如何在 SQL Server 中获取每个组/分区的最大行数？

How to get the max row number per group/partition in SQL Server?

提问by DaveBurns

I'm using SQL Server 2005. I have a payments table with payment id's, user id's, and timestamps. I want to find the most recent payment for each user. This is easy to search and find an answer for. What I also want to know though is if the most recent payment is the user's first payment or not.

我使用的是 SQL Server 2005。我有一个包含付款 ID、用户 ID 和时间戳的付款表。我想为每个用户找到最近的付款。这很容易搜索并找到答案。不过，我还想知道最近的付款是否是用户的第一笔付款。

I have the following which will number each user's payments:

我有以下内容将对每个用户的付款进行编号：

SELECT
    p.payment_id,
    p.user_id,
    ROW_NUMBER() OVER (PARTITION BY p.user_id ORDER BY p.payment_date) AS paymentNumber
FROM
    payment p

I'm not making the mental leap which then lets me then pick the highest paymentNumber per user. If I use the above as a subselect by using MAX(paymentNumber) and then grouping by user_id, I lose the payment_id which I need. But if I also add the payment_id into the group by clause, I'm back to one row per payment. I'm sure I'm overlooking the obvious. Any help?

我没有进行精神上的飞跃，然后让我选择每个用户的最高付款数量。如果我使用 MAX(paymentNumber) 将上述内容用作子选择，然后按 user_id 分组，则会丢失我需要的 payment_id。但是，如果我还将 payment_id 添加到 group by 子句中，则每次付款都会返回一行。我确定我忽略了显而易见的事情。有什么帮助吗？

回答by Chandu

Try this:

尝试这个：

SELECT a.*, CASE WHEN totalPayments>1 THEN 'NO' ELSE 'YES' END IsFirstPayment
  FROM(
                SELECT  p.payment_id,     
                                p.user_id,     
                                ROW_NUMBER() OVER (PARTITION BY p.user_id ORDER BY p.payment_date DESC) AS paymentNumber,
                                SUM(1) OVER (PARTITION BY p.user_id) AS totalPayments
                    FROM payment p 
            ) a
WHERE   paymentNumber = 1       

回答by btilly

Do the same thing again.

再做同样的事情。

SELECT
    p.payment_id,
    p.user_id,
    ROW_NUMBER() OVER (PARTITION BY p.user_id ORDER BY p.payment_date) AS paymentNumber,
    ROW_NUMBER() OVER (PARTITION BY p.user_id ORDER BY p.payment_date DESC) AS reversePaymentNumber,
FROM
    payment p

Now the most recent payment has reversePaymentNumber 1, and the number of payments will be paymentNumber.

现在最近的付款有 reversePaymentNumber 1，付款次数将为 paymentNumber。

回答by Ben Thul

; with cte as (
SELECT
    p.payment_id,
    p.user_id,
    ROW_NUMBER() OVER (PARTITION BY p.user_id ORDER BY p.payment_date desc) AS paymentNumber
FROM
    payment p
) select * from cte where paymentNumber = 1

回答by DForck42

a less cool way i suppose

我想一个不太酷的方式

; with maxp as
(
    select
        p.user_id,
        max(p.payment_date) as MaxPaymentDate
    from payment p
    group by p.userid
),
nump as
(
    select
        p.payment_id,     
        p.user_id,     
        p.payment_date,
        ROW_NUMBER() OVER (PARTITION BY p.user_id ORDER BY p.payment_date) AS paymentNumber 
    FROM payment p
),
a as
(
select
    nump.payment_id,
    nump.user_id,
    nump.paymentNumber
    case when maxp.MaxPaymentDate is null then 'Old' else 'New' end as NewState
from nump
    left outer join maxp
        on nump.user_id=maxp.user_id
            and nump.payment_date=maxp.MaxPaymentDate
)

select
*
from a
where NewState='New'

回答by Abhishek Mukherjee

SELECT * FROM (
        SELECT ROW_NUMBER() OVER(PARTITION BY OS.ContactId ORDER BY OS.Date ASC) AS FirstRow#,
        ROW_NUMBER() OVER(PARTITION BY OS.ContactId ORDER BY OS.Date DESC) AS LastRow#,
        OS.Contactid,CONVERT(VARCHAR,OS.Date,106) 'Purchase Month',
        OS.ProductId 'MyCII Subscription/Directory', OS.Charges 'Amount(INR)',OS.Date 'RAWDate'
        FROM tblOnlineServices OS
        WHERE Date IS NOT NULL AND Contactid IN('C000013112','C000010859') 
    ) FirstPurchase 
    WHERE FirstRow# = 1 OR LastRow# = 1
    ORDER BY Contactid, RAWDate

回答by MarioVW

How about this?

这个怎么样？

SELECT
    p.user_id,
    MAX(p.payment_date) as lastPayment,
    CASE COUNT(p.payment_id) WHEN 1 THEN 1 ELSE 0 END as isFirstPayment
FROM
    payment p
GROUP BY
    p.user_id