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
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;
其他回答
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。
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;
会是这样的:
SELECT t1.firstname t1.lastname t1.address FROM list t1
INNER JOIN list t2
WHERE
t1.id < t2.id AND
t1.address = t2.address;
这里的大多数答案不适用于有多个重复结果和/或有多个列要检查重复的情况。当你在这种情况下，你可以使用这个查询来获得所有重复的id:
SELECT address, email, COUNT(*) AS QUANTITY_DUPLICATES, GROUP_CONCAT(id) AS ID_DUPLICATES
FROM list
GROUP BY address, email
HAVING COUNT(*)>1;
如果希望将每个结果作为一行列出，则需要更复杂的查询。这是我发现有效的方法:
CREATE TEMPORARY TABLE IF NOT EXISTS temptable AS (
SELECT GROUP_CONCAT(id) AS ID_DUPLICATES
FROM list
GROUP BY address, email
HAVING COUNT(*)>1
);
SELECT d.*
FROM list AS d, temptable AS t
WHERE FIND_IN_SET(d.id, t.ID_DUPLICATES)
ORDER BY d.id;
Powerlord的答案确实是最好的，我建议再做一个改变:使用LIMIT来确保db不会超载:
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
LIMIT 10
如果没有WHERE和when连接，使用LIMIT是一个好习惯。从小值开始，检查查询有多重，然后增加限制。
