Question

我使用 Struts Spring和Hibernate集成与mysql＆amp;进行在线考试项目Eclipse kepler。在registration.jsp页面中提交值时，我试图将这些值存储在同一数据库中的两个不同的表（user_details，address）中。我可以将它们存储在DB中，但是我无法获取user_id，这是地址表的外键。 user_id是user_details表中的主键。除了地址表中的user_id外，所有其他字段都填充了正确的值。我试图在地址表中使用它。但我无法做到这一点。我已经附上了我现在正在使用的代码，

user.hbm.xml

<hibernate-mapping>
    <class name="UserDetails" table="user_details">
        <id name="user_id" type="int" column="user_id" >
            <generator class="identity">

            </generator>
        </id>
        <property name="first_name" type="string">
            <column name="first_name"/>
        </property>
        <property name="last_name" type="string">
            <column name="last_name"/>
        </property>
        <property name="email" type="string">
            <column name="email"/>
        </property>
        <property name="password" type="string">
            <column name="password"/>
        </property>
        <property name="gender" type="string">
            <column name="gender"/>
        </property>
        <property name="dob" type="int">
            <column name="dob"/>
        </property>
        <property name="phone" type="int">
            <column name="phone"/>
        </property>
        <property name="experience" type="float">
            <column name="experience"/>
        </property>

         <set name="addr" table="address"  
                inverse="true" lazy="true" fetch="select" cascade = "save-update">
            <key>
                <column name="user_id" not-null="false"  />
            </key>
            <one-to-many class="UserAddress"  />
        </set>
    </class>
</hibernate-mapping>

useraddress.hbm.xml

<hibernate-mapping> 
    <class name="UserAddress" table="address">
        <id name="address_id" type="int" column="address_id">
            <generator class="identity"/>
        </id>
        <property name="addr_line1" type="string">
            <column name="addr_line_1"/>
        </property>
        <property name="addr_line2" type="string">
            <column name="addr_line_2"/>
        </property>
        <property name="addr_line3" type="string">
            <column name="addr_line_3"/>
        </property>
        <property name="city" type="string">
            <column name="city"/>
        </property>
        <property name="zipcode" type="int">
            <column name="zipcode"/>
        </property>
        <property name="state" type="string">
            <column name="state"/>
        </property>
        <property name="country" type="string">
            <column name="country"/>
        </property>

        <many-to-one name="user_detail" class="UserDetails" fetch="select">
        <column name="user_id" not-null="false"></column>
        </many-to-one>      
    </class>
</hibernate-mapping>

UserDetails.java

public class UserDetails {

        @Id
        @GeneratedValue(strategy=GenerationType.AUTO)
        //@OneToMany (mappedBy="user_details", cascade = CascadeType.ALL)
        @OneToMany (cascade = { CascadeType.PERSIST, CascadeType.MERGE}, mappedBy="user_detail")
        public int user_id; //primary key
        private String first_name;
        private String last_name;
        private String email;
        private String password;
        private String gender;
        private int dob;
        private int phone;
        private float experience;
        private Set<UserAddress> addr;//set name

    //getters and setters created

UserAddress.java

public class UserAddress {

@Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private int address_id; //primary key
    @ManyToOne(fetch=FetchType.EAGER, targetEntity=UserDetails.class)
    @JoinColumn(name="user_id")
    private UserDetails user_detail;
        private String addr_line1;
        private String addr_line2;
        private String addr_line3;
        private String city;
        private int zipcode;
        private String state;
        private String country;
        //getters and setters created

我认为我在hibernate映射部分缺少一些东西，因为我能够存储除user_id之外的其他地址表值。如果有人有兴趣使用完整的代码

如何从一个表中获取数据＆amp;使用hibernate映射插入另一个

0 个答案: