mysql 如何使用别名显示两个计数之间的差异?

c86crjj0  于 2023-02-28  发布在  Mysql
关注(0)|答案(4)|浏览(111)

我正在创建显示数据库中不同事物的总计的自定义视图,我还想显示差异。
例如:

SELECT
   (SELECT COUNT(*) FROM `documents`) AS `doc_count`,
   (SELECT COUNT(*) FROM `contacts`) AS `user_count`,
   (`doc_count` - `user_count`) AS `difference`;

用这种方法使用别名时出现错误。有没有办法在不重复select count(*)查询的情况下编写此查询?

cld4siwp

cld4siwp1#

您可以用另一个查询 Package 这两个查询:

SELECT doc_count, user_count, doc_count - user_count AS difference
FROM   ((SELECT COUNT(*) FROM `documents`) AS doc_count,
        (SELECT COUNT(*) FROM `contacts`) AS user_count) t
jdgnovmf

jdgnovmf2#

不,您不能在查询的同一级别使用别名,必须使用整个表达式或使用子选择

SELECT
   (SELECT COUNT(*) FROM `documents`) AS `doc_count`,
   (SELECT COUNT(*) FROM `contacts`) AS `user_count`,
   ((SELECT COUNT(*) FROM `documents`) - (SELECT COUNT(*) FROM `contacts`)) AS `difference`;
2exbekwf

2exbekwf3#

以下是获得您所需结果的“变通方法”:

SELECT C.doc_count
    ,C.user_count
    ,C.doc_count - C.user_count AS `difference`
FROM (SELECT
         (SELECT COUNT(*) FROM `documents`) AS `doc_count`
         ,(SELECT COUNT(*) FROM `contacts`) AS `user_count`) C

但是我不确定这种查询的性能...
希望这对你有帮助

bpsygsoo

bpsygsoo4#

我会将这些语句移到from子句中,并使用cross join

SELECT d.doc_count, u.user_count, (d.doc_count - u.user_count) as difference
FROM (SELECT COUNT(*) as doc_count FROM `documents`) d CROSS JOIN
     (SELECT COUNT(*) as user_count FROM `contacts`) u;

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