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;
其他回答
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;
我试着用最好的答案来回答这个问题，但它还是把我弄糊涂了。实际上，我只需要在表的一个字段上。下面这个链接的例子对我来说效果很好:
SELECT COUNT(*) c,title FROM `data` GROUP BY title HAVING c > 1;
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
SELECT firstname, lastname, address FROM list
WHERE
Address in
(SELECT address FROM list
GROUP BY address
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;
