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
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;
其他回答
SELECT *
FROM (SELECT address, COUNT(id) AS cnt
FROM list
GROUP BY address
HAVING ( COUNT(id) > 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;
另一个解决方案是使用表别名，如下所示:
SELECT p1.id, p2.id, p1.address
FROM list AS p1, list AS p2
WHERE p1.address = p2.address
AND p1.id != p2.id
在这种情况下，您真正要做的是获取原始的列表表，从中创建两个假装的表——p1和p2，然后在地址列上执行连接(第3行)。第4行确保相同的记录不会在结果集中多次出现(“重复重复”)。
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。
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;
