Adobe Lightroom spend 5.7 to 6 without the use of CC

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• without the use of union or union of all the

  Hello experts

  I have the following data below

  create the table t_me

  (

  first_one varchar (1000).

  second_one varchar (1000).

  third_one varchar (1000).

  week_needed varchar (1000)

  );

  insert into t_me values ('A', 'B', 'C', ' week 1 "');

  insert into t_me values ('A', 'B', 'C', ' week 1 "');

  insert into t_me values ('B', 'B', 'C', ' week 1 "');

  insert into t_me values ('C', 'A', 'C', ' week 2 "");

  insert into t_me values ('C', 'A', 'C', "3 week");

  insert into values of t_me ('B', 'C', 'B', "3 week");

  output desired without the use of union

  the characters first_one, second_one, third_one week_needed

  Week 1             A:                2               0              0

  Week 1             B:               1               3              0

  Week 1             C:               0                0              3

  Week 2             A:               0                1              0

  Week 2             C:               1                0              1

  Week 3             A:               0                 1             0

  Week 3             B:               1                 1              0

  Week 3             C:               1                 0              2

  Any help is appreciated. Thank you

  Hello

  If you want something that works in your version, you should tell what version it is, especially if this version is 12 years old.

  Here's a way to make the unpivot operator and rotate in Oracle 9:

  WITH cntr AS

  (

  SELECT LEVEL AS col_num

  OF the double

  CONNECT BY LEVEL<=>

  )

  unpivoted_data AS

  (

  SELECT week_needed

  col_num

  CASE col_num

  WHEN 1 THEN first_one

  WHEN 2 THEN second_one

  WHEN 3 THEN third_one

  AS END characters

  OF t_me

  CROSS JOIN cntr

  )

  SELECT week_needed, characters

  , COUNT (CASE WHEN col_num = 1, 1 END) AS first_one

  , COUNT (CASE WHEN col_num = END 2 THEN 1) AS second_one

  , COUNT (CASE WHEN col_num = END of the 3 THEN 1) AS third_one

  Of unpivoted_data

  GROUP BY week_needed, characters

  ORDER BY week_needed, characters

  ;

• without the use of grouping sets

  Dear expert;

  
  

  I need help in the creation of this report without the use of the grouping sets. Any help is appreciated

  Week                                                          hours

  John                                                            24

  Tim                                                             1

  Mark                                                           16

  Adam                                                          10

  Subtotal (30/06/2013-06/07/2013) 66

  Adam                                                           7

  Subtotal (07/07/2013-08/07/2013) 7

  Month Total                                                   73

  Any help is appreciated. Thank you.

  see examples of data below. Thank you

  create table info_mech
  (
  
  req_name varchar(1000),
  beginwork_date date,
  endwork_date date
  );
  
  insert into info_mech values ('John', to_date('07/01/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/01/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('John', to_date('07/02/2013 10:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/02/2013 19:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('John', to_date('07/05/2013 11:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/05/2013 21:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Tim', to_date('07/02/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/02/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Tim', to_date('07/04/2013 09:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/04/2013 18:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  
  insert into info_mech values ('Mark', to_date('07/01/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/01/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Mark', to_date('07/02/2013 10:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/02/2013 19:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Adam', to_date('07/05/2013 11:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/05/2013 21:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Adam', to_date('07/08/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/05/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  
  
  
  

  Hello

  It's a good start.  Is this really what you want to GROUP BY?

  Group of (next_day (beginwork_date, 'Sunday') - 7 - next_day (beginwork_date, "Saturday))

  Looks like always, which evaluates to-6 or - 13, according to was looking for beginwork_date is a Saturday or not.  If you want something which returns a different value for each week?  I would like to use

  TRUNC (beginwork_date + 1, 'IW')

  which returns the Monday of the week even as beginwork_date.  (Always ISO weeks start on Mondays.  The 'magic number' 1 in the above expression reflects the fact that your startw week 1 day earlier, on Sunday.)

  UNION, you try will work.  Finally, you want 3 different levels of rollup, so eventually you will need a Union of 3 tracks, each with a GROUP BY clause to different.  Here's a different approach, which gets 3 copies of the same data and totals in a single GROUP BY:

  WITH cntr AS

  (

  SELECT LEVEL AS grouping_level

  OF the double

  CONNECT BY LEVEL<=>

  )

  unaggregated_data AS

  (

  SELECT THE CHECK BOX

  WHEN c.grouping_level<=>

  THEN TRUNC (i.beginwork_date + 1, 'IW')

  Week END AS

  CASE

  WHEN c.grouping_level = 1

  THEN i.req_name

  END AS req_name

  24 * (i.endwork_date - i.beginwork_date)

  As schedules

  c.grouping_level

  Info_mech I have

  CROSS JOIN cntr c

  WHERE i.beginwork_date > = DATE ' 01-07-2013'

  AND i.endwork_date<  date="">

  )

  SELECT CASE grouping_level

  WHEN 1

  THEN req_name

  WHEN 2

  THEN "subtotal (' |)" To_char (week - 1, ' MM/DD/YYYY' -)

  || To_char (week + 6, MM/DD/YYYY")')

  ANOTHER "Grand Total".

  END AS name_or_week

  The amount (in hours) AS total_hours

  Of unaggregated_data

  GROUP BY week

  grouping_level

  req_name

  ORDER a week

  grouping_level

  req_name

  ;

  This assumes that, on a given line of the table, beginwork_date is in the same week as endwork_date.  The sample data you posted include 1 row when this was not true:

  insert into info_mech values ('Adam', to_date (July 8, 2013 08:00 ',' ' DD-MM-YYYY HH24:MI:SS), to_date (July 5, 2013 15:00 ',' ' DD-MM-YYYY HH24:MI:SS));))

  Not only are the two dates in different weeks, but beginwork_date is later than endwork_date.  I assumed that was a typo and changed endwork_date to July 8.  With this change, the result I got is:

  NAME_OR_WEEK TOTAL_HOURS

  ------------------------------------ -----------

  Adam                                          10

  John                                          26

  Mark                                          16

  Tim                                           16

  Subtotal (30/06/2013-07/07/2013) 68

  Adam                                           7

  Subtotal (07/07/2013-14/07/2013) 7

  Grand Total 75

  The last line is the total of all hours for all dates included.  In the query above, this means that the month of July 2013, because I've hardcoded the dates in the WHERE clause.  You can climatiqueLes WHERE clause for what you want, or omitted completely.

  Note how much more complicated the query above is that of the solution of GROUPING SETS below:

  WITH got_week AS

  (

  SELECT req_name

  TRUNC (beginwork_date + 1, 'IW') AS the week

  24 * (endwork_date - beginwork_date) AS hours

  Of info_mech

  WHERE beginwork_date > = DATE ' 01-07-2013'

  AND endwork_date<  date="">

  )

  SELECT THE CHECK BOX

  WHEN you GROUP (req_name) = 0

  THEN req_name

  WHEN you GROUP (week) = 0

  THEN "subtotal (' |)" To_char (week - 1, ' MM/DD/YYYY' -)

  || To_char (week + 6, MM/DD/YYYY")')

  ANOTHER "Grand Total".

  END AS name_or_week

  The amount (in hours) AS total_hours

  OF got_week

  GROUP BY GROUPING SETS ((semaine, req_name))

  (week)

  , ()

  )

  ORDER a week

  req_name

  ;

  GROUPING SETS will also be more effective.

• adjust audio speed without the use of effect - the quantification in real time?

  Don't know how to explain it, but that's probably why I have difficulty finding an answer

  In other audio programs you can adjust the speed or the duration of a part of a track by grabbing a handful and make it longer or shorter.

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  thatguyjay wrote:

  Maybe it's simply not a function, you can reconcile in hearing?

  As you describe, no. Next audition happens to what either as it's the tool for automatic alignment of speech for the clips, but which is intended for ADR applications where you want to align a track of dialogue precisely with an original replacement.

  And that's the point, really - hearing is changing software aimed rather at the audio and video post-production, and is not really software music creation at all - this is why it has no MIDI and a lot of the things that go with. There has been a lot of discussions about this in the past, but Adobe has made quite clear that they follow the money with the development, and music is not just where it is. Is not to say that he cannot do any good music production - I use it a lot for the classic edition - but if I wanted to in machine - creationof music, I would definitely watch Reaper, or something like that, because they have much more appropriate facilities for this.

• Updated MDS in SOA, without the use of AIA

  Hello

  I want to update MDS in SOA. I have not installed of AIA.

  Is it possible to do?

  Thank you

  Jehanne

  Rouillard,

  Yes, you can do it without the AIA. Here are the options,

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  Hello
  
  If we are able to access the data without using a cursor as shown below in a way:
  
  
  CREATE OR REPLACE PROCEDURE tmr_exception
  IS
  x LINK_STATUS. COUNTRY_CODE % TYPE;
  
  BEGIN
  SELECT COUNTRY_CODE LINK_STATUS x;
  
  EXCEPTION
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  dbms_output.put_line (' too many lines).
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  dbms_output.put_line ("' another problem");
  
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  (2) A SELECT... INTO opens a cursor. He has simply done implicitly. Just like

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    LOOP
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  Opens an implicit cursor.

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  Justin

  Published by: Justin Cave on October 9, 2010 20:17

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• Adobe Lightroom CC license expired after subscribed to the plan of photography

  Hello Adobe community, I use Adobe Creative Cloud solutions since February.

  I subscribed the photography annual, paid monthly Plan.

  From what I could understand he should give me access to lightroom and photoshop without limit, I use the mobile lightroom without any problem at all, but in my office, I'm having some trouble with expiration dates. I downloaded applications via the tab cloud creative but they started after the evaluation period, and now that 30 days have passed, I can't enter my mac in lightroom. Any help is welcome, thanks!

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  Seems that it was a mistake hosts files, they gave me with a cleaning tool that solved the problem. Thank you!

• I can't without the use of pc to mac for my cc adobe without problem?

  I have been using adobe cc on a pc but I want to start using it on a mac. There should be no problem if I just disconnect on the pc, right?   I have a Creative Cloud Student and Teacher Edition subscription

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• What is the best way to correct the registry errors without the use of a registry cleaner in XP?

  I am sorry to raise this issue that so many people who have asked me.  I read pages and pages of discussions on the MS forum on this subject, and without exception, all the experts said Microsoft registry cleaners are completely unnecessary and that junk in the registers has no performance impact at all.  Instead, they say, to do x, y and z.  I am running Windows XP on my desk for five years, and gradually, he got so slow it's almost unbearable.  It takes 10-15 minutes just to start, and sometimes it takes 2 minutes or more just to open a window of browser (IE8) or go to the previous page by using the arrow to the left.  Before you tell me all the things it could be, please let me say that:

  1. my PC has not always been this slow!  It is getting worse.

  2. all my Windows software is up to date.  I have automatic updates on.

  3. I have a lot of disk space, almost 75% of free space on a hard drive of 80 GB.

  4. my PC is free of viruses, spyware and Trojans.  I have McAfee antivirus running and I also sometimes run Malwarebytes and Trojan Killer, because they always find things that don't have McAfee.

  5. I defragment my hard drive on a regular basis.

  6. I regularly delete all my temporary files by using the Windows Disk Cleanup utility.

  7. I have very little, if any "Add-ons".  No toolbar Yahoo or Google (or Bing!).

  8. I removed nearly every program that I start, so that they are not running in the background.

  9. I uninstalled all the programs that I don't use.

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  I can't disagree with a lot of things have already been said.

  I'm willing to accept that your machine does actually takes 10 to 15 minutes on the clock. It's completely absurd.

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  Many "experts" actually expect to, and do, re - install windows every 6-12 months!  I think it's a bit exaggerated, but they swear by it.

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• FlexVPN without the use of certificates

  Hi all

  Is there a way that we can use the Anyconnect VPN with FlexVPN clinets without authentication of certificate based (as in the old Cisco VPN clients using Group key)?

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  Shamal,

  Take a look at the doc I wrote previously:

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  The RFC for IKEv2 mentions that if you use EAP, you MUST use certificate authentication.

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  Note that you'll need to authenticate and register the trustpoint on your head of bed of this as if it were external devices.

  M.