2010-12-12 146 views
4

使用Play!框架,我有以下兩種模式:完整性約束違規

@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(); 

    } 
} 

當我嘗試運行下面的測試:

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); 
    } 



} 

,我得到這個錯誤:

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(); 

日誌輸出:

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, ?, ?, ?, 
?, ?, ?, ?, ?, ?, ?)] 

任何人都可以幫助我erstand錯誤意味着什麼,我做錯了什麼?

+0

我不確定這是否是問題的原因,但Hibernate需要「所有持久化類必須具有默認構造函數」(更精確,無參數構造函數)。我沒有看到它的網站和用戶。 – Ralph 2010-12-12 09:37:29

回答

6

這裏的問題是Site模型類和User模型類之間的映射。給出的錯誤是缺少一個外鍵。

如果您在用戶類中註釋掉網站列表,您的測試將通過。這樣就縮小了問題範圍。

更新:

問題是因爲當你的網站對象被保存,這將節省的子對象第一(其中包括用戶對象)。發生這種情況時,它嘗試創建對Site對象的引用(由於mappedBy參數),但尚未保存(這將在保存User對象後完成)。

因此,另一種方法是根據您有權訪問的值(例如siteId)進行映射,或者在保存後將用戶添加到站點(因此已生成ID值)

我改變你的代碼mappedBy="siteId"和試運行的罰款。

+0

對我來說很好! – 2010-12-12 10:44:37

+0

我想你的建議,現在我得到這個錯誤: – 2010-12-12 14:19:17

+0

09:17:54710 ERROR〜不成功:ALTER TABLE網站加入索引FK2753671A145791( 網站ID),加約束FK2753671A145791外鍵(網站ID)引用的用戶(我 d) 09:17:54,710錯誤〜BLOB/TEXT列'siteId'在密鑰規範中使用,沒有 密鑰長度爲 – 2010-12-12 14:20:15

2

我通過使用甲骨文「初期遞延DEFERRABLE」解決這樣的問題,直到您提交整個事務這延遲的檢查!

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