Question

我有以下表格产品和测试。

select id,pname from products;                                                     
+----+---------+                                                                   
| id | pname   |                                                                   
+----+---------+                                                                   
|  1 | prd1    |                                                                   
|  2 | prd2    |                                                                   
|  3 | prd3    |                                                                   
|  4 | prd4    |                                                                   
+----+---------+                                                       

select pname,testrunid,testresult,time from tests;                       
+--------+-----------+------------+-------------+                                  
| pname  | testrunid | testresult | time        |                                  
+--------+-----------+------------+-------------+                                  
| prd1   |       800 | PASS       | 2017-10-02  |                                  
| prd1   |       801 | FAIL       | 2017-10-16  |                                  
| prd1   |       802 | PASS       | 2017-10-02  |                                  
| prd1   |       803 | NULL       | 2017-10-16  |                                  
| prd1   |       804 | PASS       | 2017-10-16  |                                  
| prd1   |       805 | PASS       | 2017-10-16  |                                  
| prd1   |       806 | PASS       | 2017-10-16  |                                  
+--------+-----------+------------+-------------+

我喜欢计算产品的测试结果，如果没有可用的结果，产品只显示零。类似于下表：

+--------+------------+-----------+----------------+---------------+               
| pname  | total_pass | total_fail| pass_lastweek  | fail_lastweek |               
+--------+------------+-----------+----------------+---------------+               
| prd1   |      5     |     1     |      3         |       1       |               
| prd2   |      0     |     0     |      0         |       0       |               
| prd3   |      0     |     0     |      0         |       0       |               
| prd4   |      0     |     0     |      0         |       0       |               
+--------+------------+-----------+----------------++--------------+

我尝试了以下不同的查询，这只是针对一种产品而且不完整：

SELECT pname, count(*) as pass_lastweek FROM tests where testresult = 'PASS' AND time
>= '2017-10-11' and pname in (select pname from products) group by pname;       
+-------------+---------------+                                                 
| pname       | pass_lastweek |                                                 
+-------------+---------------+                                                 
| prd1        |          3    |                                                 
+-------------+---------------+

它看起来很基本，但我仍然无法写出来，任何想法？

Answer 1

使用条件聚合。 COUNT函数计数NULL自动为零，因此无需处理。

select p.pname,
       count(case when testresult = 'PASS' then 1 end) as total_pass,
       count(case when testresult = 'FAIL' then 1 end) as total_fail,
       count(case when testresult = 'PASS' and time >= curdate() - INTERVAL 6 DAY then 1 end) as pass_lastweek  ,
       count(case when testresult = 'FAIL' and time >= curdate() - INTERVAL 6 DAY then 1 end) as fail_lastweek  ,
from products p
left join tests t on t.pname = p.pname
group p.id, p.pname

Answer 2

通常，在分组之前，您需要LEFT JOIN第一个表和第二个表。连接将为每个产品提供一行（即使没有测试结果可以加入; INNER JOIN将排除没有相关测试的产品）+每个测试结果的额外行（超出第一个）。然后你可以将它们分组。

SELECT products.*, tests.* FROM products 
LEFT JOIN tests ON products.pname = tests.pname 
GROUP BY products.id

另外，我强烈建议在测试表中使用product_id列，而不是使用pname（如果products.pname发生更改，则整个数据库都会中断，除非您还可以为每个测试结果更新pname字段。一般查询将如下所示：

SELECT products.*, tests.* FROM products 
LEFT JOIN tests ON products.id = tests.product_id
GROUP BY products.id

Answer 3

我使用了2个查询，第一个使用条件计数，第二个是将所有 null 值更改为0：

select pname, 
    case when total_pass is null then 0 else total_pass end as total_pass,
     case when total_fail is null then 0 else total_fail end as total_fail,
      case when pass_lastweek is null then 0 else pass_lastweek end as pass_lastweek,
       case when fail_lastweek is null then 0 else fail_lastweek end asfail_lastweek from (
    select products.pname,
           count(case when testresult = 'PASS' then 1 end) as total_pass,
           count(case when testresult = 'FAIL' then 1 end) as total_fail,
           count(case when testresult = 'PASS' and time >= current_date -7 DAY then 1 end) as pass_lastweek  ,
           count(case when testresult = 'FAIL' and time >= current_date -7 DAY then 1 end) as fail_lastweek  ,
    from products 
    left join tests  on tests.pname = products.pname
    group 1 ) t1

加入两个表并计数，如果第二个表中没有记录，则避免为零

3 个答案: