DISTINCT COUNT (*) will return a row for each unique count. What you want is COUNT(DISTINCT <expression>) : evaluates expression for each row in ... ... <看更多>

This is how I'd do it: SELECT * FROM #MyTable AS mt CROSS APPLY ( SELECT COUNT(DISTINCT mt2.Col_B) AS dc FROM #MyTable AS mt2 WHERE mt2. ... <看更多>

最简单的方法. 这个问题逻辑上很简单，SQL也很容易写出来，例如：. SELECT COUNT( ... ... <看更多>

sql (which is running COUNT(DISTINCT ...) ) and bench-count-distinct.sql (this extension). An example of results from one particular machine (CPU Intel i5-2500k ... ... <看更多>