您可以添加
SEPARATOR为关键字。实现自己
DialectResolver的关键字，并将关键字 以小写形式 添加到所得的方言中：
public class MyDialectResolver implements DialectResolver { public Dialect resolveDialect(DialectResolutionInfo info) { for (Database database : Database.values()) { Dialect dialect = database.resolveDialect(info); if (dialect != null) { dialect.getKeywords().add("separator"); return dialect; } } return null; }}对于5.2.13 / 5.3.0 之前的 Hibernate版本相同：
public class MyDialectResolver extends StandardDialectResolver { protected Dialect resolveDialectInternal(DatabasemetaData metaData) throws SQLException { Dialect dialect = super.resolveDialectInternal(metaData); dialect.getKeywords().add("separator"); return dialect; }}然后，您将不得不告诉Hibernate使用您的方言解析器。例如，在JPA中，您可以在persistence.xml中执行此操作：
<persistence> <persistence-unit> ... <property name="hibernate.dialect_resolvers" value="mypackage.MyDialectResolver"/> </persistence-unit></persistence>
其他方言中的汇总功能也是如此。例如，在Oracle中，
WITHIN缺少关键字。
还有另一种选择，它更独立于数据库（我更喜欢）。创建以下内容
SQLFunction：
public class ListAggFunction implements SQLFunction { private String pattern; public ListAggFunction() { this("LISTAGG(DISTINCT {path}, {separator}) WITHIN GROUP(ORDER BY {orderByPath})"); } public ListAggFunction(String pattern) { this.pattern = pattern; } public Type getReturnType(Type firstArgumentType, Mapping mapping) throws QueryException { return StringType.INSTANCE; } public boolean hasArguments() { return true; } public boolean hasParenthesesIfNoArguments() { return true; } public String render(Type firstArgumentType, List arguments, SessionFactoryImplementor factory) throws QueryException { if (arguments.isEmpty() || arguments.size() > 3) { throw new IllegalArgumentException( "Expected arguments for 'listagg': path [, separator [, order by path]]"); } String path = (String) arguments.get(0); String separator = arguments.size() < 2 ? "''" : (String) arguments.get(1); String orderByPath = arguments.size() <= 2 ? path : (String) arguments.get(2); return StringUtils.replaceEach(this.pattern, new String[] { "{path}", "{separator}", "{orderByPath}" }, new String[] { path, separator, orderByPath }); }}您可以使用与上面的关键字相同的方式在DialectResolver中注册此功能：
if ("MySQL".equals(info.getDatabaseName()) || "H2".equals(info.getDatabaseName())) { dialect.getFunctions().put("listagg", new ListAggFunction("GROUP_CONCAt(DISTINCT {path} ORDER BY {orderByPath} SEPARATOR {separator})")); } else { dialect.getFunctions().put("listagg", new ListAggFunction()); }现在，您可以在JPQL / HQL / Criteria查询中使用此功能，而无需考虑方言的语法：
SELECT e.group, listagg(e.stringProperty, ', ') FROM Entity e GROUP BY e.group
