好的,首先我有不可变的值:
4 8 16 32 64 128 256
我有一张类似的 table :
+----+------+---------------------+-------------+
| id | full_name | club_name | y_of_birth |
+----+------+---------------------+-------------+
| 1 | Ahmed Sayed | El Ahly | 2000 |
+----+------+---------------------+-------------+
| 2 | Kareem Gaber | El Ahly | 2000 |
+----+------+---------------------+-------------+
| 3 | Maher Zein | El Ahly | 2003 |
+----+------+---------------------+-------------+
| 4 | Mohab Saeed | El Ahly | 2003 |
+----+------+---------------------+-------------+
| 5 | Kamal saber | wadi dgla | 2000 |
+----+------+---------------------+-------------+
| 6 | gamel kamel | el-nasr | 2002 |
+----+------+---------------------+-------------+
| 7 | omar galal | Cocorico | 2000 |
+----+------+---------------------+-------------+
| 8 | Kamal saber | Cocorico | 2004 |
+----+------+---------------------+-------------+
| 9 | Mohamed gad | Ismaily | 2000 |
+----+------+---------------------+-------------+
| 10 | ehab zeyad | Ismaily | 2005 |
+----+------+---------------------+-------------+
| 11 | moaz maged | Smouha | 2001 |
+----+------+---------------------+-------------+
| 12 | mazen mahmod | elmasry | 2006 |
+----+------+---------------------+-------------+
| 13 | ahmed shawky | Petroget | 2002 |
+----+------+---------------------+-------------+
| 14 | shaker ali | Petroget | 2007 |
+----+------+---------------------+-------------+
我尝试使用查询过滤数据库中的数据
select full_name,club_name from players where y_of_birth=2000
结果是5
玩家应该是这样的:
+--------------+--------------+
| full_name | club_name |
+--------------+--------------+
| Ahmed Sayed | El Ahly |
+----+------+--+--------------+
| Kareem Gaber | El Ahly |
+------+-------+--------------+
| Kamal saber | wadi dgla |
+------+-------+--------------+
| omar galal | Cocorico |
+------+-------+--------------+
| Mohamed gad | Ismaily |
+------+-------+--------------+
好的条件是:
如果结果大于 4 > 4
小于 8 <= 8
将结果放入 8
在我们的例子中,结果是 5
意思是 8 - 5 = 3
意思是把这个词迭代3次,结果应该是这样的:
+--------------+-------------+
| full_name | club_name |
+--------------+-------------+
| Ahmed Sayed | El Ahly |
+----+------+--+-------------+
| **ANY WORD** | |
+--------------+-------------+
| Kareem Gaber | El Ahly |
+------+-------+-------------+
| Kamal saber | wadi dgla |
+------+-------+-------------+
| **ANY WORD** | |
+--------------+-------------+
| omar galal | Cocorico |
+------+-------+-------------+
| Mohamed gad | Ismaily |
+------+-------+-------------+
| **ANY WORD** | |
+--------------+-------------+
注意:**ANY WORD**
之间请勿相邻到上面的例子:
+------+-------+
| **ANY WORD** |
+--------------+
| **ANY WORD** |
+--------------+
或
club_name
之间没有相邻到上面的例子:
+------+-------+
| El Ahly |
+--------------+
| El Ahly |
+--------------+
更新:
又是一个例子
它基于这些数字
4 8 16 32 64 128 256
而条件是:
- 如果查询结果 <=
4
和 >2
意思是(4 - the number of query result)
例子:如果查询结果是3
所以4 - 3 = 1
所以1
是 **ANY WORD**
的数字所以所需的输出将是这样的:
+--------------+-------------+
| full_name | club_name |
+--------------+-------------+
| Ahmed Sayed | El Ahly |
+----+------+--+-------------+
| **ANY WORD** | |
+--------------+-------------+
| Kareem Gaber | El Ahly |
+------+-------+-------------+
| Kamal saber | wadi dgla |
+--------------+-------------+
- 另一个示例与
8
if 查询结果 <=8
和 >4
意思是(8 - the number of query result)
例如:
查询结果是5
所以8 - 5 = 3
所以3
是 **ANY WORD**
的数字所以想要的输出会是这样的
+--------------+-------------+
| full_name | club_name |
+--------------+-------------+
| Ahmed Sayed | El Ahly |
+----+------+--+-------------+
| **ANY WORD** | |
+--------------+-------------+
| Kareem Gaber | El Ahly |
+------+-------+-------------+
| Kamal saber | wadi dgla |
+------+-------+-------------+
| **ANY WORD** | |
+--------------+-------------+
| omar galal | Cocorico |
+------+-------+-------------+
| Mohamed gad | Ismaily |
+------+-------+-------------+
| **ANY WORD** | |
+--------------+-------------+
4
以此类推和 16
和 32
和 64
..等直到256
.
任何帮助将不胜感激。
最佳答案
新的和改进的(第 3 版如何)使用变量并使用与 here 基本相同的技巧:
SELECT
IF(is_real, '**ANY WORD**', full_name) AS full_name,
IF(is_real, '', club_name) AS club_name
FROM
(
SELECT
full_name,
club_name,
(@row_num2:= @row_num2 + 1) AS row_num
FROM
(
SELECT p3.*
FROM
(
SELECT
p2.*,
(@row_num := @row_num + 1) AS row_num
FROM
(
SELECT *
FROM players AS p1
WHERE y_of_birth = 2000
) AS p2
CROSS JOIN
(
SELECT
@row_num := 0,
@count := (SELECT COUNT(*) FROM players WHERE y_of_birth = 2000)
) AS vars
ORDER BY club_name
) AS p3
ORDER BY row_num % FLOOR(@row_num / 2), row_num
) AS p4
CROSS JOIN
(
SELECT
@row_num2 := -1,
@extra := GREATEST(2, POW(2, CEIL(LOG2(@count)))) - @count) AS vars
) AS data
LEFT JOIN
(
(SELECT 1 AS is_real)
UNION ALL
(SELECT 0 AS is_real)
) AS filler
ON
MOD(row_num, FLOOR(@count / @extra)) = 0 AND
row_num / FLOOR(@count / @extra) < @extra
ORDER BY row_num, is_real
对于您提供的示例数据,这会产生如下内容:
+--------------+-----------+
| full_name | club_name |
+--------------+-----------+
| Ahmed Sayed | El Ahly |
| **ANY WORD** | |
| Mohamed gad | Ismaily |
| **ANY WORD** | |
| omar galal | Cocorico |
| **ANY WORD** | |
| Kareem Gaber | El Ahly |
| Kamal saber | wadi dgla |
+--------------+-----------+
这应该适用于任何大小的结果;只需将条件 (y_of_birth = 2000
) 更改为您想要的任何条件。我升级到 MySQL 5.6 来测试这个(实际上结果有一点点不同)。
基本技巧是使用 UNION
创建具有静态值的两行表(在本例中为 1
和 0
)然后 LEFT JOIN
将其加入实际结果的次数以填充到 2 的幂。这意味着我们已经计算了结果中每一行的数量(称为 row_num
) 以便我们可以正确地制定连接条件。最后,这会每隔这么多行产生一个重复行;最后一点是通过检查我们是在真实的还是假的(1
或 0
) 行。
这应该可以防止来自同一支球队的球员彼此相邻,除非因为一支球队的球员太多而无法做到这一点;有关如何执行此操作的更多信息,请参阅上面的链接。基本思路是按俱乐部排序,然后从该列表的前半部分和后半部分交替挑选。
最后一个技巧是找出要加入虚拟行的数量和位置。在尝试了几件事之后,我意识到这实际上非常简单:只需加入每一行,直到我们达到所需的虚拟行数(@extra
)。但是,这会将所有虚拟行打包在结果的顶部;要将它们分散得更多(不是完全分散,而是分散得更多),计算我们需要添加一个的频率(FLOOR(@count/@extra)
),然后每隔那么多行放置一个(ON
条件的第一部分)直到添加足够的内容(第二部分)。
关于mysql - SQL - 在具有许多条件的每列之间打印许多单词,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31259849/