Mini bucket
Thank you for your time and this is
A
There are a number of things we need to do before we can find out the underlying cause.
1. remove Kaspersky and use the built-in Defender
2 - update these drivers
BcmBusCtr_64.sys 31/05/2010 04:03:33
drxvi314_64.sys 2010-05-31 04:07:52
CLVirtualDrive.sys 2011-12-26 08:26:47
SFEP.sys 06/18/2012 03:11:46
IntcDAud.sys 19/06/2012 09:40:51
Search Google for the name of the driver
Compare the Google results with what is installed on your system to determine on what device/program, it belongs to
Visit the web site of the manufacturer of the hardware or program to get the latest drivers (do not use Windows Update or Device Manager Update Driver function).
If there are difficulties to locate them, post back with questions and someone will try and help you find the right program
The most common drivers can be found on these pages:
http://www.carrona.org/dvrref.php
http://SysNative.com/drivers/
Driver manufacturer links are on this page: http://www.carrona.org
If you continue to plant I would update to win 8.1
Tags: Windows
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Hi all
I'm using Oracle 11 g release 2.
I want to dynamically generate buckets of he in the graph.
For example, turnover is between 1 and 30. I want the number of buckets to be always 5. If I need to be between 10-30 from 1 to 30, I want evenly distributed in all 5. It's 10,15,20,25,30.
If my series 20-30 and number of buckets is 5 then 22,24,26,28,30.
Can it be done in a single query.
Basically, I am trying to identify the buckets in the charts. It will be always 5 buckets, but based on the lower and upper limit, the bucket must be obtained. Please let me know how to achieve this.
Thanks in advance,
jaggyamHello.
Here is an example of the use of my request thank you:
WITH d AS ( SELECT 1 id, 10 val FROM dual UNION SELECT 2 id, 13 val FROM dual UNION SELECT 3 id, 15 val FROM dual UNION SELECT 4 id, 17 val FROM dual UNION SELECT 5 id, 18 val FROM dual UNION SELECT 6 id, 20 val FROM dual UNION SELECT 7 id, 23 val FROM dual UNION SELECT 8 id, 30 val FROM dual UNION SELECT 9 id, 35 val FROM dual UNION SELECT 10 id,40 val FROM dual UNION SELECT 11 id,42 val FROM dual UNION SELECT 12 id,43 val FROM dual UNION SELECT 13 id,46 val FROM dual UNION SELECT 14 id,49 val FROM dual UNION SELECT 15 id,50 val FROM dual ), aux_intervals AS ( select (SELECT MAX(val) FROM d) - (level - 1) * FLOOR(((SELECT MAX(val) FROM d) - (SELECT MIN(val) FROM d))/(:number_buckets-1)) int from dual connect by level <= :number_buckets ), intervals AS ( SELECT NVL(LAG(int) OVER (ORDER BY int),0) low, int up FROM aux_intervals ) SELECT low, up, COUNT(*) cnt FROM d,intervals i WHERE d.val > low AND d.val <= up GROUP BY low,up ORDER BY 1; :number_buckets = 5 LOW UP CNT ---------------------- ---------------------- ---------------------- 0 10 1 10 20 5 20 30 2 30 40 2 40 50 5
I hope this helps.
Kind regards.
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Hi all
I am facing a very difficult situation and I am not able to write a query.
consider the following data
I'm writing a query that gave two values, it will give you the sum of the values in a particular bucket.SELECT 16 num, 17 num2 , 3 num3, 22 num4, 10 num5 FROM dual UNION ALL SELECT 9 num, 15 num2 , 21 num3, 2 num4, 24 num5 FROM dual UNION ALL SELECT 1 num, 8 num2 , 14 num3, 20 num4, 25 num5 FROM dual UNION ALL SELECT 5 num, 6 num2 , 7 num3, 13 num4, 19 num5 FROM dual UNION ALL SELECT 18 num, 4 num2 , 5 num3, 11 num4, 12 num5 FROM dual [http://djmannynyc.com/Capture.JPG] [http://djmannynyc.com/Capture2.JPG] [http://djmannynyc.com/Capture3.JPG]
for example, please see the attachments, highlight in yellow.
If the value 15 and 19 is given, I want to find the sum total of all the values of yellow (capture.jpg).
If the value 8 and 21 is given, I want to find the sum total of all the values in yellow (capture2.jpg)
If the value 8 and 14 is given, I want to find the sum total of all values in yellow (capture3.jpg)
Basically, when it is given to two values, I draw a square or rectangle of starting (first number) number of end and sum up all the values
can someone write a query for this scenario. I've heard this can be done with the analytical functions, but do not know how to use it.
IM using oracle 10g and 11gYou would need an extra column in your data to expressly order the lines, since there is no way to guarantee the data is queried in order, planned otherwise.
I have created an example of a control key table:
SQL> create table t as 2 SELECT 1 order_key, 16 num, 17 num2 , 3 num3, 22 num4, 10 num5 FROM dual UNION ALL 3 SELECT 2 order_key, 9 num, 15 num2 , 21 num3, 2 num4, 24 num5 FROM dual UNION ALL 4 SELECT 3 order_key, 1 num, 8 num2 , 14 num3, 20 num4, 25 num5 FROM dual UNION ALL 5 SELECT 4 order_key, 5 num, 6 num2 , 7 num3, 13 num4, 19 num5 FROM dual UNION ALL 6 SELECT 5 order_key, 18 num, 4 num2 , 5 num3, 11 num4, 12 num5 FROM dual; Table created SQL> SQL> with t2 as ( 2 select 1 as x, order_key as y, num from t union all 3 select 2 as x, order_key as y, num2 from t union all 4 select 3 as x, order_key as y, num3 from t union all 5 select 4 as x, order_key as y, num4 from t union all 6 select 5 as x, order_key as y, num5 from t) 7 -- pivot data then find coordinates 8 select sum(t2.num) selected_sum 9 from (select min(x) keep (dense_rank first order by x, y) min_x, 10 min(x) keep (dense_rank last order by x, y) max_x, 11 min(y) keep (dense_rank first order by y, x) min_y, 12 min(y) keep (dense_rank last order by y, x) max_y 13 from t2 14 where t2.num in (15, 6)) lim 15 cross join t2 16 where t2.x between min_x and max_x 17 and t2.y between min_y and max_y; SELECTED_SUM ------------ 29
You can link any 2 values in the IN clause.
I guess that your data does not allow repeated values. That would cause the SQL at the bust.
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Application of distribution for the bucket a bound column
Hello
I need connect chain for the table data and assign a number of unique to those bucket.
Here is the example (Col1 and Col2 are the table columns and bucket is aquery column):
| Col1 | Col2. Bucket |
| A1 | B1 | 1.
| A1 | B2 | 1.
| A2 | B2 | 1.
| A3 | B3 | 2.
| A4 | B3 | 2.
| A4 | B4 | 2.
| A5 | B2 | 1.
In the above example you can see that there is a string A1-> B2-> A2 and A5, so they have the same number of bucket 1. It is also interesting to note that all lines containing A1 in Col1 will reside in the same compartment, and therefore A1-> B1 is assigned also bucket 1.
Another example is a String->->-> B4 A4 B3 A3 and therefore have bucket number 2.
I tried the hierarchical queries, but wasn't able to get the desired result.
Can anyone suggest an approach or a query?
Thanks in advance.
Rgds,
AKS.Hi, Aks,
We don't know what you want.
Do you want to assign the bucket such that all rows that have the same value of col1 or col2 the same value receive the same number of bucket?Maybe you want something like this:
WITH got_relatives AS ( SELECT col1 , CONNECT_BY_ROOT col2 AS relative FROM table_x CONNECT BY NOCYCLE col1 = PRIOR col1 OR col2 = PRIOR col2 ) , got_bucket AS ( SELECT col1 , DENSE_RANK () OVER ( ORDER BY MIN (relative) ) AS bucket FROM got_relatives GROUP BY col1 ) SELECT x.* , b.bucket FROM table_x x JOIN got_bucket b ON x.col1 = b.col1 ;
I hope that answers your question.
If not, post CREATE TABLE and INSERT to your sample data and results statements you like from these data (otherwise what you've posted already).
Explain, using specific examples, how you get these results from these data.Always tell what version of Oracle you are using. This is particularly important with CONNECT BY queries, because each version since Oracle 7 has had major improvements in this area.
You will answer sooner if you always include this information whenever you have a question.
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What are the factors or business cases that should be considered in the choice of planning that we should go?
Any help would be appreciated
My Client is interested in the above 2 methods
Thank you
MahendraIt of a broad question and is not easy to answer.
One major difference is your philosophy.
Kanban (if properly implemented) is pull based system.
Min - max is always anxious that is it looks at your current onhand, future demand, future supply and then comes up with a purchase requisition.But there is no hard rule.
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Kanban is a bit more complicated. So if your organization is not not mature enough or requires a simpler solution, min - max will work better.Sandeep Gandhi
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If your computer does not meet requirements to install the above, it can still meet the specifications to install Lion.
You can buy Lion to the Apple store. The cost is $19.99 (as it was before) plus taxes. It's a download. You will receive an email with a redemption code that you then use on the Mac App Store for download Lion. Save a copy of this installer to your download folder because Setup deletes itself at the end of the installation.
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