1.開發DataModel
在app/moders 下新建User.java
package models;import java.util.*;import javax.persistence.*;import play.db.jpa.*;@Entitypublic class User extends Model { public String email; public String passWord; public String fullname; public String isAdmin; public User(String email, String password, String fullname) { this.email = email; this.password = password; this.fullname = fullname; }}@Entity標識是一個JPA entity，繼承自play.db.jpa.Model，提供了JPA實現
類的字段，會自動映射到DB表中，默認表明是"User"，如果要修改表明，在類上添加標簽"@Table(name="blog_user")"
2.測試
運行
>play test yape
或在Eclipse中運行，Test Yet Another Blog Engine
訪問 http://localhost:9000/@tests， 進入測試模式
選擇Test, Start執行，成功會標記為綠色，失敗會有提示
3.寫測試用例
修改 /test/BasicTest.java
@Testpublic void createAndRetrieveUser() { //Create a new user and save it new User("alex@Gmail.com", "####", "Alex").save(); //Retrieve the user with email address User user = User.find("byEmail", "alex@gmail.com").first(); //Test assertNotNull(user); assertEquals("Alex", user.fullname); }創建User，查找User，進行斷言
User繼承自Model，提供了save/find等方法
User.java添加connect方法
public static User connect(String email, String passowrd) { return find("byEmailAndPassword", email, passowrd).first();}
添加測試用例
@Testpublic void tryConnectAsUser() { // Create a new user and save it new User("bob@gmail.com", "####", "Bob").save(); // Test assertNotNull(User.connect("bob@gmail.com", "####")); assertNull(User.connect("bob@gmail.com", "$$$$")); assertNull(User.connect("tom@gmail.com", "####"));}
..
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疑難解答
