Question

我试图找到每天分配给员工的新任务。小提琴是here

员工可以在4个任务A，B，C和C之间循环。 D. 1月1日，他可能会被分配任务，1月2日他可能会被分配任务＆＃39; B＆＃39; 1月5日任务＆＃39; C＆＃39;第6 Jan任务＆＃39; D＆＃39; ＆安培; 1月14日他说他被分配了任务＆＃39; A＆＃39;再次。我希望看到日期＆amp;与任何特定日期相比，新分配的任务与前一天相比。

我需要的SQL输出应该只显示他在给定日期分配的新任务。如果给定日期不存在数据，则new的定义与前一天或最后一个条目有关。

以下是理想的外观 -

更新的SQL

FIDDLE HERE

SELECT   task_date, 
         employee_name, 
         Group_concat(task_name) 
FROM     ( 
                         SELECT DISTINCT a.task_date, 
                                         a.employee_name, 
                                         CASE 
                                                         WHEN b.employee_name IS NOT NULL
                                                         AND             c.employee_name IS NULL THEN NULL
                                                         ELSE a.task_name
                                         END       AS task_name 
                         FROM            forgerock AS a 
                         LEFT OUTER JOIN forgerock AS b 
                         ON              a.employee_name = b.employee_name = 'A'
                         AND             a.task_date >= '2015-01-02' 
                         AND             a.task_date <= '2015-01-04' 
                         AND             b.task_date >= '2015-01-02' 
                         AND             b.task_date <= '2015-01-04' 
                         AND             a.task_date - 1 = b.task_date
                         AND             a.region = b.region = 'USA' 
                         LEFT OUTER JOIN forgerock AS c 
                         ON              a.employee_name = c.employee_name = 'A'
                         AND             a.task_date >= '2015-01-02' 
                         AND             a.task_date <= '2015-01-04' 
                         AND             c.task_date >= '2015-01-02' 
                         AND             c.task_date <= '2015-01-04' 
                         AND             a.task_date - 1 = c.task_date
                         AND             a.task_name <> c.task_name 
                         AND             a.region = b.region = 'USA' 
                         ORDER BY        a.task_date, 
                                         a.employee_name, 
                                         a.task_name) AS temp 
GROUP BY task_date, 
         employee_name

Answer 1

select task_date,employee_name,GROUP_CONCAT(task_name) from(
select distinct a.task_date,a.employee_name
,case when b.employee_name is not null and c.employee_name is null
then null
else a.task_name end as task_name
from ForgeRock as a left outer join ForgeRock as b
on a.employee_name = b.employee_name and a.task_date-1 = b.task_date 
left outer join ForgeRock as c
on a.employee_name = c.employee_name and a.task_date-1 = c.task_date 
and a.task_name <> c.task_name
order by a.task_date,a.employee_name,a.task_name) as temp
group by task_date,employee_name

添加条件：

select task_date,employee_name,GROUP_CONCAT(task_name) from(
select distinct a.task_date,a.employee_name
,case when b.employee_name is not null and c.employee_name is null
then null
else a.task_name end as task_name
from ForgeRock as a left outer join ForgeRock as b
on a.employee_name = b.employee_name and a.task_date-1 = b.task_date and a.task_date between  '2015-01-02' AND '2015-01-04' and b.task_date between  '2015-01-02' AND '2015-01-04' 
left outer join ForgeRock as c
on a.employee_name = c.employee_name and a.task_date-1 = c.task_date and a.task_date between  '2015-01-02' AND '2015-01-04' and c.task_date between  '2015-01-02' AND '2015-01-04'
and a.task_name <> c.task_name
where a.region = 'USA'  and a.task_date between '2015-01-02' AND '2015-01-04'
order by a.task_date,a.employee_name,a.task_name) as temp

group by task_date,employee_name

MySQL - 仅为每天选择新条目

1 个答案: