我有一个查询，可以生成一个人在他的时间表上注册的假期。 时间表中的每一天都是单独的记录，但如果您休假一周（2011年12月26日至2011年12月30日），则必须通过查询在一个区块（1行）中表示。
这是我的查询有点简化的问题：
SELECT -1 ord_li_pers_plan_id
,ts.person_id person_id
,-1 order_line_id
,ts.timesheet_dt start_dt
,ts.timesheet_dt end_dt
,'Vacation' project
,null color
,1013011 planning_type
FROM timesheets ts
JOIN persons pers ON ts.person_id = pers.person_id
JOIN person_holidays per_hol ON
ts.person_holiday_id = per_hol.person_holiday_id
JOIN v_holiday_types hoty_peho ON
per_hol.holiday_type_id = hoty_peho.holiday_type_id
WHERE ts.person_holiday_id IS NOT NULL
AND per_hol.holiday_type_id IN (SELECT holiday_type_id FROM v_holiday_types)
AND ts.person_id = :p_person_id
这是我现在得到的输出：
-1 11182 -1 30-DEC-11 30-DEC-11 Vacation 1013011
-1 11182 -1 29-DEC-11 29-DEC-11 Vacation 1013011
-1 11182 -1 28-DEC-11 28-DEC-11 Vacation 1013011
-1 11182 -1 27-DEC-11 27-DEC-11 Vacation 1013011
-1 11182 -1 26-DEC-11 26-DEC-11 Vacation 1013011
-1 11182 -1 31-OCT-11 31-OCT-11 Vacation 1013011
-1 11182 -1 02-SEP-11 02-SEP-11 Vacation 1013011
-1 11182 -1 29-JUL-11 29-JUL-11 Vacation 1013011
-1 11182 -1 22-JUL-11 22-JUL-11 Vacation 1013011
-1 11182 -1 25-APR-11 25-APR-11 Vacation 1013011
更新：
相反，查询应输出以下内容：
-1 11182 -1 26-DEC-11 30-DEC-11 Vacation 1013011
-1 11182 -1 31-OCT-11 31-OCT-11 Vacation 1013011
-1 11182 -1 02-SEP-11 02-SEP-11 Vacation 1013011
-1 11182 -1 29-JUL-11 29-JUL-11 Vacation 1013011
-1 11182 -1 22-JUL-11 22-JUL-11 Vacation 1013011
-1 11182 -1 25-APR-11 25-APR-11 Vacation 1013011
注意第一行。它现在代表整个假期。
更新2
我的客户提出了另一个想法。此查询链接到时间表，其中包含指定人员的所有项目。项目日期最初存储为一个块（每个日期范围的1条记录，例如01/01/2012 - 10/01/2012 =＆gt; 1条记录）。现在，客户希望将所有日期存储为单独的记录。这并不难，而且有效。
我现在面临的问题是，由于ord_li_pers-plan_id是PK，因此group by不再起作用。
附件是查询。这是一个相当大的。
SELECT person_id ,
order_line_id ,
MIN(start_dt) start_dt ,
MAX(end_dt) end_dt ,
project ,
color ,
planning_type,
ord_li_pers_plan_id
FROM
(SELECT ord_li_pers_plan_id ,
person_id ,
order_line_id ,
start_dt ,
end_dt ,
project ,
color ,
planning_type ,
SUM(gap) OVER (PARTITION BY person_id ORDER BY start_dt) contiguous_grp
FROM
(SELECT ord_li_pers_plan_id ,
person_id ,
order_line_id ,
start_dt ,
end_dt ,
project ,
color ,
planning_type ,
CASE
WHEN lag(end_dt) over(PARTITION BY person_id ORDER BY start_dt) + 1 >= start_dt
THEN 0
ELSE 1
END gap
FROM
(SELECT ord_li_pers_plan_id ,
person_id ,
order_line_id ,
start_dt ,
end_dt ,
project ,
color ,
planning_type
FROM
(SELECT op.ord_li_pers_plan_id ord_li_pers_plan_id ,
p.person_id person_id ,
p.order_line_id order_line_id ,
op.start_dt start_dt ,
op.end_dt end_dt ,
pl$planning.prep_tooltip(NVL2(ord.end_customer_id, end_cus.name, cus.name)
|| ' - '
|| NVL2(ord_li.project_cd,ord_li.project_cd,ord.project_cd)
|| ' - '
|| func_tp.name
|| ' - '
|| ROUND((con_tp.contract_tp / 5), 2)
||' - '
|| bl$gen_codes.Name_by_Code_Id (op.planning_type)) project ,
olc.color color ,
op.planning_type planning_type
FROM order_line_person_planning op
JOIN order_line_persons p
ON p.ORDER_LINE_PERSON_ID = op.ORDER_LINE_PERSON_ID
JOIN order_lines ord_li
ON ord_li.order_line_id = p.order_line_id
JOIN orders ord
ON ord.order_id = ord_li.order_id
LEFT JOIN order_line_colors olc
ON olc.order_line_id = ord_li.order_line_id
JOIN customers cus
ON ord.customer_id=cus.customer_id
LEFT JOIN customers end_cus
ON ord.end_customer_id=end_cus.customer_id
LEFT JOIN v_contract_types con_tp
ON ord.contract_type_id=con_tp.contract_type_id
JOIN v_function_types func_tp
ON ord_li.function_id=func_tp.function_id
UNION
SELECT -1 ord_li_pers_plan_id ,
ts.person_id person_id ,
-1 order_line_id ,
ts.timesheet_dt start_dt ,
ts.timesheet_dt end_dt ,
'Vacation' project ,
'#99FF33' color ,
-1 planning_type
FROM hrm_iadvise.timesheets ts
JOIN hrm_iadvise.persons pers
ON ts.person_id = pers.person_id
JOIN hrm_iadvise.person_holidays per_hol
ON ts.person_holiday_id = per_hol.person_holiday_id
JOIN hrm_iadvise.v_holiday_types hoty_peho
ON per_hol.holiday_type_id = hoty_peho.holiday_type_id
WHERE ts.person_holiday_id IS NOT NULL
UNION
SELECT -1 ord_li_pers_plan_id ,
per_hol.person_id person_id ,
-1 order_line_id ,
hol.dt start_dt ,
hol.dt end_dt ,
'Vacation' project ,
'#99FF33' color ,
-1 planning_type
FROM holidays hol
JOIN person_holidays per_hol
ON per_hol.holiday_type_id = hol.holiday_type_id
JOIN countries coty
ON coty.country_id = hol.country_id
WHERE coty.country_cd = 150
)
)
)
)
GROUP BY person_id ,
order_line_id ,
project ,
color ,
planning_type ,
contiguous_grp,
ord_li_pers_plan_id;
第一个查询获取人员分配到的所有项目 第二个查询获得一个人已经注册的所有假期 第三个查询所有国家法定假日
提前致谢
答案 0 :(得分：2)
您可以使用分析。数据只读一次：
SQL> WITH your_query AS (
2 SELECT 11182 person_id, to_date('30-DEC-11') start_dt,
3 to_date('30-DEC-11') end_dt FROM dual
4 UNION ALL SELECT 11182,to_date('29-DEC-11'),to_date('29-DEC-11') FROM DUAL
5 UNION ALL SELECT 11182,to_date('28-DEC-11'),to_date('28-DEC-11') FROM DUAL
6 UNION ALL SELECT 11182,to_date('27-DEC-11'),to_date('27-DEC-11') FROM DUAL
7 UNION ALL SELECT 11182,to_date('26-DEC-11'),to_date('26-DEC-11') FROM DUAL
8 UNION ALL SELECT 11182,to_date('31-OCT-11'),to_date('31-OCT-11') FROM DUAL
9 UNION ALL SELECT 11182,to_date('02-SEP-11'),to_date('02-SEP-11') FROM DUAL
10 UNION ALL SELECT 11182,to_date('29-JUL-11'),to_date('29-JUL-11') FROM DUAL
11 UNION ALL SELECT 11182,to_date('22-JUL-11'),to_date('22-JUL-11') FROM DUAL
12 UNION ALL SELECT 11182,to_date('25-APR-11'),to_date('25-APR-11') FROM DUAL
13 )
14 SELECT person_id, MIN(start_dt) start_dt, MAX(end_dt) end_dt
15 FROM (SELECT person_id, start_dt, end_dt,
16 SUM(gap) over(PARTITION BY person_id
17 ORDER BY start_dt) contiguous_grp
18 FROM (SELECT person_id, start_dt, end_dt,
19 CASE WHEN lag(end_dt) over(PARTITION BY person_id
20 ORDER BY start_dt) + 1 >= start_dt
21 THEN 0 ELSE 1
22 END gap
23 FROM (SELECT * FROM your_query)))
24 GROUP BY person_id, contiguous_grp
25 ORDER BY 1, 2 DESC;
PERSON_ID START_DT END_DT
---------- --------- ---------
11182 26-DEC-11 30-DEC-11
11182 31-OCT-11 31-OCT-11
11182 02-SEP-11 02-SEP-11
11182 29-JUL-11 29-JUL-11
11182 22-JUL-11 22-JUL-11
11182 25-APR-11 25-APR-11
答案 1 :(得分：1)
您需要使用分层查询来实现此目的。使用这样的东西：
SELECT
person_id,
person_holiday_id,
MIN(BEGIN),
END
FROM
(
SELECT
person_id,
person_holiday_id,
connect_by_root(timesheet_dt) AS BEGIN,
timesheet_dt AS END,
connect_by_isleaf AS is_leaf
FROM
timesheets
CONNECT BY
PRIOR TRUNC(timesheet_dt) + 1 = TRUNC(timesheet_dt)
AND PRIOR person_id = person_id
)
WHERE
is_leaf = 1
GROUP BY
person_id,
person_holiday_id,
END
;
每个用户每个连续休假期间返回一行。将其放在视图中并使用它代替您在问题中发布的查询中的timesheets表。
请注意：
timesheets中的每一行都用作分层查询的根行，可能用作叶子行和中间行。比较运行子查询并将它返回的行数与完整查询返回的行数进行比较时，可以看到这一点。timesheet_dt上放置索引并确保它永远不包含时间部分，那么您可以提高性能 - 然后您可以摆脱TRUNC函数。timesheets表，则此声明将返回的最长休假期为星期一至星期五的一周。如果这是一个问题，则需要调整CONNECT BY子句。但是，这不能解决公共假日没有进入timesheets表......