分區值更改的row_number

Question

我有這段sql代碼，它根據表__working中的某些值查找row_number，表__working加入查找表__Eval中

;WITH r 
 AS (
    select 

    w.uid,
    t.TypeId,
    --weight
    ROW_NUMBER () OVER (PARTITION BY w.uid ORDER BY  DIFFERENCE(t.val1, w.val1) +  DIFFERENCE(t.val2, w.val2) + DIFFERENCE(t.val3, w.val3) + DIFFERENCE(t.val4, w.val4)  DESC) as Score
    ,w.account

    from __Working w
        join __eval t on w.val1 like t.val1 and IsNull(w.val4, '') like t.val4 and IsNull(w.val2, '') like t.val2 and IsNull(w.val3, '') like t.val3

)

select * from r     where   r.account = 1 and score = 1

這將返回一個typeId = 1

但是如果我這樣寫

   ;WITH r 
     AS (
        select 

        w.uid,
        t.TypeId,
        --weight
        ROW_NUMBER () OVER (PARTITION BY w.uid ORDER BY  DIFFERENCE(t.val1, w.val1) +  DIFFERENCE(t.val2, w.val2) + DIFFERENCE(t.val3, w.val3) + DIFFERENCE(t.val4, w.val4)  DESC) as Score
        ,w.account

        from __Working w
            join __eval t on w.val1 like t.val1 and IsNull(w.val4, '') like t.val4 and IsNull(w.val2, '') like t.val2 and IsNull(w.val3, '') like t.val3
            where   r.account = 1
    )

    select * from r     where   r.account = 1 and score = 1

它返回TypeId =2。我希望，如果在__working中跨不同帳戶使用多個UID，但我沒有。 我在這里想念什么？

Answer 1

哦，這是一種不穩定的怪異現象。 您的row_number()表達式為：

 ROW_NUMBER() OVER (PARTITION BY w.uid
                    ORDER BY  DIFFERENCE(t.val1, w.val1) +
                              DIFFERENCE(t.val2, w.val2) +
                              DIFFERENCE(t.val3, w.val3) +
                              DIFFERENCE(t.val4, w.val4)  DESC
                   ) as Score

問題是多個行的ORDER BY鍵具有相同的值。 不同的調用可以任意選擇多個行中的哪一個是第一行，第二行，依此類推。

規范的解決方案是包括某種唯一鍵，以便這種鍵是穩定的：

 ROW_NUMBER() OVER (PARTITION BY w.uid
                    ORDER BY  (DIFFERENCE(t.val1, w.val1) +
                               DIFFERENCE(t.val2, w.val2) +
                               DIFFERENCE(t.val3, w.val3) +
                               DIFFERENCE(t.val4, w.val4)
                              ) DESC,
                              ??  -- perhaps typeId
                   ) as Score

但是，我可能會建議一個更艱巨的解決方案。 接受關系可能存在的事實，並使用rank()和dense_rank()進行識別。 然后， 明確找出在打平領帶的情況下該怎么做-也許所有事情對您來說都同樣有趣，或者您可能還有其他打平領帶的方法。