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
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。
其他回答
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;
这样不是更简单吗?
SELECT *
FROM tc_tariff_groups
GROUP BY group_id
HAVING COUNT(group_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;
这还将显示有多少重复项，并将在没有连接的情况下对结果进行排序
SELECT `Language` , id, COUNT( id ) AS how_many
FROM `languages`
GROUP BY `Language`
HAVING how_many >=2
ORDER BY how_many DESC
SELECT *
FROM (SELECT address, COUNT(id) AS cnt
FROM list
GROUP BY address
HAVING ( COUNT(id) > 1 ))
