Question

这个问题非常简单我出于某种原因无法获得正确的结果来显示重复的记录

Table   : Paypal_ipn_orders
id                              payer_email
1                               susan@gmail.com
2                               ryan@gmail.com   
3                               susan@gmail.com
4                               steve@gmail.com
5                               steve@gmail.com

SELECT id, COUNT( payer_email ) `tot`
FROM paypal_ipn_orders
GROUP BY payer_email
HAVING `tot` >1

示例输出

id       tot
1         2
4         2

预期产出

id       payer_email 
1        susan@gmail.com
3        susan@gmail.com
4        steve@gmail.com
5        steve@gmail.com

我该如何实现？

Answer 1

SELECT id, payer_email
FROM paypal_ipn_orders
WHERE payer_email IN (
    SELECT payer_email
    FROM paypal_ipn_orders
    GROUP BY payer_email
    HAVING COUNT(id) > 1
)

sqlfiddle

Answer 2

IN在我的情况下（180秒）太慢了

所以我使用JOIN代替（0.3秒）

SELECT i.id, i.payer_email
FROM paypal_ipn_orders i
INNER JOIN (
 SELECT payer_email
    FROM paypal_ipn_orders 
    GROUP BY payer_email
    HAVING COUNT( id ) > 1
) j ON i.payer_email=j.payer_email

Answer 3

从表格中获取所有重复行的列表：

Select * from TABLE1 where PRIMARY_KEY_COLUMN NOT IN ( SELECT PRIMARY_KEY_COLUMN
FROM TABLE1 
GROUP BY DUP_COLUMN_NAME having (count(*) >= 1))

Answer 4

这是一个简单的例子：

select <duplicate_column_name> from <table_name> group by <duplicate_column_name> having count(*)>=2

它肯定会起作用。：）

Answer 5

如果我想选择一个或多个不相同的列值或者两个行数据可能相同，则上面的答案将无效

对于Ex。我也想选择用户名，出生日期。但在数据库中，用户名不是重复但出生日期会重复，那么此解决方案将无效。

为此使用此解决方案需要在同一桌/

上进行自我加入

SELECT  
    distinct(p1.id),  p1.payer_email , p1.username, p1.birth_date

FROM 
    paypal_ipn_orders AS p1 

INNER JOIN paypal_ipn_orders AS p2 

ON p1.payer_email=p2.payer_email

WHERE 

p1.birth_date=p2.birth_date

以上查询将返回具有相同email_id和相同出生日期的所有记录

Answer 6

使用此代码

javax.net.ssl.SSLException: Connection closed by peer 
at com.android.org.conscrypt.NativeCrypto.SSL_do_handshake(Native Method)
at com.android.org.conscrypt.OpenSSLSocketImpl.startHandshake(OpenSSLSocketImpl.java:406)
at okhttp3.internal.io.RealConnection.connectTls(RealConnection.java:188)
at okhttp3.internal.io.RealConnection.connectSocket(RealConnection.java:145)
at okhttp3.internal.io.RealConnection.connect(RealConnection.java:108)
at okhttp3.internal.http.StreamAllocation.findConnection(StreamAllocation.java:188)
at okhttp3.internal.http.StreamAllocation.findHealthyConnection(StreamAllocation.java:127)
at okhttp3.internal.http.StreamAllocation.newStream(StreamAllocation.java:97)
at okhttp3.internal.http.HttpEngine.connect(HttpEngine.java:289)
at okhttp3.internal.http.HttpEngine.sendRequest(HttpEngine.java:241)
at okhttp3.RealCall.getResponse(RealCall.java:240)
at okhttp3.RealCall$ApplicationInterceptorChain.proceed(RealCall.java:198)
at okhttp3.logging.HttpLoggingInterceptor.intercept(HttpLoggingInterceptor.java:203)
at okhttp3.RealCall$ApplicationInterceptorChain.proceed(RealCall.java:187)
at okhttp3.RealCall.getResponseWithInterceptorChain(RealCall.java:160)
at okhttp3.RealCall.access$100(RealCall.java:30)
at okhttp3.RealCall$AsyncCall.execute(RealCall.java:127)
at okhttp3.internal.NamedRunnable.run(NamedRunnable.java:33)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
at java.lang.Thread.run(Thread.java:841)

Answer 7

SELECT id, payer_email FROM paypal_ipn_orders
WHERE payer_email IN (
    SELECT payer_email FROM papypal_ipn_orders GROUP BY payer_email HAVING COUNT(*) > 1)

Answer 8

这对我来说最快

SELECT
    primary_key
FROM
    table_name
WHERE
    primary_key NOT IN (
        SELECT
            primary_key
        FROM
            table_name
        GROUP BY
            column_name
        HAVING
            COUNT(*) = 1
    );

Answer 9

我认为这种方式更简单。输出显示ID和付款人的电子邮件，其中付款人的电子邮件在此表中的多个记录中。结果按ID排序。

    SELECT id, payer_email
    FROM paypal_ipn_orders
    WHERE COUNT( payer_email )>1
    SORT BY id;

Answer 10

与this answer类似，但我使用的是临时表：

CREATE TEMPORARY TABLE duplicates (
    SELECT payer_email
    FROM paypal_ipn_orders
    GROUP BY payer_email
    HAVING COUNT(id) > 1
);
SELECT id, payer_email
FROM paypal_ipn_orders AS p
INNER JOIN duplicates AS d ON d.payer_email=p.payer_email;

Answer 11

SELECT * FROM `table` t1 join `table` t2 WHERE (t1.name=t2.name) && (t1.id!=t2.id)

Answer 12

尝试此查询：

SELECT id, COUNT( payer_email ) `tot`
FROM paypal_ipn_orders
GROUP BY id
HAVING `tot` >1

有帮助吗？

在mysql中只选择并显示重复的记录

12 个答案: