Question

我需要使用PHP和MySQL按照列值过滤表中的值。我正在解释下面的表格和代码。

db_images：

image_id     member_id      subcat_id   from_day   to_day

  1            220           56           1         3

  2           220            56           1          3


  3           220            56           1         1

  4          120             22            1         5

  5          120             22            2         4

我在下面解释我的问题。

$qry=mysqli_query($connect,"select * from db_iamges group by member_id,subcat_id order by image_id desc");

使用我的查询我得到一条如下记录。

image_id     member_id      subcat_id   from_day   to_day

    3           220             56         1          1

在这里，我需要to_day应该始终具有更高的价值。如果存在相同的member_id and subcat_id，则to_day将始终为更高的值，而from_day将始终为更小的值。预期的输出应如下所示。

image_id     member_id      subcat_id   from_day   to_day

   1            220             56          1        3

   4            120             22          1        5

Answer 1

看起来你有语法问题，因为如果你复制粘贴，就把＆＃34; db_iamges＆＃34 ;.我做了一张桌子：

mysql> select * from prueba1;
+----------+-----------+-----------+----------+--------+
| image_id | member_id | subcat_id | from_day | to_day |
+----------+-----------+-----------+----------+--------+
|        1 |       220 |        56 |        1 |      3 |
|        2 |       220 |        56 |        1 |      3 |
|        3 |       220 |        56 |        1 |      1 |
|        4 |       120 |        22 |        1 |      5 |
|        5 |       120 |        22 |        2 |      4 |
|        6 |       120 |        22 |        2 |      9 |
|        7 |       120 |        22 |        2 |      2 |
+----------+-----------+-----------+----------+--------+
7 rows in set (0.00 sec)

和

mysql> select image_id, member_id, subcat_id, min(from_day), max(to_day) from prueba1 group by member_id, subcat_id order by image_id asc;
+----------+-----------+-----------+----------+-------------+
| image_id | member_id | subcat_id | from_day | max(to_day) |
+----------+-----------+-----------+----------+-------------+
|        1 |       220 |        56 |        1 |           3 |
|        4 |       120 |        22 |        1 |           9 |
+----------+-----------+-----------+----------+-------------+
2 rows in set (0.00 sec)

它正在运作

编辑：已更新，因为我不了解您的主要问题。

Answer 2

Select
  T.*
From db_images t
Inner join (
  Select member_id, subcat_id, max(to_day) to_day
  db_images group by member_id,subcat_id
) t2 on t.member_id = t2.member_id
And t.Subcat_id = t2.subcat_id
And t.to_day = t2.to_day;

Answer 3

使用max（列名）获得更高的值。

select image_id, member_id,subcat_id,from_day, max(to_day) as to_day from from db_iamges group by member_id,subcat_id order by image_id desc

Answer 4

试试这个

select t1.* from db_images t1 inner join(select max(member_id) as 
mid,max(subcat_id) as cid,min(from_day) as fdy from db_images group 
by member_id) t2 on t1.member_id = t2.mid and t2.cid = t1.subcat_id
and t2.fdy = t1.from_day group by member_id order by image_id asc;

check here

如何使用PHP和MySQL按照列值从表中获取数据

4 个答案: