按mysql中的范围分组

Question

Table:   
new_table                                                    
user_number  | diff                  
     2       |  0                      
     1       |  28  
     2       |  32  
     1       |  40  
     1       |  53  
     1       |  59  
     1       |  101  
     1       |  105  
     2       |  108  
     2       |  129  
     2       |  130    
     1       |  144  


            |(result)
            v

range  | number of users  
0-20   |  2  
21-41  |  3  
42-62  |  1  
63-83  |  2  
84-104 |  1  
105-135|  0  
136-156|  3


select t.range as [range], count(*) as [number of users]  
from (  
  select case    
    when diff between 0 and 20 then ' 0-20'  
    when diff between 21 and 41 then ' 21-41'  
    when diff between 42 and 62 then ' 42-62'  
    when diff between 63 and 83 then ' 63-83'  
    when diff between 84 and 104 then ' 84-104'  
    when diff between 105 and 135 then ' 105-135'  
    else '136-156'   
     end as range  
  from new_table) t  
group by t.diff  

Error:

You have an error in your SQL syntax, near '[range], count(*) as [number of users]  
from (  
  select case  
    when' at line 1  

Answer 1

以下是按范围分组的通用代码,因为执行case语句变得非常麻烦.

函数'floor'可用于查找范围的底部(不是波希米亚使用的'round'),并添加金额(下例中为19)以查找范围的顶部.切记不要重叠范围的底部和顶部!

mysql> create table new_table (user_number int, diff int);
Query OK, 0 rows affected (0.14 sec)

mysql>  insert into new_table values (2, 0), (1, 28), (2, 32), (1, 40), (1, 53),
        (1, 59), (1, 101), (1, 105), (2, 108), (2, 129), (2, 130), (1, 144);
Query OK, 12 rows affected (0.01 sec)
Records: 12  Duplicates: 0  Warnings: 0

mysql> select concat(21*floor(diff/21), '-', 21*floor(diff/21) + 20) as `range`,
       count(*) as `number of users` from new_table group by 1 order by diff;
+---------+-----------------+
| range   | number of users |
+---------+-----------------+
| 0-20    |               1 |
| 21-41   |               3 |
| 42-62   |               2 |
| 84-104  |               1 |
| 105-125 |               2 |
| 126-146 |               3 |
+---------+-----------------+
6 rows in set (0.01 sec)

Answer 2

这是一个适用于任何大小差异的解决方案:

select
  concat(21 * round(diff / 21), '-', 21 * round(diff / 21) + 20) as `range`,
  count(*) as `number of users`
from new_table
group by 1
order by diff;

这是一些可测试的代码及其输出:

create table new_table (user_number int, diff int);
insert into new_table values (2, 0), (1, 28), (2, 32), (1, 40), (1, 53), (1, 59), (1, 101), (1, 105), (2, 108), (2, 129), (2, 130), (1, 144); 
-- run query, output is: 
+---------+-----------------+
| range   | number of users |
+---------+-----------------+
| 0-20    |               1 |
| 21-41   |               1 |
| 42-62   |               2 |
| 63-83   |               2 |
| 105-125 |               3 |
| 126-146 |               2 |
| 147-167 |               1 |
+---------+-----------------+

Answer 3

Mysql作为关键字的分隔符使用反引号"`",而不是方括号(如sql server)

Answer 4

如果你有常规范围,更快的解决方案是在div函数的帮助下进行分组.

例如:

select diff div 20 as range, sum(user_number)
from new_table
group by diff div 20;

在这种情况下,范围表示为单个数字,您必须知道它们的含义:0 = 0-19,1 = 20-39,2 = 40-59,...

如果你需要不同的范围,可以使用不同的分隔符,或者从diff中减去一些数字.例如"(diff - 1)div 10"给出范围1-10,11-20,21-30,......