选择具有的独特

Question

我希望做一个选择statment，在连续选择所有不同的项目，并提供每个结果

SELECT DISTINCT [Column 16] FROM [tab] 

的合计总数鲜明TSQL和多数民众赞成我的TSQL走多远

Answer 1

您是否在寻找

SELECT [Column 16], COUNT(*) 
FROM [tab] 
GROUP BY [Column 16] 
--WITH ROLLUP (if you need to the total) 
ORDER BY COUNT(*) 

您可以附加DESC关键字查询的结束，即ORDER BY COUNT(*) DESC如果你想订购的结果集以升序排列（ASC）（升序为默认值，您可以省略）。

Answer 2

当你GROUP BY，你会得到一个清晰的[列16]。然后你可以使用一个聚合函数。

SELECT [Column 16], Sum ([Column To Sum]) Total_T1 
FROM [tab] 
Group By [Column 16] 

如果你想通过多列群体，他们应该出现在选择，并在GROUP BY

SELECT [Column 16], [Column 17], Sum ([Column To Sum]) Total_T1 
FROM [tab] 
Group By [Column 16], [Column 17] 

您也可以指定多个集合（这些没有出现在group by子句中

SELECT [Column 16], [Column 17], 
     Sum ([Column To Sum]) Total_T1, 
     Count ([Column To Sum]) NumOf_T1, 
     Avg ([Column To Sum]) Avg_T1, 
     Min ([Column To Sum]) Min_T1, 
     Max ([Column To Sum]) Max_T1 
FROM [tab] 
Group By [Column 16], [Column 17]