SQL DENSE_RANK и PARTITION BY - PullRequest
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SQL DENSE_RANK и PARTITION BY

2 голосов
/ 03 июля 2019

Я не полностью понимаю результаты запроса, перечисленные ниже: 

CREATE TABLE #tmpAccountsTable (ID INT PRIMARY KEY, AccountID INT, 

AccountName varchar(20), CompanyID int, CompanyName varchar(50))
INSERT INTO #tmpAccountsTable (ID, AccountID, AccountName, CompanyID, CompanyName)
SELECT 1 AS ID, 1 AS AccountID, 'ABC Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 2 AS ID, 1 AS AccountID, 'ABC Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 3 AS ID, 2 AS AccountID, 'DEF Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 4 AS ID, 3 AS AccountID, 'GHI Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 5 AS ID, 4 AS AccountID, 'JKL Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 6 AS ID, 1 AS AccountID, 'ABC Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 7 AS ID, 1 AS AccountID, 'ABC Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 8 AS ID, 1 AS AccountID, 'ABC Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName UNION
SELECT 9 AS ID, 2 AS AccountID, 'DEF Account' AS AccountName, 1 AS CompanyID, 'Stark Enterprise' AS CompanyName

SELECT DENSE_RANK() OVER (PARTITION BY AccountID ORDER BY CompanyName) as AccountRANK, * FROM #tmpAccountsTable

DROP TABLE #tmpAccountsTable

При выполнении вышеизложенного я получаю следующий набор результатов:

Result Set

Из моего понимания SELECT DENSE_RANK() OVER (PARTITION BY AccountID ORDER BY CompanyName) as AccountRANK, * FROM #tmpAccountsTable должен быть создан раздел для всех одинаковых AccountId, и они должны были получить одно и то же значение DENSE_RANK(), но, как видно, это не так.

Что мне не хватает?

1 Ответ

4 голосов
/ 03 июля 2019

Если вы хотите отдельное значение для каждого AccountId, то оно должно быть частью ORDER BY, а не PARTITION BY: 

SELECT DENSE_RANK() OVER (ORDER BY AccountID, CompanyName) as AccountRANK, *
FROM #tmpAccountsTable;

PARTITION BY используется для сброса счетчика, поэтому каждый AccountId начинается с 1. Затем упорядочиваются CompanyNames. Поскольку все они имеют одинаковое значение, результат равен 1.

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