Php 是否可以创建一个包含3个表和条件语句的触发器？

我的数据库中有3个表，它们都是相互依赖的

有一个项目表

+-----------+---------------------+-------------------------------------------------+
| id        | project name        |status (1 confirmed / 2 in progress / 3 done )   |
+-----------+---------------------+-------------------------------------------------+
|   233     | project 1           |1                                                |
+-----------+---------------------+-------------------------------------------------+
|  234      | project 2           | 3                                               |
+-----------+---------------------+-------------------------------------------------+

表2是这一类别

+-----------+---------------------+-------------------+-------------------+-------------------+ 
| id        | category            | confirmed_counter | inprogress_counter|done_counter       |
+-----------+---------------------+-------------------+-------------------+-------------------+
|         1 | c++                 |      2            |3                  | 1                 |
+-----------+---------------------+-------------------+-------------------+-------------------+
|         2 | php                 |      3            |4                  | 6                 |
+-----------+---------------------+-------------------+-------------------+-------------------+

表3是分类对象

+-----------+---------------------+------------+ 
| id        | project_id          | category_id|
+-----------+---------------------+------------+
|         1 | 233                 |      2     |
+-----------+---------------------+------------+
|         2 | 244                 |      3     |
+-----------+---------------------+------------+

我想创建一个触发器，在项目状态更改时更新类别的计数器

当管理员确认一个项目时，它的状态将在类别表中更改为1

已确认\u计数器应递增
但首先我必须查看类别对象，看看该项目的类别是什么，然后更新这些类别的计数器字段

$c = $project->get_categorys();
///// it returns array of that project category ids ( 2 ,3 ,4)
$sql = update `categorys` set `confirmed_counter` = confirmed_counter+1 where `id` in (2,3,4);

当一个项目正在进行时

$sql = update `categorys` set `confirmed_counter` = confirmed_counter-1 , `inprogress_counter` = inprogress_counter+1 where `id` in (2,3,4);

当项目完成时

$sql = update `categorys` set `inprogress_counter` = inprogress_counter-1 , `done_counter` = done_counter+1 where `id` in (2,3,4);

而且也不总是这样
项目在进行中时可能会被重置

因此，在更新计数器之前，有两个重要步骤

首先，我必须从第三个表中获取项目的类别id

然后，我必须根据项目的当前和以前的状态增加和减少计数器

是否可以创建触发器来执行此操作

//////////////////////////////////////////////////编辑///////////////////////////////////

有没有可能在代码中使用OLD来指定它，而不只是假设它是透水状态？
喜欢改变

  SET
    confirmed_counter = 
      CASE NEW.status
        WHEN 1 THEN confirmed_counter + 1
        WHEN 2 THEN confirmed_counter - 1
        ELSE confirmed_counter
      END,

到

我试着这样使用它，但最后一条语句似乎总是被执行的
是的，这是可能的-

CREATE TRIGGER trigger1
  AFTER UPDATE
  ON project
  FOR EACH ROW
BEGIN
  UPDATE category c
    JOIN category_objects co
      ON co.category_id = c.id
  SET
    confirmed_counter = 
      CASE NEW.status
        WHEN 1 THEN confirmed_counter + 1
        WHEN 2 THEN confirmed_counter - 1
        ELSE confirmed_counter
      END,
    inprogress_counter =
      CASE NEW.status
        WHEN 1 THEN inprogress_counter
        WHEN 2 THEN inprogress_counter + 1
        ELSE inprogress_counter - 1
      END,
    done_counter =
      CASE NEW.status
        WHEN 1 THEN done_counter
        WHEN 2 THEN done_counter
        ELSE done_counter + 1
      END
  WHERE
    co.project_id = NEW.id;
END


关于第二个问题的例子：

...
confirmed_counter =
  CASE
    WHEN NEW.status = 1 THEN confirmed_counter + 1
    WHEN OLD.status = 1 THEN confirmed_counter - 1
    ELSE confirmed_counter
END
....

您可以使用CASE statant的第二个版本-CASE WHEN[condition]然后result[WHEN[condition]然后result…][ELSE result]END。更多信息-
...
confirmed_counter =
  CASE
    WHEN NEW.status = 1 THEN confirmed_counter + 1
    WHEN OLD.status = 1 THEN confirmed_counter - 1
    ELSE confirmed_counter
END
....