假设我有一个Hibernate / JPA实体,如下所示:
@Entity
public class FooEntity {
...
@Type(type = "hstore")
HashMap<String, String> tags;
}
...而hstore
Type是来自this资源的简单UserType实现。
有没有办法在类似于这个Pseudocode的JPQL查询中访问hstore:
SELECT f FROM FooEntity f WHERE f.tags CONTAINS KEY(:key)
答案 0 :(得分:2)
您也可以简单地创建一个Hibernate org.hibernate.usertype.UserType。你扩展了那个班级;我们自己实现的一个例子:
public class HstoreUserType implements UserType {
/**
* PostgreSQL {@code hstore} field separator token.
*/
private static final String HSTORE_SEPARATOR_TOKEN = "=>";
/**
* {@link Pattern} used to find and split {@code hstore} entries.
*/
private static final Pattern HSTORE_ENTRY_PATTERN = Pattern.compile(String.format("\"(.*)\"%s\"(.*)\"", HSTORE_SEPARATOR_TOKEN));
/**
* The PostgreSQL value for the {@code hstore} data type.
*/
public static final int HSTORE_TYPE = 1111;
@Override
public int[] sqlTypes() {
return new int[] { HSTORE_TYPE };
}
@SuppressWarnings("rawtypes")
@Override
public Class returnedClass() {
return Map.class;
}
@Override
public boolean equals(final Object x, final Object y) throws HibernateException {
return x.equals(y);
}
@Override
public int hashCode(final Object x) throws HibernateException {
return x.hashCode();
}
@Override
public Object nullSafeGet(final ResultSet rs, final String[] names,
final SessionImplementor session, final Object owner)
throws HibernateException, SQLException {
return convertToEntityAttribute(rs.getString(names[0]));
}
@SuppressWarnings("unchecked")
@Override
public void nullSafeSet(final PreparedStatement st, final Object value, final int index,
final SessionImplementor session) throws HibernateException, SQLException {
st.setObject(index, convertToDatabaseColumn((Map<String,Object>)value), HSTORE_TYPE);
}
@SuppressWarnings("unchecked")
@Override
public Object deepCopy(final Object value) throws HibernateException {
return new HashMap<String,Object>(((Map<String,Object>)value));
}
@Override
public boolean isMutable() {
return true;
}
@Override
public Serializable disassemble(final Object value) throws HibernateException {
return (Serializable) value;
}
@Override
public Object assemble(final Serializable cached, final Object owner)
throws HibernateException {
return cached;
}
@Override
public Object replace(final Object original, final Object target, final Object owner)
throws HibernateException {
return original;
}
private String convertToDatabaseColumn(final Map<String, Object> attribute) {
final StringBuilder builder = new StringBuilder();
for (final Map.Entry<String, Object> entry : attribute.entrySet()) {
if(builder.length() > 1) {
builder.append(", ");
}
builder.append("\"");
builder.append(entry.getKey());
builder.append("\"");
builder.append(HSTORE_SEPARATOR_TOKEN);
builder.append("\"");
builder.append(entry.getValue().toString());
builder.append("\"");
}
return builder.toString();
}
private Map<String, Object> convertToEntityAttribute(final String dbData) {
final Map<String, Object> data = new HashMap<String, Object>();
if (dbData != null) {
final StringTokenizer tokenizer = new StringTokenizer(dbData, ",");
while(tokenizer.hasMoreTokens()) {
final Matcher matcher = HSTORE_ENTRY_PATTERN.matcher(tokenizer.nextToken().trim());
if(matcher.find()) {
data.put(matcher.group(1), matcher.group(2));
}
}
}
return data;
}
}
现在你可以在Entity bean中使用它,如下所示:
@Entity
@Table(name="YourEntityBeanTable")
@TypeDefs({
@TypeDef(name = "hstore", typeClass = HstoreUserType.class)
})
public class YourEntityBean {
.....
@Type(type = "hstore")
@Column(name= "an_hstore_column", columnDefinition = "hstore")
private Map<String, String> anHStoreColumn = new HashMap<>();
}
答案 1 :(得分:1)
Hibernate在许多数据库中提供了一个通用的查询抽象,因此非SQL语法很难被抽象出来。
如果你真的需要那些,我会去寻找id并使用它们来获取Hibernate实体。
如果您只对投影感兴趣,那么本机查询是您的最佳选择。