Question

我有三种不同类型的事务表，它们保存3种不同类型的事务的数据。我必须显示每个数据，但问题是我必须混合这些行并根据日期排序。例如，我有transac1，transac2，transac1和用户表。 transac1就像

id txn_id user_id field_a date
1   223    1       23   12/12/12
2   r23    1       33   12/12/12

transac2就像

id txn_id user_id field_b date
1   nne    1       53     12/12/12
2   wr3    1       93     11/12/12

transac3就像

id txn_id user_id field_c  date
1   g4t3    1       73     12/12/12
2   d3ts    1       83     12/12/12

当我从这些数据中获取数据时，数据应该像

id row_id txn_id user_id field_a field_b field_c table_name date
1  2   wr3        1       0       93      0      transac_2   11/12/12
2  1   223        1       23      0       0      transac_1   12/12/12
3  2   r23        1       33      0       0      transac_1   12/12/12
4  1   nne        1       0       53      0      transac_2   12/12/12
5  1   g4t3       1       0       0       73     transac_3   12/12/12
6  2   dt3s       1       0        0      83     transac_3   12/12/12

数据应根据此日期进行排序，表格将分别与其他表格连接。任何身体知道我该怎么做。顺便说一句，我正在使用Codeigniter框架，如果它可能与CI的查询构建器，它将不胜感激。感谢

Answer 1

试试这个解决方案：

此解决方案适用于 WITH ROW NUMBER

SELECT  @rownum := @rownum + 1 rownum , t.*
 FROM ( 
                 select id,txn_id,user_id,field_a, 0 as field_b, 0 as field_c,
                     'transac_1' as table_name , date from transac_1
               UNION ALL 
                 select id,txn_id,user_id,0 as field_a, field_b, 0 as field_c,
                     'transac_2' as table_name , date from transac_2
               UNION ALL
                 select id,txn_id,user_id,0 as field_a, 0 as field_b, field_c,
                     'transac_3' as table_name , date from transac_3 
    )t , (SELECT @rownum := 0) r Order by t.date desc

此解决方案适用于 WITHOUT ROW NUMBER

 SELECT  t.*
     FROM ( 
                     select id,txn_id,user_id,field_a, 0 as field_b, 0 as field_c,
                         'transac_1' as table_name , date from transac_1
                   UNION ALL 
                     select id,txn_id,user_id,0 as field_a, field_b, 0 as field_c,
                         'transac_2' as table_name , date from transac_2
                   UNION ALL
                     select id,txn_id,user_id,0 as field_a, 0 as field_b, field_c,
                         'transac_3' as table_name , date from transac_3 
        )t Order by t.date desc

我希望这个解决方案对您有所帮助。如果您需要其他任何内容，请告诉我，否则此解决方案对您无效。

Answer 2

创建一个变量以生成新序列：

确保使用union all而非union（因为工会将丢弃一式两份）

set @var:=;
select @var:=@var+1, * from ( 
select id as row_id, ..... from table_1
 union all 
 select id as row_id, ..... from table_2
) as alais

Answer 3

试试这个

Rextester Sample

select @rownum:=@rownum+1 as id, t.* from
(
select id as row_id,txn_id,user_id,field_a,0 as field_b,0 as field_c,
'transac_1' as table_name,date
from transac1
union all
select id as row_id,txn_id,user_id,0 as field_a,field_b,0 as field_c,
'transac_2' as table_name,date
from transac2
union all
select id as row_id,txn_id,user_id,0 as field_a,0 as field_b,field_c,
'transac_3' as table_name,date
from transac3
) t , (SELECT @rownum:=0) r
order by t.date;

输出：

+----+--------+--------+---------+---------+---------+---------+------------+---------------------+
| id | row_id | txn_id | user_id | field_a | field_b | field_c | table_name |        date         |
+----+--------+--------+---------+---------+---------+---------+------------+---------------------+
|  1 |      2 | wr3    |       1 |       0 |      93 |       0 | transac_2  | 12.11.2012 00:00:00 |
|  2 |      2 | d3ts   |       1 |       0 |       0 |      83 | transac_3  | 12.12.2012 00:00:00 |
|  3 |      1 | nne    |       1 |       0 |      53 |       0 | transac_2  | 12.12.2012 00:00:00 |
|  4 |      1 | 223    |       1 |      23 |       0 |       0 | transac_1  | 12.12.2012 00:00:00 |
|  5 |      1 | g4t3   |       1 |       0 |       0 |      73 | transac_3  | 12.12.2012 00:00:00 |
|  6 |      2 | r23    |       1 |      33 |       0 |       0 | transac_1  | 12.12.2012 00:00:00 |
+----+--------+--------+---------+---------+---------+---------+------------+---------------------+

从两个以上的表中选择，每个表都没有关系

3 个答案: