角色出现在字符串的第一个位置和第二个位置的频率是多少？

Question

我试图回答这样一个问题：在使用MySQL的SQL查询

How often does a character occur in the first position versus the second position of a string?

。

但是我得到一个语法错误。

代码：

SELECT 
    onechar, 
    ASCII(onechar) as asciival, 
    COUNT(*) as cnt, 
    SUM(CASE WHEN pos = 1 THEN 1 ELSE 0 END) as pos_1, 
    SUM(CASE WHEN pos = 2 THEN 1 ELSE 0 END) as pos_2 
FROM (
    (SELECT 
     SUBSTRING(`city`, 1, 1) as onechar, 
     1 as pos 
    FROM `orders` 
    WHERE LEN(`city` >= 1) 

    UNION ALL 

    (SELECT 
     SUBSTRING(`city`, 2, 1) as onechar, 
     2 as pos 
    FROM `orders` 
    WHERE LEN(`city` >= 2) 
    ) 
GROUP BY onechar 
ORDER BY onechar 

错误：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'GROUP BY onechar ORDER BY onechar LIMIT 0, 30' at line 1

试了几种方法都没有成功。

任何人都可以告诉我这个问题？

Answer 1

圆括号看起来不正确，查询缺少派生表的别名。另外，因为mysql会将布尔值计算为1或0，所以可以简化您的sum语句。试试这个：

SELECT onechar, ASCII(onechar) as asciival, COUNT(*) as cnt, 
SUM(pos = 1) as pos_1, 
SUM(pos = 2) as pos_2 
FROM (
    SELECT SUBSTRING(`city`, 1, 1) as onechar, 1 as pos 
    FROM `orders` WHERE LENGTH(`city`) >= 1 
    UNION ALL 
    SELECT SUBSTRING(`city`, 2, 1) as onechar, 2 as pos 
    FROM `orders` WHERE LENGTH(`city`) >= 2 
) t 
GROUP BY onechar 
ORDER BY onechar 

Answer 2

您错过了一些右括号。

FROM (
    (SELECT 
     SUBSTRING(`city`, 1, 1) as onechar, 
     1 as pos 
    FROM `orders` 
    WHERE LEN(`city` >= 1) 
    ) 
    UNION ALL 

    (SELECT 
     SUBSTRING(`city`, 2, 1) as onechar, 
     2 as pos 
    FROM `orders` 
    WHERE LEN(`city` >= 2) 
    ) 
)