如何使用 Hibernate 在 HSQLDB 表的日期类型列中插入日期？

Question

我正在尝试在 HSQLDB 的表的列中插入日期类型条目。问题是我正在使用 Hibernate 来执行此操作，有人可以帮我怎么做吗？

表格如下：

CREATE TABLE TEST(
WORKINGDATE DATE NOT NULL
)

Bean文件如下：

import java.util.Date;

public class Test {

    private Date workingDate;


    public Date getWorkingDate() {
        return workingDate;
    }

    public void setWorkingDate(Date workingDate) {
        this.workingDate = workingDate;
    }

    @Override
    public String toString() {
        return "Test [workingDate=" + workingDate + "]";
    }

}

XML 映射如下：

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
        "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class name="bean.hibpackage.Test" table="TEST">

        <property name="workingDate" column="WORKINGDATE" type="date" />

    </class>
</hibernate-mapping>

现在的测试场景是：

import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
import org.hibernate.cfg.Configuration;

import bean.hibpackage.Test;

public class TestHibernate2 {

    /**
     * @param args
     * @throws ParseException 
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Test test = new Test();

        String date_s = "2015-02-22";

        {
        //How to format the above given string to a date type so that the
        //program will work
        }

        test.setWorkingDate(?);

        Session ses = null;

        try {
            SessionFactory sessions = null;
            sessions = new Configuration().configure().buildSessionFactory();

            ses = sessions.openSession();

            Transaction tran = ses.beginTransaction();

            ses.save(test);

            tran.commit();
        } finally {
            ses.close();
        }

    }

}

Answer 1

您只需要解析Date 字符串表示：

String date_s = "2015-02-22";
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
test.setWorkingDate(sdf.parse(date_s));