Question

我可以从一个表中的列执行以下SQL Server选择的不同（或非重复名称），如下所示：

SELECT COUNT(DISTINCT [Name]) FROM [MyTable]

但是，如果我有多个表（所有这些表都包含名为[Name]的名称字段），我需要知道两个或多个表中非重复名称的数量。

如果我运行这样的事情：

SELECT COUNT(DISTINCT [Name]) FROM [MyTable1], [MyTable2], [MyTable3]

我收到错误，“不明确的列名称'名称'”。

PS。所有三个表[MyTable1]，[MyTable2]，[MyTable3]都是之前选择的产物。

Answer 1

澄清后，使用：

  SELECT x.name, COUNT(x.[name])
    FROM (SELECT [name]
            FROM [MyTable]
          UNION ALL
          SELECT [name]
            FROM [MyTable2]
          UNION ALL
          SELECT [name]
            FROM [MyTable3]) x
GROUP BY x.name

如果我理解正确，请使用：

  SELECT x.name, COUNT(DISTINCT x.[name])
    FROM (SELECT [name]
            FROM [MyTable]
          UNION ALL
          SELECT [name]
            FROM [MyTable2]
          UNION ALL
          SELECT [name]
            FROM [MyTable3]) x
GROUP BY x.name

UNION将删除重复项; UNION ALL不会，而且速度更快。

Answer 2

编辑：看到最近的评论后不得不改变。

这会给你你想要的吗？这会在合并所有表格中的行后为每个人提供一个计数。

SELECT [NAME], COUNT(*) as TheCount
FROM
    (
     SELECT [Name] FROM [MyTable1]
     UNION ALL
     SELECT [Name] FROM [MyTable2]
     UNION ALL
     SELECT [Name] FROM [MyTable3]
     ) AS [TheNames]
GROUP BY [NAME]

Answer 3

这是另一种方式：

SELECT x.name, SUM(x.cnt)
FROM ( SELECT [name], COUNT(*) AS cnt
       FROM [MyTable]
       GROUP BY [name]
     UNION ALL
       SELECT [name], COUNT(*) AS cnt
       FROM [MyTable2]
       GROUP BY [name]
     UNION ALL
       SELECT [name], COUNT(*) AS cnt
       FROM [MyTable3]
       GROUP BY [name]
     ) AS x
GROUP BY x.name

Answer 4

如果每个表的列数不同，例如：

表1有3列，
table2有2列，
表3有1列

您想计算不同列名的不同值的数量，在 AthenaSQL 中，对我有用的是使用computed: { brandImg() { let brandImg = ''; let thisBrand = this.$store.state.ui.theme.brand; switch (thisBrand) { case 'brandOne': return '/img/iconOne.png'; case 'brandTwo': return '/img/iconTwo.png'; case 'brandThree': return '/img/iconThree.png'; default: console.warn('not configured`); break; } } }，因为您的输出将只有一行，这将是1种组合：

CROSS JOIN

将返回具有一行及其计数的表：

SELECT * FROM (
SELECT COUNT(DISTINCT name1) as amt_name1,
       COUNT(DISTINCT name2) as amt_name2,
       COUNT(DISTINCT name3) as amt_name3,
FROM table1 ) t1
CROSS JOIN
(SELECT COUNT(DISTINCT name4) as amt_name4,
        COUNT(DISTINCT name5) as amt_name5,
        MAX(t3.amt_name6) as amt_name6
 FROM table2
 CROSS JOIN
 (SELECT COUNT(DISTINCT name6) as amt_name6
  FROM table3) t3) t2

从几个表中选择COUNT（DISTINCT [name]）

4 个答案: