如何做嵌套SQL select count

问题描述

我查询的系统不允许使用 DISTINCT ，因此我的替代方法是执行 GROUP BY 以接近结果

我想要的查询的意图如下：

  SELECT 
 SUM（column1）AS column1，
 SUM（column2）AS column2，
 COUNT（DISTINCT（column3））AS column3 
 FROM table 
  pre> 
 
 
，我会认为我需要一些类型的嵌套查询，沿着这一行
  SELECT 
 SUM（column1）AS column1，
 SUM（column2）AS column2，
 COUNT（SELECT column FROM table GROUP BY column）AS column3 
 FROM table 
  
但它没有工作。 
解决方案
您使用的语法错误 COUNT（DISTINCT）。  DISTINCT 部分是一个关键字，而不是一个函数。根据，这应该可以工作：
  SELECT 
 SUM（column1）AS column1，
 SUM（column2）AS column2，
 COUNT（DISTINCT column3）AS column3 
 FROM table 
  
但是，请阅读文档。 BigQuery的 COUNT（DISTINCT）的实现有点不寻常，显然是为了更好地扩展大数据。如果您尝试计数大量不同的值，那么您可能需要指定第二个参数（和，您有一个固有的缩放问题）。
 
 
 
 更新：
 
 
 
如果您有大量不同的 column3 要计数，并且您想要确切的计数，则您可以执行连接，而不是在选择列表（BigQuery似乎不允许）中放置子查询：
  SELECT * 
 FROM（
 SELECT 
 SUM（column1）AS column1，
 SUM（column2）AS column2 
 FROM table 
）
 CROSS JOIN（
 SELECT count（*）AS column3 
 FROM（
 SELECT column3 
 FROM table 
 GROUP BY column3 
）
）
  
 更新2 ：
 
 
 
不是加入两个单行表将是非常昂贵的，但@FelipeHoffa让我更多地思考这个，我意识到我错过了一个更简单的解决方案：
  SELECT 
 SUM（column1）AS column1，
 SUM（column2）AS column2，
 COUNT AS column3 
 FROM（
 SELECT 
 SUM（column1）AS column1，
 SUM（column2）AS column2 
 FROM table 
 GROUP BY column3 
）
  
这个计算 column1的小计和 column2 值，按 column3 分组，然后计算和总计所有小计行。感觉很正确。
 
i'm querying a system that won't allow using DISTINCT, so my alternative is to do a GROUP BY to get near to a result
my desired query was meant to look like this,
SELECT
SUM(column1) AS column1,
SUM(column2) AS column2,
COUNT(DISTINCT(column3)) AS column3
FROM table
for the alternative, i would think i'd need some type of nested query along the lines of this,
SELECT
SUM(column1) AS column1,
SUM(column2) AS column2,
COUNT(SELECT column FROM table GROUP BY column) AS column3
FROM table
but it didn't work. Am i close?
 解决方案 
You are using the wrong syntax for COUNT(DISTINCT).  The DISTINCT part is a keyword, not a function.  Based on the docs, this ought to work:
SELECT
  SUM(column1) AS column1,
  SUM(column2) AS column2,
  COUNT(DISTINCT column3) AS column3
FROM table
Do, however, read the docs.  BigQuery's implementation of COUNT(DISTINCT) is a bit unusual, apparently so as to scale better for big data.  If you are trying to count a large number of distinct values then you may need to specify a second parameter (and you have an inherent scaling problem).
Update:
If you have a large number of distinct column3 values to count, and you want an exact count, then perhaps you can perform a join instead of putting a subquery in the select list (which BigQuery seems not to permit):
SELECT *
FROM (
    SELECT
      SUM(column1) AS column1,
      SUM(column2) AS column2
    FROM table
  )
  CROSS JOIN (
      SELECT count(*) AS column3
      FROM (
          SELECT column3
          FROM table
          GROUP BY column3
        )
    )
Update 2:
Not that joining two one-row tables would be at all expensive, but @FelipeHoffa got me thinking more about this, and I realized I had missed a simpler solution:
SELECT
  SUM(column1) AS column1,
  SUM(column2) AS column2,
  COUNT(*) AS column3
FROM (
    SELECT
      SUM(column1) AS column1,
      SUM(column2) AS column2
    FROM table
    GROUP BY column3
  )
This one computes a subtotal of column1 and column2 values, grouping by column3, then counts and totals all the subtotal rows.  It feels right.
                        
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