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
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”)。
为什么不直接INNER JOIN表本身呢?
SELECT a.firstname, a.lastname, a.address
FROM list a
INNER JOIN list b ON a.address = b.address
WHERE a.id <> b.id
如果地址可能存在两次以上，则需要DISTINCT。
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;
最快重复项删除查询过程:
/* create temp table with one primary column id */
INSERT INTO temp(id) SELECT MIN(id) FROM list GROUP BY (isbn) HAVING COUNT(*)>1;
DELETE FROM list WHERE id IN (SELECT id FROM temp);
DELETE FROM temp;
SELECT firstname, lastname, address FROM list
WHERE
Address in
(SELECT address FROM list
GROUP BY address
HAVING count(*) > 1)
