Question

我有一个存储过程，它从我的数据库中获取提交条目以及每个条目引发了多少。我想要获得＆＃34; Rank＆＃34;提交的数量基于所提出的金额与所有其他提交的数量。

第二次回归应该是等级1，因为它的总增加量比第一次应该是等级2的记录要高。我接近了吗？

我的sp：

 SELECT A.[id],
             A.[petName],
             A.[petCaption],
             B.[petType],
             C.[FirstName] as ownerFirstName,
             C.[LastName] as ownerLastName,
             D.[imageName],

             (
                SELECT CONVERT(varchar(20),sum(transactionAmount), 1) as totalRaised,
                RANK() OVER (ORDER BY sum(transactionAmount) DESC) AS Rank
                FROM petContestTransactions
                WHERE submissionID = A.[id] and paymentType = 'donation'
                FOR    XML PATH ('transactionDetails'), TYPE, ELEMENTS
             )
      FROM petContestSubmissions as A
      JOIN petContestTypes as B
      ON A.[petType] = B.[id]
      JOIN EmpTable as C
      ON A.[empID] = C.EmpID
      JOIN petContestImages as D
      ON A.[image] = D.[submissionID]
      JOIN petContestTransactions as E
      ON E.[submissionID] = A.[id]
      WHERE E.[transactionStatus] = 'completed' and E.[paymentType] = 'submission' 
      FOR    XML PATH ('submission'), TYPE, ELEMENTS, ROOT ('root');

返回XML：

<root>
  <submission>
    <id>1</id>
    <petName>Nala</petName>
    <petCaption>Shes a wonder pup!</petCaption>
    <petType>Dog</petType>
    <ownerFirstName>Carl</ownerFirstName>
    <ownerLastName>H</ownerLastName>
    <imageName>nalaHUS123.png</imageName>
    <transactionDetails>
      <totalRaised>130.00</totalRaised>
      <Rank>1</Rank>
    </transactionDetails>
  </submission>
  <submission>
    <id>2</id>
    <petName>Simba</petName>
    <petCaption>Shes a wonder pup!</petCaption>
    <petType>Cat</petType>
    <ownerFirstName>Carl</ownerFirstName>
    <ownerLastName>H</ownerLastName>
    <imageName>simbaHUS123.png</imageName>
    <transactionDetails>
      <totalRaised>250.00</totalRaised>
      <Rank>1</Rank>
    </transactionDetails>
  </submission>
</root>

Answer 1

SELECT A.[id],
    A.[petName],
    A.[petCaption],
    B.[petType],
    C.[FirstName] as ownerFirstName,
    C.[LastName] as ownerLastName,
    D.[imageName],
    RANK() OVER (ORDER BY sum(F.totalRaised) DESC) AS Rank
FROM petContestSubmissions as A
JOIN petContestTypes as B
ON A.[petType] = B.[id]
JOIN EmpTable as C
ON A.[empID] = C.EmpID
JOIN petContestImages as D
ON A.[image] = D.[submissionID]
JOIN petContestTransactions as E
ON E.[submissionID] = A.[id]
OUTER APPLY
(
    SELECT sum(transactionAmount) as totalRaised,
    FROM petContestTransactions pt
    WHERE pt.submissionID = A.[id] and pt.paymentType = 'donation'
) F
WHERE E.[transactionStatus] = 'completed' and E.[paymentType] = 'submission' 
FOR    XML PATH ('submission'), TYPE, ELEMENTS, ROOT ('root');

我使用外部申请获得totalRaised然后你可以直接排名。

在SQL中获得结果排名

1 个答案: