如何在 SELECT 语句中重命名和引用 COUNT(*)?

How to re-name and reference COUNT(*) in a SELECT statement?

我正在尝试在 SELECT 语句中使用 COUNT(*)。但是,我需要重命名它并能够在 WHERE 子句中引用它。

我试过使用 AS,也试过省略 AS,因为根据 oracle 页面似乎没有必要:https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions032.htm

尝试将新标识符括在引号中,但这也不起作用。

这可行,但是为计数提供一个 oracle 生成的名称,这并不理想,而且我不知道如何引用每一行的计数:

SELECT
    school_name,
    (SELECT COUNT(*)
    FROM liason_to
    WHERE school_name = s.school_name)
FROM school s;

这是我尝试过但行不通的方法:

SELECT
    school_name,
    (SELECT COUNT(*) AS numLiasons
    FROM liason_to
    WHERE school_name = s.school_name)
FROM school s
WHERE numLiasons > 0;

它没有生成列名 "numLiasons",最后的 where 子句不知道 numLiasons 是什么,所以失败了。

您可以使用连接和分组来避免子查询..并且您可以将您喜欢的 anme 分配为别名

SELECT s.school_name, COUNT(*) as my_count 
FROM school s 
INNER JOIN liason_to l on s.school_name = l.school_name 
GROUP BY s.school_name 

使用您的代码,您可以简单地在 (subselected ) 列分配一个别名

SELECT
    school_name,
    (SELECT COUNT(*)
    FROM liason_to
    WHERE school_name = s.school_name) as my_name 
FROM school s;

无论如何要过滤一个聚合结果你可以使用但是 count(*) 请记住,这仅适用于非空行,因此通常 count(*) > 0

您不能在 WHERE 子句中引用别名,但您可以这样做:

SELECT
    t.school_name,
    t.numLiasons
FROM (
  SELECT  
    s.school_name,
    (
      SELECT COUNT(*) 
      FROM liason_to
      WHERE school_name = s.school_name
    ) AS numLiasons
  FROM school s
) t
WHERE t.numLiasons > 0;

您可以通过简单地在列后面加上一个名称来为列添加别名,中间可以选择使用关键字 AS。它与您已经对表格所做的基本相同。

SELECT school_name,
       (SELECT count(*)
               FROM liason_to l
               WHERE l.school_name = s.school_name) AS numliasons
       FROM school s;

或者干脆

SELECT school_name,
       (SELECT count(*)
               FROM liason_to l
               WHERE l.school_name = s.school_name) numliasons
       FROM school s;

但是您不能在 WHERE 子句中使用别名(别名是在 WHERE 子句中的条件选择了记录之后发生的)。你必须重复这个表达式。

SELECT school_name,
       (SELECT count(*)
               FROM liason_to l
               WHERE l.school_name = s.school_name) numliasons
       FROM school s
       WHERE (SELECT count(*)
                     FROM liason_to l
                     WHERE l.school_name = s.school_name) > 0;