Question

我的数据库中有这个表：

╔════╦═══════╦═══════╗
║ id ║ Name  ║ Price ║
╠════╬═══════╬═══════╣
║  1 ║ ciao  ║   123 ║
║  2 ║ ciao  ║    55 ║
║  3 ║ bye   ║    43 ║
║  4 ║ hello ║    12 ║
║  5 ║ ciao  ║     1 ║
║  6 ║ ciao  ║    77 ║
╚════╩═══════╩═══════╝

..我想创建一个显示两列的新视图：

1）＆＃34; ciao＆＃34;或者＆＃34;不是ciao＆＃34;

2）上一表中有多少行是＆＃34; ciao＆＃34;或者＆＃34;不是ciao＆＃34;，像这样：

╔════╦══════════╦═══════╗
║ id ║   Name   ║ Count ║
╠════╬══════════╬═══════╣
║  1 ║ ciao     ║     4 ║
║  2 ║ not ciao ║     2 ║
╚════╩══════════╩═══════╝

我试图找到一个解决方案，但我无法按照＆＃34;而不是ciao＆＃34;值：

SELECT Name, COUNT(*) 
FROM mytable
WHERE Name = "Ciao"
GROUP BY Name

Answer 1

您需要将'ciao'以外的所有内容映射到'not ciao'，如下所示：

SELECT CASE WHEN Name!='ciao' THEN 'not ciao' ELSE Name END as Name, COUNT(*) 
FROM mytable
GROUP BY CASE WHEN Name!='ciao' THEN 'not ciao' ELSE Name END

Answer 2

这可能有用（未经测试）：

SELECT CASE
         WHEN name = 'ciao' THEN 'ciao'
         ELSE 'not ciao'
       END as NAME,
       COUNT(*)
FROM   table
GROUP  BY CASE
            WHEN name = 'ciao' THEN 'ciao'
            ELSE 'not ciao'
          END

MySQL中的OR

SELECT CASE
         WHEN name = 'ciao' THEN 'ciao'
         ELSE 'not ciao'
       END as NAME1,
       COUNT(*)
FROM   table
GROUP  BY NAME1

Answer 3

最简单的方法：

SELECT Name, COUNT(*) 
FROM
(select case when name <> "Ciao" then "Not Ciao"
                  else "Ciao" end as name
 from mytable)
GROUP BY Name

使用GROUP BY语句的SQL查询

3 个答案: