MySQL -在列中包含特定字符串的行

Question

我希望得到4个结果，这些结果通常在4行中返回，然后在2行中返回。

我有一个疑问：

SELECT meta_key, meta_value, t.post_date_gmt FROM (SELECT * 
FROM dr_posts
JOIN dr_postmeta ON post_id = ID
WHERE post_author = 2
AND post_type ='donation') AS t WHERE meta_key = '_donation_amount' OR meta_key= '_anonymous_donation'

返回：

=======================================================
|       meta-key       |  meta-value  | post_date_gmt |
=======================================================
| _anonymous_donation  |       0      |   2015-03-01  |
|   _donation_amount   |       10     |   2015-03-01  |
| _anonymous_donation  |       0      |   2015-03-02  |
|   _donation_amount   |       15     |   2015-03-02  |
_______________________________________________________

我真正想要的(列名并不重要--只是结果)：

=============================================================
| _anonymous_donation  |  _donation_amount  | post_date_gmt |
=============================================================
|           0          |         10         |   2015-03-01  |
|           0          |         15         |   2015-03-02  |
_____________________________________________________________

怎么做？我读到我需要使用PIVOT，但我不确定这是否是我真正需要的。

Answer 1

我认为这个查询应该做您想做的事情：

select 
    max(case when meta_key =  '_anonymous_donation' then meta_value end) as '_anonymous_donation',
    max(case when meta_key =  '_donation_amount' then meta_value end) as '_donation_amount',
    post_date_gmt
from (SELECT * 
    FROM dr_posts
    JOIN dr_postmeta ON post_id = ID
    WHERE post_author = 2
    AND post_type ='donation'
    ) AS t
WHERE meta_key = '_donation_amount' OR meta_key= '_anonymous_donation'
group by post_date_gmt

Answer 2

我想这可能会有帮助

select min(meta-value) as  _anonymous_donation,
       max(meta-value) as _donation_amount,
       post_date_gmt  
from table
group by post_date_gmt