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这是我的查询，它有一些连接，除了 GROUP_CONCAT(p.product_id) 和 SUM(p.price) 部分外，所有连接都运行良好。

SELECT ts.name as main_name, tp.class_id, ts.step_number, GROUP_CONCAT(p.product_id) as product_id, SUM(p.price) as price
from template as t 
JOIN template_step as ts on ts.template_id=t.template_id 
JOIN template_product as tp on tp.template_id=ts.template_id 
JOIN product as p on p.product_id=tp.product_id 
JOIN product_description as pd on pd.product_id=p.product_id 
where t.template_id = '59' 
group by tp.class_id, ts.step_number 
ORDER by ts.step_number, tp.class_id

问题是返回 product_ids 的元素和 sum 字段重复

这是我的查询数据

Networking  1   1    88,156,151,275,48,101,274,133,154,125,135,148,63,63    3070.0000
Networking  2   1    275,235,164,274,154,124,169,148,62,98,62,277,191,270   3695.0000
Networking  3   1    92,98,216,181,133,187,272,154,274,148,126,62,62,165    4970.0000
Back Office 1   2    63,88,156,151,275,48,101,274,133,154,125,135,148,63    3070.0000
Back Office 2   2    275,235,164,274,154,124,169,148,62,98,62,277,191,270   3695.0000
Back Office 3   2    62,165,92,98,216,181,133,187,272,154,274,148,126,62    4970.0000
Data Back   1   3    148,63,63,88,156,151,275,48,101,274,133,154,125,135    3070.0000
Data Back   2   3    270,275,235,164,274,154,124,169,148,62,98,62,277,191   3695.0000
Data Back   3   3    62,62,165,92,98,216,181,133,187,272,154,274,148,126    4970.0000
Kitchen     1   4    135,148,63,63,88,156,151,275,48,101,274,133,154,125    3070.0000

每个类应该只返回 1 或 2 个 product_id。如果我可以提供任何其他信息来帮助其他人帮助我，我可以提供任何东西..db结构..ext ..

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1 回答 1

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GROUP_CONCAT()不会删除重复的行，除非您使用DISTINCT：

GROUP_CONCAT([DISTINCT] expr [,expr ...]
             [ORDER BY {unsigned_integer | col_name | expr}
                 [ASC | DESC] [,col_name ...]]
             [SEPARATOR str_val])
于 2011-06-27T19:05:10.553 回答