我有以下 Hibernate forumla 查询，我可以在 mysql workbanch 中执行。

select group_concat(distinct t.column_1_name SEPARATOR ', ') from table_name t and t.fk_record_id = record_id

在使用 Hibernate 执行此查询时，hibernate 将父表附加到 SEPRATOR 关键字，如下面的查询所示。

select group_concat(distinct t.column_1_name parent_table.SEPARATOR ', ') from table_name t and t.fk_record_id = record_id

这里 hibernate 没有将 SEPRATOR 视为关键字。有人对此有任何想法吗？

最佳答案

您可以添加 SEPARATOR 作为关键字。实现您自己的 DialectResolver 并将关键字小写 添加到生成的方言中:

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 版本也是如此:

public class MyDialectResolver extends StandardDialectResolver {

    protected Dialect resolveDialectInternal(DatabaseMetaData metaData) throws SQLException {
        Dialect dialect = super.resolveDialectInternal(metaData);
        dialect.getKeywords().add("separator");
        return dialect;
    }

}

然后您必须告诉 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>

这同样适用于其他方言中的聚合函数。例如在 Oracle 中缺少 WITHIN 关键字。

还有另一种选择，它更独立于数据库(我更喜欢这种选择)。创建以下 SQLFunction:

public class ListAggFunction implements SQLFunction {

    /**
     * The pattern that describes how the function is build in SQL.
     *
     * Replacements:
     * {path} - is replaced with the path of the list attribute
     * {separator} - is replaced with the separator (defaults to '')
     * {orderByPath} - is replaced by the path that is used for ordering the elements of the list
     */
    private String pattern;

    /**
     * Creates a new ListAggFunction definition which uses the ANSI SQL:2016 syntax.
     */
    public ListAggFunction() {
        this("LISTAGG(DISTINCT {path}, {separator}) WITHIN GROUP(ORDER BY {orderByPath})");
    }

    /**
     * Creates a new ListAggFunction definition which uses a database specific syntax.
     *
     * @param pattern  The pattern that describes how the function is build in SQL.
     */
    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 });
    }

}

你可以像上面的关键字一样在DialectResolver中注册这个函数:

 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 查询中使用此函数，而无需考虑方言的语法:

 SELECT e.group, listagg(e.stringProperty, ', ') FROM Entity e GROUP BY e.group

关于mysql - SEPARATOR 关键字在 Hibernate 公式中无法正常工作，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32688660/