如何将org.postgresql.geometric.PGpoint映射到Hibernate类型 [英] How to map org.postgresql.geometric.PGpoint to Hibernate Type
问题描述
我将我的应用程序从spring jdbc迁移到JPA + Hibernate，并且使用的数据库是postgres。如果我使用spring jdbc，DB中的其中一个表的点数据类型转换为org.postgresql.geometric.PGpoint。我不知道如何将它映射到Hibernate类型。有人可以让我知道如何将点数据类型映射到休眠类型。
首先，我认为 GEOMETRY 数据类型由支持，但如果他们不是，你总是可以定义一个自定义的Hibernate类型和一个自定义的Hibernate方言。
当管理 POINT 包含地理点的列。
我创建了一个扩展 PostgreSQL9Dialect 的 PostgisDialect 类，在这里注册新的数据类型
public PostgisDialect（）{
registerColumnType（Types.BINARY，地理）;
}
在这种情况下，您可以将类型注册为geometry
然后，定义一个类，它实现 UserType UnsupportedOperationException
public class PointType implements UserType {
private static final Type [] PROPERTY_TYPES = new Type [] {
StringType.INSTANCE};
public String [] getPropertyNames（）{
return new String [] {point}; }
public Type [] getPropertyTypes（）{
return PROPERTY_TYPES;
}
public class returnedClass（）{
return Point.class;
$ b $ public boolean equals（Object o，Object o1）throws HibernateException {
if（（o instanceof Point&& o1 instanceof Point）== false）
返回false;
Point p1 =（Point）o;
Point p2 =（Point）o1;
boolean equal =（（p1.getX（）== p2.getX（））&&（p1.getY（）== p2.getY（）））;
回报相等;
$ b public Object nullSafeGet（ResultSet rs，String [] strings，SessionImplementor si，Object o）throws HibernateException，SQLException {
//从列中获取数据并使用BinaryParser将其转换为Point的方法
BinaryParser bp = new BinaryParser（）;
尝试{
String binaryString = rs.getString（strings [0]）;
返回bp.parse（binaryString）;
}
catch（Exception ex）{return null;}
$ b $ public void nullSafeSet（PreparedStatement ps，Object o，int i，SessionImplementor si）抛出HibernateException，SQLException {
Point p =（Point）o;
if（p！= null）{
BinaryWriter bw = new BinaryWriter（）;
ps.setObject（i，bw.writeBinary（p））;
}
public Object deepCopy（Object o）throws HibernateException {
Point p =（Point）o;
Point newPoint = null;
if（p！= null）{
newPoint = new Point（p.x，p.y）;
newPoint.setSrid（p.getSrid（））;
}
返回newPoint;
}
public boolean isMutable（）{
return true;
public int [] sqlTypes（）{
return new int [] {Types.BINARY};
}
}
一些快速注释：nullSafeSet和nullSafeGet分别使用BinaryWriter / BinaryParser对象向/从数据库写入和读取值。
一旦定义了所有这些，是你如何注释你的模型类，以便它使用你的自定义类型
@Column（name =point）
@Type（type =eu.enricorampazzosoluzioni.incidentpredicter.dataCopier.hibernate.types.PointType）
private point point;
最后但并非最不重要的是，您需要告诉Hibernate使用您的自定义方言。如果你使用Spring来定义会话工厂，你可以通过 hibernateProperties
pre $
< property name =hibernateProperties>
<道具>
< prop key =hibernate.dialect> eu.enricorampazzosoluzioni.incidentpredicter.dataCopier.hibernate.dialect.PostgisDialect< / prop>
< /道具>
< / property>
I'm migrating my application from spring jdbc to JPA + Hibernate and the DB used is postgres. One of my table in the DB has point data type which translates to org.postgresql.geometric.PGpoint if i use spring jdbc. I don't how to map this to Hibernate type. Can some one please let me know how do I map the point data type to a hibernate type.
First and foremost, I think GEOMETRY data types are supported by Hibernate Spatial, but if they are not, you can always define a custom Hibernate Type and a custom Hibernate Dialect.
I had a similar problem when managing a POINT column which contained geography points.
I created a PostgisDialect class which extended PostgreSQL9Dialect, where you register the new data type in this way
public PostgisDialect() {
registerColumnType(Types.BINARY, "geography");
}
in your case, you would register the type as "geometry"
then, you define a GeometryType class which implements UserType
Oddly enough, the process of writing a custom Hibernate Type is not one of the most documented features, so I will paste here what I wrote to define my PointType. For the other methods in the interface, I let them throw UnsupportedOperationException
public class PointType implements UserType{
private static final Type[] PROPERTY_TYPES = new Type[] {
StringType.INSTANCE };
public String[] getPropertyNames() {
return new String[] {"point"}; }
public Type[] getPropertyTypes() {
return PROPERTY_TYPES;
}
public Class returnedClass() {
return Point.class;
}
public boolean equals(Object o, Object o1) throws HibernateException {
if((o instanceof Point && o1 instanceof Point) == false)
return false;
Point p1 = (Point) o;
Point p2 = (Point) o1;
boolean equal = ((p1.getX() == p2.getX()) && (p1.getY() == p2.getY()));
return equal;
}
public Object nullSafeGet(ResultSet rs, String[] strings, SessionImplementor si, Object o) throws HibernateException, SQLException {
// the method which gets the data from the column and converts it to a Point using BinaryParser
BinaryParser bp = new BinaryParser();
try{
String binaryString = rs.getString(strings[0]);
return bp.parse(binaryString);
}
catch(Exception ex){ return null;}
}
public void nullSafeSet(PreparedStatement ps, Object o, int i, SessionImplementor si) throws HibernateException, SQLException {
Point p = (Point) o ;
if(p!=null){
BinaryWriter bw = new BinaryWriter();
ps.setObject(i,bw.writeBinary(p));
}
public Object deepCopy(Object o) throws HibernateException {
Point p = (Point) o;
Point newPoint = null;
if(p!=null){
newPoint = new Point(p.x, p.y);
newPoint.setSrid(p.getSrid());
}
return newPoint;
}
public boolean isMutable() {
return true;
}
public int[] sqlTypes() {
return new int[]{Types.BINARY};
}
}
some quick notes: nullSafeSet and nullSafeGet writes and reads the values to/from the database, respectively, using the BinaryWriter/BinaryParser objects.
Once you have defined all this, this is how you annotate your model class so that it uses your custom type
@Column(name="point")
@Type(type="eu.enricorampazzosoluzioni.incidentpredicter.dataCopier.hibernate.types.PointType")
private Point point;
Last, but not least, you need to tell Hibernate to use your custom dialect. If you use Spring to define your session factory, you can define it via hibernateProperties
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">eu.enricorampazzosoluzioni.incidentpredicter.dataCopier.hibernate.dialect.PostgisDialect</prop>
</props>
</property>
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