SELECT t.*,(select count(*) from city as tt where tt.name=t.name) as count FROM `city` as t where (select count(*) from city as tt where tt.name=t.name) > 1 order by count desc

用你的表格替换城市。 将name替换为字段名

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。

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;

    SELECT *
    FROM (SELECT  address, COUNT(id) AS cnt
    FROM list
    GROUP BY address
    HAVING ( COUNT(id) > 1 ))

要快速查看重复的行，可以运行一个简单的查询

在这里，我正在查询表，并列出所有重复的行相同的user_id, market_place和sku:

select user_id, market_place,sku, count(id)as totals from sku_analytics group by user_id, market_place,sku having count(id)>1;

要删除重复的行，必须决定要删除哪一行。例如id较低(通常较旧)或其他日期信息。在我的情况下，我只是想删除较低的id，因为较新的id是最新的信息。

首先仔细检查是否正确的记录将被删除。在这里，我正在选择将被删除的副本中的记录(通过唯一id)。

select a.user_id, a.market_place,a.sku from sku_analytics a inner join sku_analytics b where a.id< b.id and a.user_id= b.user_id and a.market_place= b.market_place and a.sku = b.sku;

然后我运行delete查询来删除dupes:

delete a from sku_analytics a inner join sku_analytics b where a.id< b.id and a.user_id= b.user_id and a.market_place= b.market_place and a.sku = b.sku;

备份，双重检查，验证，验证备份，然后执行。

select `cityname` from `codcities` group by `cityname` having count(*)>=2

这是你问的类似的问题，它是200%的工作和简单。 享受! !