Question

我有5个包含以下属性的表，

+--------------------+-------------+------+-----+---------+-------+
| Field              | Type        | Null | Key | Default | Extra |
+--------------------+-------------+------+-----+---------+-------+
| actor_id           | int(11)     | YES  | MUL | NULL    |       |
| activity_object_id | int(11)     | YES  | MUL | NULL    |       |
| interest_level     | tinyint(4)  | YES  |     | 10      |       |
| feed_view          | smallint(6) | YES  |     | 0       |       |
| quick_view         | smallint(6) | YES  |     | 0       |       |
| page_view          | smallint(6) | YES  |     | 0       |       |
| fullscreen_view    | smallint(6) | YES  |     | 0       |       |
| reserved1          | int(11)     | YES  |     | NULL    |       |
| reserved2          | int(11)     | YES  |     | NULL    |       |
| reserved3          | int(11)     | YES  |     | NULL    |       |
| created_at         | datetime    | YES  |     | NULL    |       |
| updated_at         | datetime    | YES  |     | NULL    |       |
+--------------------+-------------+------+-----+---------+-------+

我们如何创建新的临时表，它是所有5个表值的总和。 activity_object_id 是唯一的，一个表可能包含activity_object_id，而另一个表可能不包含。

table1有一个active_object_id说'gowthamkey'，table2有相同的键'gowthamkey'，table3可能没有'gowthamkey'。所以我想将所有表值总结到新表中，以便它有一个键'gowthamkey'，其中值是以下的总和：

feed_view，quick_view，page_view，fullscreen_view，reserved1，reserved2，reserved3 ，除了actor_id，interest_level，created_at，updated_at。

根据@bummer回答，这是我的查询：

CREATE TABLE actor_activity_object_stats_temp_7_days_12 AS 
select actor_id, activity_object_id, interest_level, SUM(feed_view) AS feed_view, SUM(quick_view) AS quick_view, SUM(fullscreen_view) as fullscreen_view 
from (
    select * from actor_activity_object_stats_temp_2016_04_29 
    union all, 
    select * from actor_activity_object_stats_temp_2016_04_30 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_01 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_02 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_03 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_04 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_05 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_06 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_07 
    union all, 
    select * from actor_activity_object_stats_temp_2016_05_08 ) AS X
group by activity_object_id

Answer 1

从$this->db->select('*'); $this->db->from('yourtablename'); $this->db->order_by("column_name", "ASC"); // or lower case asc // $this->db->limit(1); limits the number of results. $query = $this->db->get(); return $query->result();开始，然后使用UNION添加同一SUM()所有表格中的所有值。

activity_object_id

Answer 2

我相信这可以解决你的问题。

create table new_table
as
select * from table1
union
select * from table2
union
select * from table3
union
select * from table4
union
select * from table5)

这里有sql fiddle来说明。

将所有5个表值相加以形成新表？

2 个答案: