为什么我的嵌套 POJO return 从 Play 中的表单中为空! 1.x
Why does my nested POJO return null from a form in Play! 1.x
游戏中! Framework 1.5.1,为什么我得到 thingy.Owner 的空值?自动绑定不应该处理这个吗?
用户class
package models;
@Entity
@Table(name="objtest_user")
public class User extends Model
{
@Required
public String username;
@Password
@Required
public String password;
public String fullname;
public User(String username, String password, String fullname)
{
this.username = username;
this.password = password;
this.fullname = fullname;
}
@Override
public String toString()
{
return this.fullname;
}
}
和这个引用用户 class
的东西 class
package models;
import java.util.*;
import javax.persistence.*;
import play.db.jpa.*;
import play.data.validation.*;
@Entity
public class Thingy extends Model
{
@Required
public String Name;
@ManyToOne
public User Owner;
public Thingy(String name, User owner)
{
this.Name = name;
this.Owner = owner;
}
@Override
public String toString()
{
return Name;
}
}
和这个模板表单
#{extends 'main.html' /}
#{set title:'Home' /}
<p>Current user = ${currentUser}</p>
#{form @saveThingy(), id:'saveThingy'}
<input type="text" id="thingy.Name" name="thingy.Name"/>
<input type="hidden" id="thingy.Owner" name="thingy.Owner" value="${currentUser}"/>
<input type="submit" id="Save" value="Save"/>
#{/form}
控制器方法
public static void saveThingy(Thingy thingy)
{
System.out.println("Name = " + thingy.Name);
System.out.println("Owner = " + thingy.Owner);
thingy.save();
}
尝试更改以下行
<input type="hidden" id="thingy.Owner" name="thingy.Owner" value="${currentUser}"/>
到
<input type="hidden" id="thingy.Owner" name="thingy.Owner.id" value="${currentUser.id}"/>
如果您查看文档 (https://www.playframework.com/documentation/1.2.x/controllers#params),并查找 JPA 对象绑定部分,它谈到要求子对象具有 ID。当它找到一个对象的 ID 时播放,它将通过 JPA/Hibernate.
加载相关实体
游戏中! Framework 1.5.1,为什么我得到 thingy.Owner 的空值?自动绑定不应该处理这个吗?
用户class
package models;
@Entity
@Table(name="objtest_user")
public class User extends Model
{
@Required
public String username;
@Password
@Required
public String password;
public String fullname;
public User(String username, String password, String fullname)
{
this.username = username;
this.password = password;
this.fullname = fullname;
}
@Override
public String toString()
{
return this.fullname;
}
}
和这个引用用户 class
的东西 class package models;
import java.util.*;
import javax.persistence.*;
import play.db.jpa.*;
import play.data.validation.*;
@Entity
public class Thingy extends Model
{
@Required
public String Name;
@ManyToOne
public User Owner;
public Thingy(String name, User owner)
{
this.Name = name;
this.Owner = owner;
}
@Override
public String toString()
{
return Name;
}
}
和这个模板表单
#{extends 'main.html' /}
#{set title:'Home' /}
<p>Current user = ${currentUser}</p>
#{form @saveThingy(), id:'saveThingy'}
<input type="text" id="thingy.Name" name="thingy.Name"/>
<input type="hidden" id="thingy.Owner" name="thingy.Owner" value="${currentUser}"/>
<input type="submit" id="Save" value="Save"/>
#{/form}
控制器方法
public static void saveThingy(Thingy thingy)
{
System.out.println("Name = " + thingy.Name);
System.out.println("Owner = " + thingy.Owner);
thingy.save();
}
尝试更改以下行
<input type="hidden" id="thingy.Owner" name="thingy.Owner" value="${currentUser}"/>
到
<input type="hidden" id="thingy.Owner" name="thingy.Owner.id" value="${currentUser.id}"/>
如果您查看文档 (https://www.playframework.com/documentation/1.2.x/controllers#params),并查找 JPA 对象绑定部分,它谈到要求子对象具有 ID。当它找到一个对象的 ID 时播放,它将通过 JPA/Hibernate.
加载相关实体