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.

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

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;

要查找Employee中的name列中有多少记录是重复的，下面的查询很有用;

Select name from employee group by name having count(*)>1;