如果要删除具有多个字段的重复行，首先将它们取消为唯一不同的行指定的新唯一键，然后使用group by命令删除具有相同新唯一键的重复行:

Create TEMPORARY table tmp select concat(f1,f2) as cfs,t1.* from mytable as t1;
Create index x_tmp_cfs on tmp(cfs);
Create table unduptable select f1,f2,... from tmp group by cfs;

一个非常晚的贡献…万一这能帮助到以后的任何人…我有一个任务是在一个银行应用程序中找到匹配的交易对(实际上是账户到账户转账的双方)，以识别每个账户间转账交易的“从”和“到”，所以我们最终得到了这个:

SELECT 
    LEAST(primaryid, secondaryid) AS transactionid1,
    GREATEST(primaryid, secondaryid) AS transactionid2
FROM (
    SELECT table1.transactionid AS primaryid, 
        table2.transactionid AS secondaryid
    FROM financial_transactions table1
    INNER JOIN financial_transactions table2 
    ON table1.accountid = table2.accountid
    AND table1.transactionid <> table2.transactionid 
    AND table1.transactiondate = table2.transactiondate
    AND table1.sourceref = table2.destinationref
    AND table1.amount = (0 - table2.amount)
) AS DuplicateResultsTable
GROUP BY transactionid1
ORDER BY transactionid1;

The result is that the DuplicateResultsTable provides rows containing matching (i.e. duplicate) transactions, but it also provides the same transaction id's in reverse the second time it matches the same pair, so the outer SELECT is there to group by the first transaction ID, which is done by using LEAST and GREATEST to make sure the two transactionid's are always in the same order in the results, which makes it safe to GROUP by the first one, thus eliminating all the duplicate matches. Ran through nearly a million records and identified 12,000+ matches in just under 2 seconds. Of course the transactionid is the primary index, which really helped.

假设您的表名为TableABC，您想要的列是Col, T1的主键是key。

SELECT a.Key, b.Key, a.Col 
FROM TableABC a, TableABC b
WHERE a.Col = b.Col 
AND a.Key <> b.Key

与上面的答案相比，这种方法的优点是它给出了Key。

SELECT ColumnA, COUNT( * )
FROM Table
GROUP BY ColumnA
HAVING COUNT( * ) > 1

SELECT varchar_col
FROM table
GROUP BY varchar_col
HAVING COUNT(*) > 1;

感谢@novocaine的精彩回答，他的解决方案对我很有效。我稍微改变了它，以包括一个百分比的循环值，这在我的例子中是需要的。以下是修改后的版本。它将百分比减少到小数点后两位。如果你把2改成0，它就不会显示小数，改成1，它就会显示一位小数，以此类推。

SELECT GROUP_CONCAT(id), name, COUNT(*) c, 
COUNT(*) OVER() AS totalRecords, 
CONCAT(FORMAT(COUNT(*)/COUNT(*) OVER()*100,2),'%') as recurringPecentage
FROM table
GROUP BY name
HAVING c > 1