B*****g 发帖数: 34098 | 1 大家都不参加包子贴?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
写一个SQL,找出next N business days (N>0)。
假设已有fucntion isBusinessDay(date), return 1 for business day and 0 for
non-business day.
可以用任何数据库,不许假设M天之内必有N个business days。
第一个打出来的发双簧包,答案符合ANSI的包子加倍。 |
y****w 发帖数: 3747 | 2 holiday算哪几天?
【在 B*****g 的大作中提到】 : 大家都不参加包子贴? : ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~` : 写一个SQL,找出next N business days (N>0)。 : 假设已有fucntion isBusinessDay(date), return 1 for business day and 0 for : non-business day. : 可以用任何数据库,不许假设M天之内必有N个business days。 : 第一个打出来的发双簧包,答案符合ANSI的包子加倍。
|
B*****g 发帖数: 34098 | 3 isBusinessDay, haha
【在 y****w 的大作中提到】 : holiday算哪几天?
|
a9 发帖数: 21638 | 4 beijing做dba做傻了,这种题目为什么要放到sql里来做?
【在 B*****g 的大作中提到】 : 大家都不参加包子贴? : ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~` : 写一个SQL,找出next N business days (N>0)。 : 假设已有fucntion isBusinessDay(date), return 1 for business day and 0 for : non-business day. : 可以用任何数据库,不许假设M天之内必有N个business days。 : 第一个打出来的发双簧包,答案符合ANSI的包子加倍。
|
y****w 发帖数: 3747 | 5 教recursive sql。
【在 a9 的大作中提到】 : beijing做dba做傻了,这种题目为什么要放到sql里来做?
|
B*****g 发帖数: 34098 | 6 Tom Kyte’s:
You should do it in a single SQL statement if at all possible.
If you cannot do it in a single SQL Statement, then do it in PL/SQL.
If you cannot do it in PL/SQL, try a Java Stored Procedure.
If you cannot do it in Java, do it in a C external procedure.
If you cannot do it in a C external routine, you might want to seriously think about why it is you need to do it…
一些问题developer不用SQL不是因为SP更好,而是因为developer写不出SQL
【在 a9 的大作中提到】 : beijing做dba做傻了,这种题目为什么要放到sql里来做?
|
B*****g 发帖数: 34098 | 7 就你明白,p
【在 y****w 的大作中提到】 : 教recursive sql。
|
g***l 发帖数: 18555 | |
g***l 发帖数: 18555 | 9 首先要有一个FUNCTION判断是不是BUSINESS DATE,别忘了HOLIDAY
create function [dbo].[udf_is_valid_business_date]
(
@date datetime
)
returns int
as
begin
declare
@is_holiday int,
@is_valid_business_date int
select @is_holiday =
case
-- memorial day (last mon of may)
when datepart(yy,@date) = 2009 and datepart(mm,@date) = 5 and
datepart(dd,@date) = 25 then 1
when datepart(yy,@date) = 2010 and datepart(mm,@date) = 5 and
datepart(dd,@date) = 31 then 1
when datepart(yy,@date) = 2011 and datepart(mm,@date) = 5 and
datepart(dd,@date) = 30 then 1
when datepart(yy,@date) = 2012 and datepart(mm,@date) = 5 and
datepart(dd,@date) = 28 then 1
when datepart(yy,@date) = 2013 and datepart(mm,@date) = 5 and
datepart(dd,@date) = 27 then 1
when datepart(yy,@date) = 2014 and datepart(mm,@date) = 5 and
datepart(dd,@date) = 26 then 1
-- thanksgiving (last thu of nov plus 1 day after that)
when datepart(yy,@date) = 2008 and datepart(mm,@date) = 11 and
datepart(dd,@date) in (27, 28) then 1
when datepart(yy,@date) = 2009 and datepart(mm,@date) = 11 and
datepart(dd,@date) in (26, 27) then 1
when datepart(yy,@date) = 2010 and datepart(mm,@date) = 11 and
datepart(dd,@date) in (25, 26) then 1
when datepart(yy,@date) = 2011 and datepart(mm,@date) = 11 and
datepart(dd,@date) in (24, 25) then 1
when datepart(yy,@date) = 2012 and datepart(mm,@date) = 11 and
datepart(dd,@date) in (29, 30) then 1
when datepart(yy,@date) = 2013 and datepart(mm,@date) = 11 and
datepart(dd,@date) in (28, 29) then 1
-- labor day (1st mon of Sep)
when datepart(yy,@date) = 2008 and datepart(mm,@date) = 9 and
datepart(dd,@date) = 1 then 1
when datepart(yy,@date) = 2009 and datepart(mm,@date) = 9 and
datepart(dd,@date) = 7 then 1
when datepart(yy,@date) = 2010 and datepart(mm,@date) = 9 and
datepart(dd,@date) = 6 then 1
when datepart(yy,@date) = 2011 and datepart(mm,@date) = 9 and
datepart(dd,@date) = 5 then 1
when datepart(yy,@date) = 2012 and datepart(mm,@date) = 9 and
datepart(dd,@date) = 3 then 1
when datepart(yy,@date) = 2013 and datepart(mm,@date) = 9 and
datepart(dd,@date) = 2 then 1
-- 4th of July (every year)
when datepart(mm,@date) = 7 and datepart(dd,@date) = 4 then 1
-- Xmas (25 dec every year)
when datepart(mm,@date) = 12 and datepart(dd,@date) = 25 then 1
-- New Years (1 jan every year)
when datepart(mm,@date) = 1 and datepart(dd,@date) = 1 then 1
--
else 0
end
-- Everytime we encouter a weekend day or holiday, we must add one day
to the shipping estimate
if( (datepart(dw, @date) = 1) or (datepart(dw, @date) = 7) or (@is_
holiday = 1) )
begin
select
@is_valid_business_date = 0
end
else
begin
select
@is_valid_business_date = 1
end
return @is_valid_business_date
end |
B*****g 发帖数: 34098 | 10 不会真有人写SQL吧,哈哈
【在 g***l 的大作中提到】 : 我有现成的,哈哈
|
|
|
a9 发帖数: 21638 | 11 这种东西,我一般就写到应用里了。
think about why it is you need to do it…
【在 B*****g 的大作中提到】 : Tom Kyte’s: : You should do it in a single SQL statement if at all possible. : If you cannot do it in a single SQL Statement, then do it in PL/SQL. : If you cannot do it in PL/SQL, try a Java Stored Procedure. : If you cannot do it in Java, do it in a C external procedure. : If you cannot do it in a C external routine, you might want to seriously think about why it is you need to do it… : 一些问题developer不用SQL不是因为SP更好,而是因为developer写不出SQL
|
g***l 发帖数: 18555 | 12 我只到了2013,不过还能优化,按星期几走就不需要LIST,这段不是我写的,我写的肯
定比这个好。LOL |
B*****g 发帖数: 34098 | 13 俺的问题是已有这个function,用这个写SQL
【在 g***l 的大作中提到】 : 我只到了2013,不过还能优化,按星期几走就不需要LIST,这段不是我写的,我写的肯 : 定比这个好。LOL
|
g***l 发帖数: 18555 | 14 用个WHILE LOOP不就行了,从GETDATE()开始+1,直到IS_BUSINESS_DATE
【在 B*****g 的大作中提到】 : 俺的问题是已有这个function,用这个写SQL
|
B*****g 发帖数: 34098 | 15 sql only
【在 g***l 的大作中提到】 : 用个WHILE LOOP不就行了,从GETDATE()开始+1,直到IS_BUSINESS_DATE
|
g***l 发帖数: 18555 | 16 declare
@next_business_date datetime,
@is_business_date int
select @next_business_date = convert(datetime, convert(varchar,getdate(),101))
while (@is_business_date = 0)
begin
select @is_business_date = dbo.udf_is_valid_business_date(@next_business_date)
if (@is_business_date =0)
begin
select @next_business_date = @next_business_date + 1
end
end
select @next_business_date |
B*****g 发帖数: 34098 | 17 还是给你发个安慰包
101))
date)
【在 g***l 的大作中提到】 : declare : @next_business_date datetime, : @is_business_date int : select @next_business_date = convert(datetime, convert(varchar,getdate(),101)) : while (@is_business_date = 0) : begin : select @is_business_date = dbo.udf_is_valid_business_date(@next_business_date) : if (@is_business_date =0) : begin : select @next_business_date = @next_business_date + 1
|
c*******e 发帖数: 8624 | 18 用qualify不行吗?
【在 B*****g 的大作中提到】 : sql only
|
B*****g 发帖数: 34098 | 19 不懂,是ANSI吗?
【在 c*******e 的大作中提到】 : 用qualify不行吗?
|
c*******e 发帖数: 8624 | 20 不是
qualify row_number() over (order by xxx) <= N
【在 B*****g 的大作中提到】 : 不懂,是ANSI吗?
|
|
|
B*****g 发帖数: 34098 | 21 把完全SQL写出来吧
【在 c*******e 的大作中提到】 : 不是 : qualify row_number() over (order by xxx) <= N
|
c*******e 发帖数: 8624 | 22 sel data_dt from your_table
where data_dt > date
and biz_day_ind = 'Y'
qualify row_number() over (order by data_dt asc) <= N
order by 1 ;
假设table里面日期都是distinct的,如果不是distinct
就变成distinct,你懂的
【在 B*****g 的大作中提到】 : 把完全SQL写出来吧
|
B*****g 发帖数: 34098 | 23 木有table。从system date后数N天business day,显示出来。有table就是老问题,90
%的数据库版SQL问题可以用partition by解决。
【在 c*******e 的大作中提到】 : sel data_dt from your_table : where data_dt > date : and biz_day_ind = 'Y' : qualify row_number() over (order by data_dt asc) <= N : order by 1 ; : 假设table里面日期都是distinct的,如果不是distinct : 就变成distinct,你懂的
|
c*******e 发帖数: 8624 | 24 用sys_calendar不可以吗?
90
【在 B*****g 的大作中提到】 : 木有table。从system date后数N天business day,显示出来。有table就是老问题,90 : %的数据库版SQL问题可以用partition by解决。
|
B*****g 发帖数: 34098 | 25 算你也对,给你发一个双簧包。不过要是不是日期(比如说已有isZhiShu的function,
返回前100个质数)或者是我要100000000天(超了SYS_CALENDAR范围)你就没辙了。
【在 c*******e 的大作中提到】 : 用sys_calendar不可以吗? : : 90
|
y****9 发帖数: 144 | 26 Can I assume to find next N business day within certain number of years?
If N=10, if find next 10 biz day in 2012, and if today is Dec-23,2012, so it
won't have 10 biz days, is it sill ok for the requirement?
【在 B*****g 的大作中提到】 : 大家都不参加包子贴? : ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~` : 写一个SQL,找出next N business days (N>0)。 : 假设已有fucntion isBusinessDay(date), return 1 for business day and 0 for : non-business day. : 可以用任何数据库,不许假设M天之内必有N个business days。 : 第一个打出来的发双簧包,答案符合ANSI的包子加倍。
|
B*****g 发帖数: 34098 | 27 其实这个题出的不好,不如改成前N个质数,更实际一些。
其实不给上限比给上限就多怎么加一个限制,让那个产生很多天list的SQL到特定位置
停止
~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~
it |
y****9 发帖数: 144 | 28 Sorry, don't understand your reply.Don't know how to do it without upper
limit, but if to find next 10 biz days in 2012, I can do this as follows:
SQL>select mydate
2 from
3 (
4 select to_char(sysdate + level, 'YYYY-Mon-DD') mydate, level
5 from dual
6 where mod(to_number(to_char(sysdate + level, 'YYYYMMDD')),3) = 0 --
equivalent to isBIzDay(sysdate+level) =1
7 connect by level < to_date('2012-12-31','YYYY-MM-DD') - sysdate
8 )
9 where rownum <=10;
MYDATE
-----------
2012-Jan-09
2012-Jan-12
2012-Jan-15
2012-Jan-18
2012-Jan-21
2012-Jan-24
2012-Jan-27
2012-Jan-30
2012-Feb-02
2012-Feb-05
10 rows selected.
【在 B*****g 的大作中提到】 : 其实这个题出的不好,不如改成前N个质数,更实际一些。 : 其实不给上限比给上限就多怎么加一个限制,让那个产生很多天list的SQL到特定位置 : 停止 : ~~~~~~~~~~~~~~~ : ~~~~~~~~~~~~~~~ : it
|
B*****g 发帖数: 34098 | 29 connect by不要用level,用r*****
【在 y****9 的大作中提到】 : Sorry, don't understand your reply.Don't know how to do it without upper : limit, but if to find next 10 biz days in 2012, I can do this as follows: : SQL>select mydate : 2 from : 3 ( : 4 select to_char(sysdate + level, 'YYYY-Mon-DD') mydate, level : 5 from dual : 6 where mod(to_number(to_char(sysdate + level, 'YYYYMMDD')),3) = 0 -- : equivalent to isBIzDay(sysdate+level) =1 : 7 connect by level < to_date('2012-12-31','YYYY-MM-DD') - sysdate
|
B*****g 发帖数: 34098 | 30 总结一下,其实大家把问题想太复杂了,尤其是咱用oracle的兄弟,其实就是个简单的
SQL。我其实是想大家用recursive SQL做。
yhwang,你写个recursive sql吧,俺写的似乎太复杂
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
写一个SQL,找出next N business days (N>0)。
假设已有fucntion isBusinessDay(date), return 1 for business day and 0 for
non-business day.
可以用任何数据库,不许假设M天之内必有N个business days。
第一个打出来的发双簧包,答案符合ANSI的包子加倍。 |
|
|
y****9 发帖数: 144 | 31
Thanks. Learned sth new today.
SQL>select to_char(sysdate + level, 'YYYY-Mon-DD') mydate
2 from dual
3 where mod(to_number(to_char(sysdate + level, 'YYYYMMDD')),3) = 0 --
equivalent to isBIzday(sysdate+level) =1
4 connect by rownum <=15
5 ;
MYDATE
-----------
2012-Jan-09
2012-Jan-12
2012-Jan-15
2012-Jan-18
2012-Jan-21
2012-Jan-24
2012-Jan-27
2012-Jan-30
2012-Feb-02
2012-Feb-05
2012-Feb-08
2012-Feb-11
2012-Feb-14
2012-Feb-17
2012-Feb-20
15 rows selected.
【在 B*****g 的大作中提到】 : connect by不要用level,用r*****
|
y****w 发帖数: 3747 | 32 with t(d,n) as
(
select current date, case when dayofweek_iso(current date) < 6 then
1 else 0 end from dual
union all
select t.d + 1 day, t.n+ case when dayofweek_iso(t.d+1 day) < 6 then
1 else 0 end from t where t.n < N
)
select min(d) from t group by n order by 1;
把case when dayofweek_iso调用换成你的isbusinessday。
【在 B*****g 的大作中提到】 : 总结一下,其实大家把问题想太复杂了,尤其是咱用oracle的兄弟,其实就是个简单的 : SQL。我其实是想大家用recursive SQL做。 : yhwang,你写个recursive sql吧,俺写的似乎太复杂 : ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~` : 写一个SQL,找出next N business days (N>0)。 : 假设已有fucntion isBusinessDay(date), return 1 for business day and 0 for : non-business day. : 可以用任何数据库,不许假设M天之内必有N个business days。 : 第一个打出来的发双簧包,答案符合ANSI的包子加倍。
|
y****w 发帖数: 3747 | 33 接着写isBusinessDay吧, 过程或sql,返回一个int, 1 or 0.
用这里的federal holiday作参考,
http://en.wikipedia.org/wiki/Public_holidays_in_the_United_Stat |
B*****g 发帖数: 34098 | 34 我一开始就这么写,结果oracle啃爹了,大家看看下面为啥不work?我简化了一下
with t(d,n) as
(
select sysdate, 1
from dual
union all
select t.d+1, t.n +1
from t
where t.n < 10
)
select * from t
下面这个work,bug?
with t(d,n) as
(
select 1, 1
from dual
union all
select t.d+1, t.n +1
from t
where t.n < 10
)
select * from t
then
then
【在 y****w 的大作中提到】 : with t(d,n) as : ( : select current date, case when dayofweek_iso(current date) < 6 then : 1 else 0 end from dual : union all : select t.d + 1 day, t.n+ case when dayofweek_iso(t.d+1 day) < 6 then : 1 else 0 end from t where t.n < N : ) : select min(d) from t group by n order by 1; : 把case when dayofweek_iso调用换成你的isbusinessday。
|
y****9 发帖数: 144 | 35
then
then
Learning SQL server now :-) Based on your sql structure; the following
works in SQL Server 2005
WITH t(d,n)
AS ( SELECT getdate(), case when day(getdate())%3=0 then 1 else
0 end
UNION ALL
SELECT t.d+ 1, t.n + case when day(t.d +1)%3=0 then 1 else
0 end
FROM t
WHERE t.n < 10
)
select min(d) from t
group by n
order by 1;
-----------------------
2012-01-06 13:27:25.843
2012-01-09 13:27:25.843
2012-01-12 13:27:25.843
2012-01-15 13:27:25.843
2012-01-18 13:27:25.843
2012-01-21 13:27:25.843
2012-01-24 13:27:25.843
2012-01-27 13:27:25.843
2012-01-30 13:27:25.843
2012-02-03 13:27:25.843
(10 row(s) affected)
HOwever if i do t.n< 50, I got:
-----------------------
Msg 530, Level 16, State 1, Line 2
The statement terminated. The maximum recursion 100 has been exhausted
before statement completion.
Is recursion level configurable in SQL server?
【在 y****w 的大作中提到】 : with t(d,n) as : ( : select current date, case when dayofweek_iso(current date) < 6 then : 1 else 0 end from dual : union all : select t.d + 1 day, t.n+ case when dayofweek_iso(t.d+1 day) < 6 then : 1 else 0 end from t where t.n < N : ) : select min(d) from t group by n order by 1; : 把case when dayofweek_iso调用换成你的isbusinessday。
|
B*****g 发帖数: 34098 | 36 有没有oracle11g试一下,俺一直出问题,只好绕一个大圈做
else
else
【在 y****9 的大作中提到】 : : then : then : Learning SQL server now :-) Based on your sql structure; the following : works in SQL Server 2005 : WITH t(d,n) : AS ( SELECT getdate(), case when day(getdate())%3=0 then 1 else : 0 end : UNION ALL : SELECT t.d+ 1, t.n + case when day(t.d +1)%3=0 then 1 else
|
y****w 发帖数: 3747 | 37 maxrecursion
http://msdn.microsoft.com/en-us/library/ms181714.aspx
else
else
【在 y****9 的大作中提到】 : : then : then : Learning SQL server now :-) Based on your sql structure; the following : works in SQL Server 2005 : WITH t(d,n) : AS ( SELECT getdate(), case when day(getdate())%3=0 then 1 else : 0 end : UNION ALL : SELECT t.d+ 1, t.n + case when day(t.d +1)%3=0 then 1 else
|
y****w 发帖数: 3747 | 38 我刚去写了一下那个isbusinessday,这个好像要麻烦多了。可以开个新贴做。
90
【在 B*****g 的大作中提到】 : 木有table。从system date后数N天business day,显示出来。有table就是老问题,90 : %的数据库版SQL问题可以用partition by解决。
|
B*****g 发帖数: 34098 | 39 你们先把oracle问题解决,我现在很郁闷。其实我下一个SQL是前100个质数,假设没有
判断质数function
【在 y****w 的大作中提到】 : 我刚去写了一下那个isbusinessday,这个好像要麻烦多了。可以开个新贴做。 : : 90
|
y****9 发帖数: 144 | 40
I asked the below question in Oracle-L, let's if any useful info can be
obtained during weekend.
---
Don't understand why the below query does not give me the next 10 days
instead get past 10 days. ( also noticed without the cast it does not work
at all)
SQL> select * from v$version
2 ;
BANNER
----------------------------------------------------------------------------
----
Oracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64bit Production
PL/SQL Release 11.2.0.2.0 - Production
CORE 11.2.0.2.0 Production
TNS for Linux: Version 11.2.0.2.0 - Production
NLSRTL Version 11.2.0.2.0 - Production
SQL> with t(d,n) as
2 (
3 select cast ( sysdate as date) , 1
4 from dual
5 union all
6 select t.d + 1, t.n +1
7 from t
8 where t.n < 10
9 )
10 select * from t
11 ;
D N
------------------- ----------
2012-01-06 20:35:28 1
2012-01-05 20:35:28 2
2012-01-04 20:35:28 3
2012-01-03 20:35:28 4
2012-01-02 20:35:28 5
2012-01-01 20:35:28 6
2011-12-31 20:35:28 7
2011-12-30 20:35:28 8
2011-12-29 20:35:28 9
2011-12-28 20:35:28 10
10 rows selected.
【在 B*****g 的大作中提到】 : 你们先把oracle问题解决,我现在很郁闷。其实我下一个SQL是前100个质数,假设没有 : 判断质数function
|
|
|
B*****g 发帖数: 34098 | 41 我也试过了 to_char(to_date(t.d, 'mmddyyyy')+1, 'mmddyyyy'), 同样是递减的
没有
【在 y****9 的大作中提到】 : : I asked the below question in Oracle-L, let's if any useful info can be : obtained during weekend. : --- : Don't understand why the below query does not give me the next 10 days : instead get past 10 days. ( also noticed without the cast it does not work : at all) : SQL> select * from v$version : 2 ; : BANNER
|
y****w 发帖数: 3747 | 42 手头没oracle环境。 质数感觉要复杂,每次计算的不是上次增量,
【在 B*****g 的大作中提到】 : 你们先把oracle问题解决,我现在很郁闷。其实我下一个SQL是前100个质数,假设没有 : 判断质数function
|
v***e 发帖数: 2108 | 43 从4开始, 先找composite, 一直到N, when N - #of composite = 98
然后{2...N} minus 所有composite
【在 y****w 的大作中提到】 : 手头没oracle环境。 质数感觉要复杂,每次计算的不是上次增量,
|
y****9 发帖数: 144 | 44 Got a reply from Oracle-L
------------------------
Looks like you are hitting Bug 11840579 :-) It looks weird, but if you
reference the date column twice it returns the expected data:
SQL> with t(d,n) as
2 (
3 select cast ( sysdate as date) , 1
4 from dual
5 union all
6 select decode(d,null,d,d +1), n+1
7 from t
8 where n<10
9 )
10 select * from t;
D N
--------- ----------
07-JAN-12 1
08-JAN-12 2
09-JAN-12 3
10-JAN-12 4
11-JAN-12 5
12-JAN-12 6
13-JAN-12 7
14-JAN-12 8
15-JAN-12 9
16-JAN-12 10
10 rows selected.
Another discussion on the same problem:
https://forums.oracle.com/forums/thread.jspa?threadID=1055057
Cheers,
Mihajlo
------------------------------------
【在 y****9 的大作中提到】 : : I asked the below question in Oracle-L, let's if any useful info can be : obtained during weekend. : --- : Don't understand why the below query does not give me the next 10 days : instead get past 10 days. ( also noticed without the cast it does not work : at all) : SQL> select * from v$version : 2 ; : BANNER
|
y****9 发帖数: 144 | 45
惭愧,看不懂。数学,算法的很差。还是喜欢next N biz day SQL problem.
【在 v***e 的大作中提到】 : 从4开始, 先找composite, 一直到N, when N - #of composite = 98 : 然后{2...N} minus 所有composite
|
y****w 发帖数: 3747 | 46
I mean do it with one SQL....
【在 v***e 的大作中提到】 : 从4开始, 先找composite, 一直到N, when N - #of composite = 98 : 然后{2...N} minus 所有composite
|
B*****g 发帖数: 34098 | 47 thanks,what is oracle-l?
work
【在 y****9 的大作中提到】 : Got a reply from Oracle-L : ------------------------ : Looks like you are hitting Bug 11840579 :-) It looks weird, but if you : reference the date column twice it returns the expected data: : SQL> with t(d,n) as : 2 ( : 3 select cast ( sysdate as date) , 1 : 4 from dual : 5 union all : 6 select decode(d,null,d,d +1), n+1
|
y****w 发帖数: 3747 | 48 l==list, I believe.
【在 B*****g 的大作中提到】 : thanks,what is oracle-l? : : work
|
y****9 发帖数: 144 | 49
It is a popular mailing list among Oracle DBAs. Some big names like Jonathan
Lewis, Tanel Poder, Cary Millsap etc occasionally answer questions there.
http://www.freelists.org/post/oracle-l/Oracle-11g-CTE-recursive
BTW I just googled a sql that could give all prime numbers within 10000,
will see if it can be adapted to find next N prime number problem if get
some time
http://digitalbush.com/2010/03/16/prime-numbers-as-a-sql-query/
【在 B*****g 的大作中提到】 : thanks,what is oracle-l? : : work
|
B*****g 发帖数: 34098 | 50 这个oracle论坛帖子一个ACE问,一个ACE direcoter + 一个ACE + 一个Guru(7800点)
答,哈哈。
Jonathan
【在 y****9 的大作中提到】 : : It is a popular mailing list among Oracle DBAs. Some big names like Jonathan : Lewis, Tanel Poder, Cary Millsap etc occasionally answer questions there. : http://www.freelists.org/post/oracle-l/Oracle-11g-CTE-recursive : BTW I just googled a sql that could give all prime numbers within 10000, : will see if it can be adapted to find next N prime number problem if get : some time : http://digitalbush.com/2010/03/16/prime-numbers-as-a-sql-query/
|
|
|
y****w 发帖数: 3747 | 51 这年头用list,newsgroup的人越来越少了。
点)
【在 B*****g 的大作中提到】 : 这个oracle论坛帖子一个ACE问,一个ACE direcoter + 一个ACE + 一个Guru(7800点) : 答,哈哈。 : : Jonathan
|
B*****g 发帖数: 34098 | 52 我前几天在linkedin上加了10几个group,现在每天都要减少灌水时间去读email。那个
oracle pro没人理我,我也懒的理他们,group太多了
7800
【在 y****w 的大作中提到】 : 这年头用list,newsgroup的人越来越少了。 : : 点)
|
y****w 发帖数: 3747 | 53 I defined some filter, so all linkedin emails will pass the inbox and don't
bother any more. suppose you're also using gmail.
【在 B*****g 的大作中提到】 : 我前几天在linkedin上加了10几个group,现在每天都要减少灌水时间去读email。那个 : oracle pro没人理我,我也懒的理他们,group太多了 : : 7800
|
B*****g 发帖数: 34098 | 54 一天一封,看看他们都在干嘛,一半是招工广告
t
那个
【在 y****w 的大作中提到】 : I defined some filter, so all linkedin emails will pass the inbox and don't : bother any more. suppose you're also using gmail.
|
y****9 发帖数: 144 | 55 Oracle-L is really a very high quality mailing list,especially for Oracle
DBAs, highly recommended if you are an Oracle DBA. I subscribe it with my
working email account, and filter the oracle-l email to a special folder, so
it is easy to follow ( but I only post with my yahoo account). Anyway I
check my working email very often, so it is convenient for me to read Oracle
-L meassage at the same place while reading working emails.
【在 y****w 的大作中提到】 : 这年头用list,newsgroup的人越来越少了。 : : 点)
|
v***e 发帖数: 2108 | 56 I see. 不过这也太坑爹了吧,DBA面试就问这个?
我想Oracle DB group的人也没几个能直接写出来的。
【在 y****w 的大作中提到】 : I defined some filter, so all linkedin emails will pass the inbox and don't : bother any more. suppose you're also using gmail.
|
y****w 发帖数: 3747 | 57 还好吧 会递归不算过份 比如算这个日子 在数据库开发中树结构处理还是很多的
但算质数肯定算变态了 不如封装一个Java或c函数
【在 v***e 的大作中提到】 : I see. 不过这也太坑爹了吧,DBA面试就问这个? : 我想Oracle DB group的人也没几个能直接写出来的。
|