在表里创建函数get_count(),返回公司员工个数
时间: 2024-02-17 20:05:01 浏览: 105
好的,假设你的员工信息存在一个名为"employee"的表中,那么可以使用SQL语句实现get_count()函数,代码如下:
```sql
CREATE FUNCTION get_count()
RETURNS INT
BEGIN
DECLARE count INT;
SELECT COUNT(*) INTO count FROM employee;
RETURN count;
END
```
这个函数会查询"employee"表中的所有记录数量,并将结果存储到变量"count"中,最后将"count"作为函数返回值。你可以在SQL客户端中执行以上代码来创建该函数。
相关问题
SELECT pj.type AS type, count( pj.id ) AS project_count, count( evt.id ) AS alarm_count, ( SELECT count(id) from t_device where type=1) as camera_count, ( SELECT count(id) from t_device where type=2) as ai_box_count, ( SELECT count(id) from t_device where type=3) as radar_count, ( SELECT count(id) from t_device where type=4) as sensor_count, ( SELECT count(evt.id) FROM t_probably where project=pj.id AND status <> 1 and status is not null ORDER BY alarm_time DESC LIMIT 1 ) AS dissipate_event FROM t_project AS pj LEFT JOIN t_device AS dev ON pj.id = dev.project LEFT JOIN t_probably AS evt ON pj.id = evt.project LEFT JOIN t_organize AS org ON pj.organize = org.id ORDER BY pj.type DESC 在这段sql中,为什么project_count的数据和表里数据数量不一致
可能是因为在 JOIN 表时,有些项目(t_project)没有对应的设备(t_device)或报警(t_probably),这些项目在统计 project_count 时会被算作 0,导致最终的 project_count 数量和表中的数据数量不一致。另外,如果在 t_project 表中有重复的 id,也会导致最终的统计结果与表中的数据数量不一致。
SELECT pj.type AS type, count( pj.id ) AS project_count, count( evt.id ) AS alarm_count, ( SELECT count(id) from t_device where type=1) as camera_count, ( SELECT count(id) from t_device where type=2) as ai_box_count, ( SELECT count(id) from t_device where type=3) as radar_count, ( SELECT count(id) from t_device where type=4) as sensor_count, ( SELECT count(evt.id) FROM t_probably where project=pj.id AND status <> 1 and status is not null ORDER BY alarm_time DESC LIMIT 1 ) AS dissipate_event FROM t_project AS pj LEFT JOIN t_device AS dev ON pj.id = dev.project LEFT JOIN t_probably AS evt ON pj.id = evt.project LEFT JOIN t_organize AS org ON pj.organize = org.id ORDER BY pj.type DESC 在这段sql中,为什么project_count的数据比表里实际数据多几条
根据这段 SQL 语句,project_count 是通过对 t_project 表进行计数得出的,而 t_project 表可能存在重复的记录,因此 project_count 的数据可能比表里实际数据多几条。如果想要避免这种情况,可以在计数时使用 DISTINCT 关键字来排除重复的记录,例如:count( DISTINCT pj.id )。
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