PlayFramework完整实现一个APP（二）

1.开发DataModel

在app\moders 下新建User.java

package models;

import java.util.*;
import javax.persistence.*;
import play.db.jpa.*;本文来源gaodai$ma#com搞$代*码*网

@Entity
public 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标识是一个JPA entity，继承自play.db.jpa.Model，提供了JPA实现

类的字段，会自动映射到DB表中，默认表明是"User"，如果要修改表明，在类上添加标签"@Table(name="blog_user")"

2.测试

运行

>play test yape

或在Eclipse中运行，Test Yet Another Blog Engine

访问 http://localhost:9000/@tests， 进入测试模式

选择Test, Start执行，成功会标记为绿色，失败会有提示

3.写测试用例

修改 /test/BasicTest.java

@Test
public void createAndRetrieveUser() {
//Create a new user and save it
new User("[email protected]", "####", "Alex").save();

//Retrieve the user with email address
User user = User.find("byEmail", "[email protected]").first();

//Test
assertNotNull(user);
assertEquals("Alex", user.fullname);
}

创建User，查找User，进行断言

User继承自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();
}

添加测试用例

@Test
public void tryConnectAsUser() {
// Create a new user and save it
new User("[email protected]", "####", "Bob").save();

// Test
assertNotNull(User.connect("[email protected]", "####"));
assertNull(User.connect("[email protected]", "$$$$"));
assertNull(User.connect("[email protected]", "####"));
}

..

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