[英]get result where one column value is identical in two following rows
我想從employeeExam
表中選擇失敗的雇員,其中以下兩行的status
列等於0
。
結果應該是這樣的:
ID COURSE_ID EMPLOYEE_ID DEGREE DATE STATUS NUMOFTAKINGEXAMS
4 2 4 17 January, 15 2013 00:00:00+0000 0 2
這是我所做的:
SQL小提琴
為了進一步說明:按id
排序時,結果應僅包含彼此之間具有相同course_id
和employee_id
且status
= 0的考試數據。
請嘗試一下並發表評論:
set @sum:=0;
set @id:=0;
select distinct x.empid, x.degree, x.date, x.status
from (
select @sum:= (case when status=0
and @id = employee_id then @sum+1
else 1 end)
as sm, @id:=employee_id as empid, degree, date, status
from employeeexam
order by employee_id)x
where x.sm >= 2
;
| EMPID | DEGREE | DATE | STATUS |
------------------------------------------------------------
| 2 | 5 | January, 16 2013 00:00:00+0000 | 0 |
| 3 | 6 | January, 16 2013 00:00:00+0000 | 0 |
| 4 | 15 | January, 14 2013 00:00:00+0000 | 0 |
盡管我懷疑它是否完全擴展了您的要求,但僅顯示一部分SQL代碼即可顯示該原理。 例如,此代碼不會檢查連續失敗的嘗試,而只會檢查所有嘗試。 它不會顯示申請人已經通過考試的任何行。
因此,例如:一個人應該參加考試並獲得狀態0,0,1,0,0; 它不會顯示,因為申請人至少通過了一項考試。
SELECT course_id, employee_id, MAX(degree), status, COUNT(id) NumExamsTaken
FROM employeeExam
GROUP BY course_id, employee_id
HAVING COUNT(id) >= 2 AND SUM(status) = 0;
SELECT a.*
FROM
( SELECT x.*
, COUNT(*) rank
FROM employeeExam x
JOIN employeeExam y
ON y.course_id = x.course_id
AND y.employee_id = x.employee_id
AND y.date <= x.date
GROUP
BY id
) a
JOIN
( SELECT x.*
, COUNT(*) rank
FROM employeeExam x
JOIN employeeExam y
ON y.course_id = x.course_id
AND y.employee_id = x.employee_id
AND y.date <= x.date
GROUP
BY id
) b
ON b.course_id = a.course_id
AND b.employee_id = a.employee_id
AND b.status = a.status
AND b.rank = a.rank - 1
WHERE a.status = 0;
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