Question

我想回答这个问题：

角色出现在第一个位置的频率与字符串的第二个位置？

在Mysql上使用SQL查询。

但是我收到语法错误。

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：

您的SQL语法有错误;检查手册对应于您的MySQL服务器版本，以便使用正确的语法靠近＆＃39; GROUP BY onechar ORDER BY onechar LIMIT 0,30＆＃39;在第1行

尝试了几种方法但没有成功。

任何人都可以解释一下这个问题吗？

Answer 1

括号外观不正确，查询缺少派生表的别名。此外，因为mysql将布尔值评估为1或0，所以您可以简化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

你错过了一些结束括号。

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)
    ) 
)

字符在字符串的第一个位置与第二个位置之间出现的频率是多少？

2 个答案: