Play Framework 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.*;

@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("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);
 }

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("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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