Question

我必须从表格中获取特定的输出格式。

假设我有一个包含2列名称和值的简单表格。

表 T1

+---------------+------------------+
| Name          | Value            |
+---------------+------------------+
| stuff1        | 1                |
| stuff1        | 1                |
| stuff2        | 2                |
| stuff3        | 1                |
| stuff2        | 4                |
| stuff2        | 2                |
| stuff3        | 4                | 
+---------------+------------------+

我知道值在1-4区间。我按名称和值对其进行分组，并计算与数字相同的行数，并获取下表：

表 T2

+---------------+------------------+--------+
| Name          | Value            | Number |
+---------------+------------------+--------+
| stuff1        | 1                | 2      |
| stuff2        | 2                | 2      |
| stuff3        | 1                | 1      |
| stuff3        | 4                | 1      |
+---------------+------------------+--------+

这是我需要你帮助的部分！如果我想获得这些格式，我该怎么办？

表 T3

+---------------+------------------+--------+
| Name          | Value            | Number |
+---------------+------------------+--------+
| stuff1        | 1                | 2      |
| stuff1        | 2                | 0      |
| stuff1        | 3                | 0      |
| stuff1        | 4                | 0      |
| stuff2        | 1                | 0      |
| stuff2        | 2                | 2      |
| stuff2        | 3                | 0      |
| stuff2        | 4                | 0      |
| stuff3        | 1                | 1      |
| stuff3        | 2                | 0      |
| stuff3        | 3                | 0      |
| stuff3        | 4                | 1      |
+---------------+------------------+--------+

感谢您的任何建议！

Answer 1

您从cross join开始生成所有可能的组合，然后左键加入现有查询的结果：

select n.name, v.value, coalesce(nv.cnt, 0) as "Number"
from (select distinct name from table t) n cross join
     (select distinct value from table t) v left outer join
     (select name, value, count(*) as cnt
      from table t
      group by name, value
     ) nv
     on nv.name = n.name and nv.value = v.value;

Answer 2

主题的变化。

Gordon Linoff和Owen现有答案之间的差异。

我更喜欢GROUP BY来获取名称而不是DISTINCT。在这种情况下，这可能具有更好的性能。（见Rob Farley's still relevant article。）
为了清楚起见，我将子查询分解为一系列CTE。

我使用表T2作为问题现在标记组结果集而不是显示为子查询。

WITH PossibleValue AS (
  SELECT 1 Value 
   UNION ALL
  SELECT Value + 1 
FROM PossibleValue 
   WHERE Value < 4 
), 
Name AS (
  SELECT Name 
    FROM T1
   GROUP BY Name
),
NameValue AS (
  SELECT Name 
        ,Value
    FROM Name
         CROSS JOIN
         PossibleValue
)
SELECT nv.Name
      ,nv.Value
      ,ISNULL(T2.Number,0) Number
  FROM NameValue nv
       LEFT JOIN
       T2 ON nv.Name = T2.Name
         AND nv.Value = T2.Value

Answer 3

另一种解决方案，这次使用CTE中的表值构造函数来构建名称值组合表。

WITH value AS 
( SELECT DISTINCT t.name, v.value
    FROM T1 AS t
   CROSS JOIN (VALUES (1),(2),(3),(4)) AS v (value)
)
  SELECT v.name AS 'Name', v.value AS 'Value', COUNT(t.name) AS 'Number'
    FROM value AS v
    LEFT JOIN T1 AS t ON t.value = v.value AND t.name = v.name
GROUP BY v.name, v.value, t.name;

我需要一个特定的输出

3 个答案: