gpt4 book ai didi

java - 有没有办法用 Hibernate/JPQL 查询 PostgreSQL hstore?

转载 作者:塔克拉玛干 更新时间:2023-11-03 04:32:24 25 4
gpt4 key购买 nike

假设我有一个如下所示的 Hibernate/JPA 实体:

@Entity
public class FooEntity {

...

@Type(type = "hstore")
HashMap<String, String> tags;
}

... 和 hstore 类型是来自 this 的简单 UserType 实现资源。

有没有一种方法可以在类似于此伪代码的 JPQL 查询中访问 hstore:

SELECT f FROM FooEntity f WHERE f.tags CONTAINS KEY(:key)

最佳答案

您也可以简单地创建一个 Hibernate org.hibernate.usertype.UserType。你扩展那个类;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;
}

现在您可以像这样在实体 bean 中使用它:

@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<>();



}

关于java - 有没有办法用 Hibernate/JPQL 查询 PostgreSQL hstore?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24517120/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com