如何使用MySQL和PHP从表保持max，min列值获取值

Question

我需要一个帮助。我需要通过使用MySQL将其某些值保持为max或min来从表中获取所有记录。我正在解释下面的表格和代码。

db_images：

id        member_id        day1       day2      images

1           241             1          1         asc.png

2           241             1          2         xzc.png

3           241             2          3         ohjy.png

4          240             1           5         asd.png

这是我的表。我正在解释下面的查询。

$member_id=241
$qry=mysqli_query($connect,"select * from db_images where member_id='".$member_id."'");

在这里我可以按照条件从表中获得3条记录。但我需要在每条记录中保留min(day1) and max(day2)值。表示每条记录将根据所需条件包含day1=1 and day2=3，并且应获取相同的3条记录。请帮帮我。

Answer 1

您可以尝试使用子选择。

select minday1, maxday2, t1.id, t2.member_id, t1.images
from (select min(day1) as minday1, max(day2) as maxday2, member_id from db_images 
group by member_id) as t2, db_images  as t1
where t2.member_id = t1.member_id and t1.member_id=$member_id

Answer 2

我认为这就是你想要的

以下查询将完成此任务。

select id,coalesce((select min(day1) from db_images where member_id=241)) as day1,
coalesce((select max(day2) from db_images where member_id=241)) as day2,
images from db_images where member_id=241 

我在这里使用过coalesce()。 您可以在这里阅读有关此功能的更多信息

Mysql Docs
Oracle Docs

Answer 3

    Select A.*,B.DAY_2 From 
    (Select id,member_id,images,min(day1) as DAY_1 From db_images where member_id=241 group by member_id) as A
    Inner Join
    (Select id,member_id,images,max(day2) as DAY_2 From db_images where member_id=241 group by member_id) as B

试试这个，希望它会对你有所帮助......如果你的期望是我的附图。

Answer 4

请尝试使用此

    SELECT * , 
    (select MIN(`day1`) from `db_images` where member_id = ".$member_id.") as minDay ,    
    (select MAX(`day2`) from `db_images` where member_id= ".$member_id.") as maxDay 
    FROM `db_images` WHERE `member_id` = ".$member_id." ;