完整性约束违规

Question

使用Play！框架，我有以下两种模式：

@Entity 
public class User extends Model { 
    public String firstName; 
    public String lastName; 
    public String email; 
    public String password; 
    public boolean isAdmin; 

    @OneToMany(mappedBy="id", cascade=CascadeType.ALL) 
    public List<Site> sites; 

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

    public static User connect(String email, String password) { 
     return User.find("byEmailAndPassword",email,Crypto.passwordHash(password)).first(); 
    } 

    public static User findUser(String email) { 
     return User.find("byEmail",email).first(); 
    } 

    public static User createUser(String firstName, String lastName, String email, String password, boolean isAdmin) { 
     String pw = Crypto.passwordHash(password); 

     User u = new User(firstName, lastName, email, pw); 
     u.isAdmin = isAdmin; 
     u.save(); 
     return u; 
    } 
} 

@Entity 
public class Site extends Model { 

    public UUID siteId; 

    public String alias; 
    public String protocol; 
    public String host; 
    public String username; 
    public String password; 
    public int port; 
    public String rootPath; 

    @Lob 
    public String description; 

    @ManyToOne 
    public User user; 

    public Site(User user, String alias, String protocol, String host, String username, String password) { 
     this.user = user; 
     this.alias = alias; 
     this.protocol = protocol; 
     this.host = host; 
     this.username = username; 
     this.password = password; 
     this.port = 21; 
     this.siteId = UUID.randomUUID(); 

    } 
} 

当我尝试运行下面的测试：

public class BasicTest extends UnitTest { 

    @Before 
    public void setup() { 
     Fixtures.deleteAll(); 
    } 

    @Test 
    public void createAndRetrieveUser() { 
     new User("Jason","Miesionczek","something","something").save(); 
     User jason = User.find("byEmail", "something").first(); 

     assertNotNull(jason); 
     assertEquals("Jason", jason.firstName); 
    } 

    @Test 
    public void userSite() { 
     new User("Jason","Miesionczek","something","something").save(); 
     User jason = User.find("byEmail", "something").first(); 

     new Site(jason, "InterEditor","ftp","something","something","something").save(); 

     List<Site> sites = Site.find("byUser", jason).fetch(); 

     assertEquals(1, sites.size()); 

     Site site1 = sites.get(0); 
     assertNotNull(site1); 
     assertEquals("InterEditor", site1.alias); 
     assertNotNull(site1.siteId); 
    } 



} 

，我得到这个错误：

A javax.persistence.PersistenceException has been caught, org.hibernate.exception.ConstraintViolationException: could not insert: [models.Site] 
In /test/BasicTest.java, line 27 : 
new Site(jason, "InterEditor","ftp","something","something","something").save(); 

日志输出：

00:06:26,184 WARN ~ SQL Error: -177, SQLState: 23000 
00:06:26,184 ERROR ~ Integrity constraint violation - no parent FK2753674FD92E0A 
table: USER in statement [insert into Site (id, alias, description, host, passw 
ord, port, protocol, rootPath, siteId, user_id, username) values (null, ?, ?, ?, 
?, ?, ?, ?, ?, ?, ?)] 

任何人都可以帮助我erstand错误意味着什么，我做错了什么？

Answer 1

这里的问题是Site模型类和User模型类之间的映射。给出的错误是缺少一个外键。

如果您在用户类中注释掉网站列表，您的测试将通过。这样就缩小了问题范围。

更新：

问题是因为当你的网站对象被保存，这将节省的子对象第一（其中包括用户对象）。发生这种情况时，它尝试创建对Site对象的引用（由于mappedBy参数），但尚未保存（这将在保存User对象后完成）。

因此，另一种方法是根据您有权访问的值（例如siteId）进行映射，或者在保存后将用户添加到站点（因此已生成ID值）

我改变你的代码mappedBy="siteId"和试运行的罚款。

Answer 2

我通过使用甲骨文“初期递延DEFERRABLE”解决这样的问题，直到您提交整个事务这延迟的检查！