    }

    public User(String firstName, String lastName, String email, String password, Collection<Role> roles) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.email = email;
        this.password = password;
        this.roles = roles;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getPassword() {
        return password;
    }

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

    public Collection<Role> getRoles() {
        return roles;
    }

    public void setRoles(Collection<Role> roles) {
        this.roles = roles;
    }

    @Override
    public String toString() {

        return "User{" +
                "id=" + id +
                ", firstName='" + firstName + '\'' +
                ", lastName='" + lastName + '\'' +
                ", email='" + email + '\'' +
                ", password='" + "*********" + '\'' +
                ", roles=" + roles +
                '}';
    }
}

有我的桌子

任何想法都受到高度贊賞

更新我設法打印了由休眠生成的 SQL 查詢，它看起來像這樣：

    insert 
    into
        Users
        (email, firstName, lastName, password, id) 
    values
        (?, ?, ?, ?, ?)

Answer 1

請編輯問題並添加插入語句。

您不能在身份列中插入任何數據，因為該列由 SQL 引擎自動填充，在Insert語句中刪除 id 列並重試。

或者設置身份，但這不是個好主意。 允許將顯式值插入到表的標識列中。

SET IDENTITY_INSERT Users ON

例如 ：

USE AdventureWorks2012;  
GO  
-- Create tool table.  
CREATE TABLE dbo.Tool(  
   ID INT IDENTITY NOT NULL PRIMARY KEY,   
   Name VARCHAR(40) NOT NULL  
);  
GO  
-- Inserting values into products table.  
INSERT INTO dbo.Tool(Name)   
VALUES ('Screwdriver')  
        , ('Hammer')  
        , ('Saw')  
        , ('Shovel');  
GO  

-- Create a gap in the identity values.  
DELETE dbo.Tool  
WHERE Name = 'Saw';  
GO  

SELECT *   
FROM dbo.Tool;  
GO  

-- Try to insert an explicit ID value of 3;  
-- should return an error:
-- An explicit value for the identity column in table 'AdventureWorks2012.dbo.Tool' can only be specified when a column list is used and IDENTITY_INSERT is ON.
INSERT INTO dbo.Tool (ID, Name) VALUES (3, 'Garden shovel');  
GO  
-- SET IDENTITY_INSERT to ON.  
SET IDENTITY_INSERT dbo.Tool ON;  
GO  

-- Try to insert an explicit ID value of 3.  
INSERT INTO dbo.Tool (ID, Name) VALUES (3, 'Garden shovel');  
GO  

SELECT *   
FROM dbo.Tool;  
GO  
-- Drop products table.  
DROP TABLE dbo.Tool;  
GO