我想在MySQL数据库中取出重复的记录。这可以用:
SELECT address, count(id) as cnt FROM list
GROUP BY address HAVING cnt > 1
结果是:
100 MAIN ST 2
我想要拖动它，以便它显示复制的每一行。喜欢的东西:
JIM JONES 100 MAIN ST
JOHN SMITH 100 MAIN ST
有什么想法可以做到吗?我试图避免做第一个，然后在代码中用第二个查询查找重复。
当前回答
这还将显示有多少重复项，并将在没有连接的情况下对结果进行排序
SELECT `Language` , id, COUNT( id ) AS how_many
FROM `languages`
GROUP BY `Language`
HAVING how_many >=2
ORDER BY how_many DESC
其他回答
select `cityname` from `codcities` group by `cityname` having count(*)>=2
这是你问的类似的问题，它是200%的工作和简单。 享受! !
这将在一次表传递中选择重复项，没有子查询。
SELECT *
FROM (
SELECT ao.*, (@r := @r + 1) AS rn
FROM (
SELECT @_address := 'N'
) vars,
(
SELECT *
FROM
list a
ORDER BY
address, id
) ao
WHERE CASE WHEN @_address <> address THEN @r := 0 ELSE 0 END IS NOT NULL
AND (@_address := address ) IS NOT NULL
) aoo
WHERE rn > 1
这个查询实际上模拟了Oracle和SQL Server中的ROW_NUMBER()
详见我博客上的文章:
分析函数:SUM, AVG, ROW_NUMBER -在MySQL模拟。
SELECT date FROM logs group by date having count(*) >= 2
不是很有效，但应该可以工作:
SELECT *
FROM list AS outer
WHERE (SELECT COUNT(*)
FROM list AS inner
WHERE inner.address = outer.address) > 1;
这样不是更简单吗?
SELECT *
FROM tc_tariff_groups
GROUP BY group_id
HAVING COUNT(group_id) >1
?
