TSQL OVER clause: COUNT(*) OVER (ORDER BY a)

偶尔善良 提交于 2019-12-03 05:56:42

It gives a running total (this functionality was not implemented in SQL Server until version 2012.)

The ORDER BY defines the window to be aggregated with UNBOUNDED PRECEDING and CURRENT ROW as the default when not specified. SQL Server defaults to the less well performing RANGE option rather than ROWS.

They have different semantics in the case of ties in that the window for the RANGE version includes not just the current row (and preceding rows) but also any additional tied rows with the same value of a as the current row. This can be seen in the number of rows counted by each in the results below.

SELECT  a, 
        b,
        COUNT(*) OVER (ORDER BY a 
                         ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS  [Rows],
        COUNT(*) OVER (ORDER BY a 
                         RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS [Range],
        COUNT(*) OVER() AS [Over()]
    FROM    t;

Returns

a        b        Rows        Range       Over()
-------- -------- ----------- ----------- -----------
NULL     NULL     1           4           12
NULL     NULL     2           4           12
NULL     NULL     3           4           12
NULL     NULL     4           4           12
a        b        5           7           12
a        b        6           7           12
a        b        7           7           12
c        d        8           11          12
c        d        9           11          12
c        d        10          11          12
c        d        11          11          12
e        NULL     12          12          12

To achieve the result that you were expecting to get omit both the PARTITION BY and ORDER BY and use an empty OVER() clause (also shown above).

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