Question

我有一张桌子：

   date       Name
2017-02-21    Mark
2017-02-21    Peter
2017-02-21    John
2017-02-22    Mark
2017-02-22    Mark
2017-02-22    Mark
2017-02-22    Peter

我想要以下输出：

    date    Mark  John  Peter
2017-02-21   1      1    1
2017-02-22   3      0    1

我如何在MySQL中实现这一目标？

Answer 1

您可以使用聚合来透视数据，如下所示：

select date,
    sum(Name = 'Mark') Mark,
    sum(Name = 'John') John,
    sum(Name = 'Peter') Peter
from your_table
group by date;

它使用MySQL评估true为1而false为0的事实。

SQLFiddle

Answer 2

使用'FIELD（）'

select 
sum(field(name,'mark')) as mark,
sum(field(name,'peter')) as peter,
sum(field(name,'john')) as john 
from table_name group by date;

Answer 3

这不是您想要的，但您可以使用GROUP BY对结果进行分组，然后计算名称，如下所示：

SELECT *,
    COUNT(name) as how_many
FROM your_table
GROUP BY date, name

SQL Fiddle

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3 个答案:

SQLFiddle