如何找到在如表中如何可以找到所有的<strong><em>不同的值，其在数据列</em></strong></p> <p>的第一个字母为如开头所示声明的MySQL

Question

CREATE TABLE table1 
     (`id` int not null auto_increment primary key, `data` char(3),`subject` varchar(250)) ; 

INSERT INTO table1 
    (`id`, `data`,`subject`) 
VALUES 
    (1, 'xa','aad'), 
    (2, 'ya','aac'), 
    (3, 'yc','aae'), 
    (4, 'zb','aaf'), 
    (5, 'yb','aag'), 
    (6, 'xc','aah'), 
    (7, 'za','aai'); 

鲜明通配符：

select * from table1 where data like '%x%' limit 1; 
select * from table1 where data like '%y%' limit 1; 
select * from table1 where data like '%z%' limit 1; 

成一个单一的查询和得到如下的输出如下所示

输出

1, 'xa', 'aad' 
2, 'ya','aac' 
3, 'zb','aaf' 

sql fiddle

Answer 1

组合DISTINCT和LEFT？但不要忘记，这种方法并不使用索引。

SELECT distinct(LEFT(`data`, 1)) first_letter, id, `data`, subject 
FROM table1 
GROUP BY first_letter 

http://sqlfiddle.com/#!9/08235/5/0

Answer 2

使用UNION：

(select * from table1 where data like '%x%' limit 1) 
UNION ALL 
(select * from table1 where data like '%y%' limit 1) 
UNION ALL 
(select * from table1 where data like '%z%' limit 1) ; 

updated fiddle

需要注意的是：你必须以限制工作对于每个查询附上每个查询在()，其他明智的你将只有一行而不是三行。

另请注意，'%z%'会为您提供那些包含z的值，不仅包含第一个字符的任何位置。您需要使用LEFT(data, 1) = 'z'代替：

(select * from table1 where LEFT(data, 1) = 'x' limit 1) 
UNION ALL 
(select * from table1 where LEFT(data, 1) = 'y' limit 1) 
UNION ALL 
(select * from table1 where LEFT(data, 1) = 'z' limit 1) 

updated fiddle 2

Answer 3

使其成为单一的查询

select 
    id, 
    data, 
    LEFT(group_concat(subject order by id),LOCATE(',',group_concat(subject order by id)) - 1), 
from table1 
group by SUBSTRING(data, 1, 1) 

Answer 4

您可以使用子查询来获得第一id每组

SELECT * 
FROM table1 
WHERE id IN (
select min(id) from table1 
where data like '%x%' 
OR data like '%y%' 
OR data like '%z%' 
GROUP BY left(data, 1) 
) 

Fiddle。