如何在SELECT语句中重命名和引用COUNT(*)?

问题描述 投票:-1回答:3

我正在尝试在SELECT语句中使用COUNT(*)。但是,我需要重新命名它,并能够为WHERE子句引用它。

我尝试过使用AS,我尝试省略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

尝试将新标识符包装在引号中,但这也不起作用。

这是有效的,但是为计数提供了一个oracle生成的名称,这是不理想的,我不知道如何引用每行的计数:

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

这是我尝试但不起作用的:

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

它不会使列名称为“numLiasons”,并且末尾的where子句不知道numLiasons是什么,因此失败。

sql oracle nested-queries
3个回答
1
投票

您可以通过在其后面添加一个名称来为列添加别名,也可以在两者之间使用关键字AS。它与您对表格的处理基本相同。

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;

但是你不能在WHERE子句中使用别名(在WHERE子句中的条件选择了记录之后发生了别名)。你必须重复这一尝试。

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;

1
投票

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指定为别名

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,您只需在(子选择)列上分配别名即可

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

无论如何要过滤你可以使用的聚合结果但是对于count(*)记住这只适用于非空行,所以通常count(*)是> 0


0
投票

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;