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。
其他回答
SELECT * FROM booking WHERE DATE(created_at) = '2022-01-11' 和代码在( 从预订中选择代码 按代码分组 have COUNT(code) > )由id DESC订购
我们也可以发现重复依赖于不止一个字段。对于这些情况，您可以使用下面的格式。
SELECT COUNT(*), column1, column2
FROM tablename
GROUP BY column1, column2
HAVING COUNT(*)>1;
关键是重写这个查询，以便它可以用作子查询。
SELECT firstname,
lastname,
list.address
FROM list
INNER JOIN (SELECT address
FROM list
GROUP BY address
HAVING COUNT(id) > 1) dup
ON list.address = dup.address;
这样不是更简单吗?
SELECT *
FROM tc_tariff_groups
GROUP BY group_id
HAVING COUNT(group_id) >1
?
从列表中选择地址where address = any (Select address from (Select address, count(id) CNT from list group by address with CNT > 1) as t1)按地址排序
内部子查询返回具有重复地址的行 外层子查询返回重复地址的地址列。 外层子查询必须只返回一列，因为它被用作操作符'= any'的操作数。
