
针对上面表的数据,现在想根据age分组,并获取每个分组后的name有哪些(去重后)。
select age, GROUP_CONCAT(DISTINCT(name)) from testtable group by age ;
结果: 
如果想要增加排序:
SELECT
age,
GROUP_CONCAT(DISTINCT name) AS names,
COUNT(DISTINCT name) AS name_count
FROM
testtable
GROUP BY
age
ORDER BY
name_count DESC;
结果:
使用ES实现类似上面的SQL效果:
SearchRequest searchRequest = new SearchRequest("indexName");
SearchSourceBuilder searchSourceBuilder = new SearchSourceBuilder().size(0);
BoolQueryBuild

最低0.47元/天 解锁文章
1万+

被折叠的 条评论
为什么被折叠?



