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.

作为利维克的答案的一个变体，它可以让你找到重复结果的id，我使用了以下方法:

SELECT * FROM table1 WHERE column1 IN (SELECT column1 AS duplicate_value FROM table1 GROUP BY column1 HAVING COUNT(*) > 1)

根据levik的回答来获取重复行的id，如果服务器支持的话，可以执行GROUP_CONCAT(这将返回一个以逗号分隔的id列表)。

SELECT GROUP_CONCAT(id), name, COUNT(*) c
FROM documents
GROUP BY name
HAVING c > 1;

CREATE TABLE tbl_master
    (`id` int, `email` varchar(15));

INSERT INTO tbl_master
    (`id`, `email`) VALUES
    (1, 'test1@gmail.com'),
    (2, 'test2@gmail.com'),
    (3, 'test1@gmail.com'),
    (4, 'test2@gmail.com'),
    (5, 'test5@gmail.com');

QUERY : SELECT id, email FROM tbl_master
WHERE email IN (SELECT email FROM tbl_master GROUP BY email HAVING COUNT(id) > 1)

Select column_name, column_name1,column_name2, count(1) as temp from table_name group by column_name having temp > 1

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.