mysql - 选择日期相等且 id 相等的行并将这些行计为一个

Question

我有一个数据库，其中有 id、new_name、product_id、date

1162        DC: 10us      1049902   2013-07-19 
1163        DC: 12us      1049902   2013-07-19 
1164        DC: 30us      1049902   2013-07-19 
1165        Top           1049902   2017-07-30 
1166        A:123         202302    2013-07-21 
1167        A:255         2023025   2013-07-21 

我需要选择日期相等的行（例如每行的日期 = 2013-07-19 和每行的 product_id =（例如）1049902）并将这些行计为 1 行

所以：

   DC 1049902  = 2 (because there are two different dates that's why 2 besause it is summary (the names are differenet it doesn't matter)
   A 202302 = 1
   A 2023025 = 1

（查询后我只是 substr() 字符串）

等等...

我试着这样做：

   select new_name, COUNT(new_name) AS n from table WHERE date<='".$today."' AND date>= '".$two_weeks_ago."' GROUP BY day(date) ORDER BY n DESC 

但每一行都算我

先感谢您！

Answer 1

试试这个方法：

select new_name, product_id,COUNT(day(date)) AS n 
from table 
WHERE date<='".$today."' AND date>= '".$two_weeks_ago."' 
GROUP BY new_name, product_id 
ORDER BY n DESC 

编辑：substring(new_name,0,5)来自 OP

select substring(new_name,0,5) as new_name, product_id,COUNT(day(date)) AS n 
from table 
WHERE date<='".$today."' AND date>= '".$two_weeks_ago."' 
GROUP BY substring(new_name,0,5), product_id 
ORDER BY n DESC 

Answer 2

按 product_id 和 new_name 分组

SELECT new_name, COUNT(new_name) AS n 
FROM table 
WHERE date<='".$today."' AND date>= '".$two_weeks_ago."' 
GROUP BY new_name,product_id
ORDER BY n DESC

Answer 3

使用此查询：

SELECT `new_name`, COUNT(`new_name`) AS n 
     FROM table 
        WHERE `date`<='".$today."' AND `date` >= '".$two_weeks_ago."' 
        GROUP BY `product_id`, `date` 
        ORDER BY n DESC 

Answer 4

@Parado，检查此代码：

select brand_name, count(*) AS num 
from (select SUBSTRING_INDEX(new_name, ':', 1) as brand_name 
from table 
WHERE date<='".$american_today."' AND date>= '".$two_weeks_ago."' 
GROUP BY day(date), products_id) 
as brands  group by brand_name ORDER BY num DESC LIMIT 5

效果更好，我做到了，因为您的代码工作错误对不起，如果我没有告诉您更多信息。

谢谢）