mysql 聚合查询返回错误结果

Question

我有两个 table 中的数据需要加入，return table 之一中记录的出现次数，

Employee table 中的数据看起来像，

empId  workpatternId
    1        20

workPattern tables中的数据看起来像，

workpatternId  monday tuesday wednesday thursday friday saturday sunday
   20           ALL     ALL     ALL       ALL     NULL   NULL     ALL

下面的查询应该是 return 5，这是 ALL 的计数，但是 returns 7 相反，

SELECT empId,b.workingPatternId, COUNT(monday='ALL') +
 COUNT(tuesday='ALL') + COUNT(wednesday='ALL')+ COUNT(thursday='ALL') + 
    COUNT(friday='ALL')+ COUNT(saturday='ALL')+ COUNT(sunday='ALL') AS COUNT
      FROM workPattern b 
 join Employee e on (e.workpatternId = b.workpatternId) and e.empId = 1
         GROUP BY empId ;

查询有什么问题？

编辑

dbfiddle

Answer 1

我看不出聚合的意义，因为似乎每个 empId 在 workPattern 中只有一行。你可以这样写：

SELECT e.empId, wp.workingPatternId, 
      (wp.monday    = 'ALL') 
    + (wp.tuesday   = 'ALL') 
    + (wp.wednesday = 'ALL') 
    + (wp.thursday  = 'ALL') 
    + (wp.friday    = 'ALL') 
    + (wp.saturday  = 'ALL') 
    + (wp.sunday    = 'ALL') cnt
FROM workPattern wp
INNER Employee e on e.workpatternId = wp.workpatternId 
WHERE e.empId = 1

如果出于某种原因需要聚合，那么您需要 sum() 而不是 count()：后者计算所有 non-null 值，而 false 条件被评估为 0（不是 null，因此在您的查询中考虑了它）：

SELECT 
    SUM(
          (wp.monday    = 'ALL') 
        + (wp.tuesday   = 'ALL') 
        + (wp.wednesday = 'ALL') 
        + (wp.thursday  = 'ALL') 
        + (wp.friday    = 'ALL') 
        + (wp.saturday  = 'ALL') 
        + (wp.sunday    = 'ALL')
    ) cnt
FROM workPattern wp
INNER JOIN Employee e on e.workpatternId = wp.workpatternId 
WHERE e.empId = 1

Answer 2

这有效：

 SELECT empId,b.workingPatternId, sum(Monday='ALL') + sum(Tuesday='ALL') + sum(Wednesday='ALL')+ sum(Thursday='ALL') + sum(Friday='ALL') + sum(Saturday='ALL')+ sum(Sunday='ALL') AS COUNT
 FROM WorkPatterns b 
 JOIN Employee e 
        ON (e.workingPatternId = b.workingPatternId) 
        AND e.empId = 1   
 GROUP BY empId ;

mysql 聚合查询返回错误结果

mysql aggregate query returning wrong result

mysql

sql

inner-join

count

mysql-8.0