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• BB10 best way to create two lists view

  Hello!

  I am doing an application with two lists, something like a file browser. A single list have records and the other have files.

  When I type on a folder on the first list, a second is updated with the files.

  The playbook, I show the two lists, a narrow with folders on the left and the one with the files on the right, but this type of display is not very usable on smartphone.

  What do you think would be the best solution in this kind of scenario?

  How would you present two lists discovered?

  Page 1A records (list). Select a folder and it slides open page with files (list) 2. An action on the bottom bar would have a back button to return to the folders. Additional action buttons allow the actions to perform on the selected file.

• Can a selection list cause two lists of parent of Cascade?

  4.2.1

  THM:2

  Hello world

  I have two regions

  Region 1 has a selection list and a text called order box and category of the order. There could be several orders in a category. When I choose a selection list order, he fills the text box with the corresponding category by using dynamic "SetValue".

  Now, I also want the second region that has an another LOV orders, fills only order_ids not selected in region 1 LOV however, it should show only other orders in the same category.

  So something like select commands from P4_ORDERS1 where category and <>P4_ORDERS order_id = P4CATEGORY

  But seems to not work. I created this on Apex

  https://Apex.Oracle.com/pls/Apex/f?p=64292:4

  workspace: ryansun

  user: [email protected]

  PWD: ryansun123

  Ideally, when I select the No. 1 of the order of P4_orders, P4_ORDERS1 should show 4 for the other order_ids, that p4_orders1 will be null.

  Thank you

  Take at look at page 5 and 6.

  Page 5 the query for the LOV P5_ORDERS1 was modified to use the value of P5_ORDERS.  Not really necessary to include the value of P5_CATEGORY

  Page 6 the parent element is P6_CATEGORY for the LOV is updated when the value of P6_CATEGORY is modifed by the dynamic Action.  Dynamic Action will present the P6_ORDERS value in session state to be used by the LOV P6_ORDERS1.

  As for the original question in the title, a list of selection may have two lists of parent of cascade? The answer is Yes.  You can include 2 (separate control) in the area of the waterfall LOV article Parent (s).

  -Jeff

• How to merge two lists thanks to a rational solution alphabetically?

  Having two lists of legends he both languages and need to merge in alphabetical order, but this solution is very poor.

  (list x)

  Imperial palace, Barcelona, 2012

  CEMENTERIO, 2009

  (list a)

  Imperial palace

  Cemetery

  to get:

  Imperial palace, Barcelona, 2012

  Imperial palace

  CEMENTERIO, 2009

  Cemetery

  A very naïve method:

  · List number

  · convert numbering to text

  · in the list there add a constant to sort in the order:

  1. Imperial Palace, Barcelona, 2012

  2 Cementerio, 2009

  Imperial palace of 1A.

  Cemetery of 2A.

  and sorting:

  1. Imperial Palace, Barcelona, 2012

  Imperial palace of 1A.

  2 Cementerio, 2009

  Cemetery of 2A.

  Remove the code:

  Imperial palace, Barcelona, 2012

  Imperial palace

  CEMENTERIO, 2009

  Cemetery

  (the italics are not required)

  Thank you.
  

  Without thinking much, here's an easy way:

  1st list: convert to table (Tab1) + add a second column to the right.

  2nd list: convert to table (Tab2).

  As I understand it, there are the same number of cells.

  Copy Tab2 instead of the second column of Tab1.

  Convert the resulting table (Tab1-2) in the text

  Search: tab, to be replaced by: \r

• Merge two lists?

  I need to know if it is possible to merge the two lists using coldfusion.  I have a list downloaded from the site while they want the ability to download a second list with may have additional fields or different number of records.  Need to know if this is possible, how if it is and how the two lists will compete.  Thanks in advance for the help.

  Right, well in this case it isn't really anything to do with the merger of the lists in itself, where the confusion I think. Technically, follow these steps:

  SELECT id

  FROM MyTable

  WHERE firstname =

  AND family name =

  UPDATE mytable

  First name SET = ,

  Name = ,

  address = ,

  etc etc etc.

  WHERE id = qCheckExists.id

  It's the way you would on a technical level, operational decisions are yours. If it was John not Jon? If it was Jonathan? This isn't a programming problem, that is for you to decide - I'm not sure we can help.

  O.

• Divide an ArrayCollection collection into two lists

  I have an ArrayCollection collection I need to split into two lists based on what 'section' they in. In other words, I want all the points where the 'section' is 1 in one list and all the points where the 'section' is 2 to be in another list.

  Should I do two ArrayCollections? Filter data in a single ArrayCollection collection in the various lists? What is the best way to do this without having to store a duplicate data in memory (IE not repeat two "sections" twice)?

  Discover ListCollectionView.

  Paul

• Problem of VORowImpl - compare two lists of LINE]

  Hello world
  
  I have a huge problem right now, I am trying to compare two list VO with the same values, it's because a he tied to an advancedtable the xml page where the user can change some values and the other is the table values, but DB, the reason is in fact a validation to compare the original values from the DB with the tabla list and update only lines that users made changes in their values, here is my code for the better explain what I'm doing:
  
  {} public void saveChanges()
  OADBTransactionImpl oadbtransactionimpl = (OADBTransactionImpl) getDBTransaction ();
  Row [] rows1 = getAppVO1 () .getAllRowsInRange ();
  Row [] rows2 = getDBVO1 () .getAllRowsInRange ();
  for (int i = 0; i < rows1.length; i ++) {}
  int j = i;
  
  To compare collections
  R1: collection of the app
  R2: collection of comics
  AppVORowImpl r1 = rows1 (AppVORowImpl);
  DBVORowImpl r2 = (DBVORowImpl) rows2 [j]; <-here in the second iteration is displayed the following error message:
  "oracle.apps.fnd.framework.OAException: java.lang.ArrayIndexOutOfBoundsException: 1.
  
  string variables to check if changes in the column
  String columnAPP = r1.getId (m:System.NET.SocketAddress.ToString ());
  String columnBD = r2.getId (m:System.NET.SocketAddress.ToString ());
  
  "If" that commits the changes
  {if (! columnAPP.Equals (columnBD))}
  
  SP string variable
  StringBuilder procedureCall = new StringBuilder();
  the call to SP
  try {}
  procedureCall.append ("... stored proc...");
  OracleCallableStatement oraclecallablestatement = (OracleCallableStatement) oadbtransactionimpl.createCallableStatement (procedureCall.toString (),-1);
  oraclecallablestatement. Execute();
  getOADBTransaction () .commit ();
  
  } catch (SQLException sqlexception) {}
  System.Err.println ("SQL Exception:" + sqlexception.getMessage ());
  getOADBTransaction () .rollback ();
  throw OAException.wrapperException (sqlexception);
  } catch (Exception e) {}
  System.Err.println ("Exception:" + e.getMessage ());
  getOADBTransaction () .rollback ();
  throw OAException.wrapperException (e);
  }
  }
  
  }
  }
  
  the first iteration works fine, but the second and the futher shows an exception error, what can I do to make this method work?
  I'll be waiting for your answers, I really hope you can help me with this one
  
  Kind regards
  Mentor

  Here's the code to dump for you. Code, you should write in AmImpl and the same controller class call.

  Code AMImpl

   public void executeBothViewObjects() //Calls this method from Controller *ProcessRequest*
       {
  
          OAViewObject vo = (OAViewObject)getMainViewObjectVO1();
          OAViewObject dvo = (OAViewObject)getDBViewObjectVO1();
         if (vo != null && dvo != null)
           {
                       //1st VO
                         dvo.setWhereClauseParams(null);
                        //Set where clause if Any  dvo.setWhereClauseParam(0,xx);
                        dvo.executeQuery();   
  
                       //2nd VO
                       vo.setWhereClauseParams(null);
                        //Set where clause if Any  dvo.setWhereClauseParam(0,xx);
                       vo.executeQuery();  
  
     }
    }
  
   public void compareViewObject()   // Calling this method to compare Attribute of both View Object *Process Form Request*
    {
        OAViewObject vo = getMainViewObjectVO1();
        OAViewObject OrigVO = getDBViewObjectVO1();
  
        MainViewObjectVORowImpl rowi = null;
        DBViewObjectVORowImpl rowii = null;
  
        int fetchedRowCount = vo.getRowCount();
        int OriginalfetchedRowCount = OrigVO.getRowCount();   
  
       RowSetIterator originalSelectIter = OrigVO.createRowSetIterator("originalSelectIter");
       RowSetIterator selectIter = vo.createRowSetIterator("selectIter");
  
       if (fetchedRowCount > 0 && OriginalfetchedRowCount >0)
          {
            selectIter.setRangeStart(0);
            selectIter.setRangeSize(fetchedRowCount);
            originalSelectIter.setRangeStart(0);
            originalSelectIter.setRangeSize(OriginalfetchedRowCount);
            for (int i = 0; i < fetchedRowCount; i++)
            {
  
              rowi = (MainViewObjectVORowImpl)selectIter.getRowAtRangeIndex(i);
              rowii = (DBViewObjectVORowImpl)originalSelectIter.getRowAtRangeIndex(i);
  
             //Compare Attribute here
  
                 if((!(rowi.getRoleStartDate().equals(rowii.getRoleStartDate())) 
  
                  {
                          // Comparing Start date here of both View Object
                  }
  
               else if(((rowi.getRoleEndDate().equals(rowii.getRoleEndDate()))
                 {
                     // Comparing End date here of both View Object
                  }
            }
         }
  
  }
  

  Thank you
  -Anil
  http://oracleanil.blogspot.com/

• Safari reading list/bookmark sidebar

  Everytime I open a new window Safari I see this. This happens only on my MacBook Air, not on my iMac (the two 10.11.3 running). How can I open a new empty window without the sidebar?

  

  I should mention that this is not the mode full-screen.

• One of the two list boxes other work returns null.

  To the right,.

  I work with an earlier program not created by myself and I have a strange behavior in a list box.  There are two areas of list with the proper functioning and the other indicates a null value in the data.  When you look at the front panel, you will see two boxes on the right side.  One marked 'file for L1 Test' one 'File for L2 Test' is not working.  Even if the path in the filenames in the L2 box fail in the program and when you explore the test line, you get a "" to the probe.  No idea why one works and the other does not?

  Gary Tyrna

  It certainly seems that the control has somehow been corrupted.  Copy work on the non-working and updated the element names and it should start working.

• How can i two lists drop-down for the same column in OBIEE Dashboardprompt?

  Hello
  
  I need a report for a specific date range. I have to add 2 drop-down lists in the dashboardprompt for the same startdate and enddate column and the report must be filtered accordingly.
  
  I can't add two drop-down lists in the DATE column in the same dashboardprompt.
  
  How can I achieve this?
  
  Can anyone help?

  Hello

  You should be able to achieve the desired result by setting the operator for the column ' is between ' when you set the prompt.

  Kind regards
  Mark

• By comparing the two lists for equality

  I am trying to determine if 2 lists contain the same elements, including the times, but any order. For example:
  
  List 1: A, B, C, D, E
  List 2: A, A, B, C, D
  List 3: A, B, D, A, C
  List 4: A, B, B, C, D
  
  I consider only lists 2 and 3 are equal. I have what I think are the 3 methods of work, but I'd like your opinion on:
  
  (1) are they really correct? Do you see an obvious flaw in the algorithm.
  (2) effectiveness. I've always aspired to look at an algorithm and be able to 'see' the big - O of it.
  (3) other ideas.
  
  Also, I'm a little more confidence in the accuracy of the #1 and #2 (efficiency by derogation) and less confident on #3. I had the idea of #3 the Javadoc on the playlist hashCode(). I don't really know if it's okay.
  

  import java.util.*;
  
  public class ListCompareTest
  {
    public static void main(String[] args)
    {
      List s1 = Arrays.asList(new String[]{"A", "B", "C", "D", "E"});
      List s2 = Arrays.asList(new String[]{"A", "A", "C", "D", "E"});
      List s3 = Arrays.asList(new String[]{"A", "E", "C", "D", "A"});
      
      System.out.println("\nUsing method 1:");
      System.out.println("s1 & s2: " + isEqual(s1, s2));
      System.out.println("s2 & s3: " + isEqual(s2, s3));
      
      System.out.println("\nUsing method 2:");
      System.out.println("s1 & s2: " + isEqual2(s1, s2));
      System.out.println("s2 & s3: " + isEqual2(s2, s3));
  
      System.out.println("\nUsing method 3:");
      System.out.println("s1 & s2: " + isEqual3(s1, s2));
      System.out.println("s2 & s3: " + isEqual3(s2, s3));
    }
    
    public static boolean isEqual(List a, List b)
    {
      if (a.size() != b.size())
      {
        return false;
      }
      
      List copyB = new ArrayList(b);
      
      for (Object obj : a)
      {
        if (! copyB.remove(obj))
        {
          return false;
        }
      }
      
      return true;
    }
    
    public static boolean isEqual2(List a, List b)
    {
      if (a.size() != b.size())
      {
        return false;
      }
      
      List copyA = new ArrayList(a);
      List copyB = new ArrayList(b);
      
      Collections.sort(copyA);
      Collections.sort(copyB);
      
      for (int i = 0; i < copyA.size(); i++)
      {
        if (! copyA.get(i).equals(copyB.get(i)))
        {
          return false;
        }
      }
      
      return true;
    }
    
    public static boolean isEqual3(List a, List b)
    {
      if (a.size() != b.size())
      {
        return false;
      }
      
      List copyA = new ArrayList(a);
      List copyB = new ArrayList(b);
      
      Collections.sort(copyA);
      Collections.sort(copyB);
      
      return copyA.hashCode() == copyB.hashCode();
    }
  }

  Approach 1:
  O(n^2), because you do actually:

  for each element in A {
    for each element in B {
  

  Approach 2:
  Never use (i) get to iterate over a list. An ArrayList is very good (iteration of o (n)), but for a LinkedList is O(n^2). Use an explicit iterator (if necessary) or a foreach loop (where possible). Or, if for some reason you really want to stay with (i) get (and I don't see why you would here), then explicitly declare it as an ArrayList.

  Method 3:
  The hashCode() method will not work. Two unequal lists can easily have the same hash code.

  Unless your lists are very large, or it is very often called, approach 1 is probably good enough, and it is the most simple. In addition, it works in all cases. Will approach 2 evolve better, but only work with classes that implement Comparable (or if you provide a comparator).

• Validation between two list selection.

  I would like to validate two selection list. What I want to do is when one of the selection list is 'ATR', the selection on the other list, it must be "N/a". I'm sure you'll tell me I should use in waterfall lov parent, but the problem is that I already associated with one of them another selection list. I know that I can associate it with more than one, but it does not work. The data model allows it and change it, it will give me a lot of changes throughout the application. That's why I'm looking for another way to validate two selection list.
  
  Thank you, Bernardo.

  Hello
  I think you can also do with a validation step! In any case, it should be of type function from PL/SQL to return a Boolean.
  Validation write an expression something like that according to your needs:
  

  
  IF :P1_LIST1 = 'ATR' AND :P1_LIST2 = 'N/A' THEN
  
  RETURN TRUE;
  
  ELSE
  
  RETURN FALSE;
  
  END IF;
  
  

  Regards Garry

• Installed finishing FixIt Beta fence has two listed owners?

  This Evo n610c xp3 broke irretrievably four times the same virus of hijacking, a rat-maze of situations with occasional direct responses to situations of contrieved, immediately after the installation of Security Centre of recommended updates.  I tried to install the microsoft fix it Center Beta first, before the updates, which took another 10 minutes of .NET 2.0 installation.  Once the difficulty he Beta completed, it displays a screen with two different owners, one whose numbers match those already on this computer and the other unknown. the unknown we had .NET 4.0 and various other software that I had not installed since the last wipe and re-installation of xp.  I chose to delete.  I tried sometimes to find something on the .NET system and installation, without hope in any time-of-key.  I just wipe the number of owner and its replacement by what that either - my name maybe - in the hope of making this thing lose his obedient habits of virus.  That means two owners registration mean?  Is deleting a this correct?  The rest?  How to do the scan to work during the installation of updates Mr. begins the process of hijacking every time, and it will not scan or anything without updates?  Why does not stop analysis or filters this virus hijacking / others?  I can not yet long enough use shop for an edition of w7sp1, assuming that it will work.  Microsoft releases on these anticipates the next windows on chip which theoretically impossible to intercept viruses.  I bought a ULCPC with w6 SE on-chip that permanently frozen within 30 seconds. It is a viable portable wild Cowboy territory less than $200 or more, with enough to get most of the things; 1 GB of RAM. Speed 1 g etc. Trying to buy an old Apple with 256 RAM and upgrade somehow to run whatever it is, is too complicated for a novice. HP Compaq Evo N610c, 1.8 ghz, 512 MB of RAM (expandable to 1 GB), xp3 without updates to date (will not work).

  remove all facilities and redownload :)

• How to combine two separate lists of bookmarks?

  Under view-> sidebar-> bookmarks there are a list of L-O-N-G files and bookmarks for individual sites.

  There is ONE ENTRY for "Bookmarks Menu" and following that are TWO independent lists a set of folders and individual bookmarks, and THEN a SECOND set of records is followed by another list of individual bookmarks. None of these sets entries are duplicates of each other.

  How can I merge the two of them, while I have a set of RECORDS, followed by a set of INDIVIDUAL bookmarks - that I can sort the entire list in alphabetical order?

  Currently, when I right click on the Menu "bookmarks" I show options that include choosing "sort by name". When I choose "Sort by name" * ONLY * the first set of RECORDS & INDIVIDUAL bookmark entries are sorted in alphabetical order.

  Ideally, I would like to have the TWO lists be merged and sorted.

  How it happened (the existence of two sets of RECORDS & INDIVIDUAL bookmark lists) I don't know.

  Your help, anyone, help is greatly appreciated.

  Thank you!

  If you sort the folder Menu bookmarks, all the bookmarks below that will be sorted to the extent where a separator - a horizontal line. Remove the separators to fix everything.

• How to combine two different bookmarks lists into a single list?

  I try to combine the two lists of bookmarks. One of the IE and the other of Google Chrome. I created two HTML Firefox files for each and they are listed in my tab of Firefox bookmarks separately, but I want to combine and more written duplicate or manually delete the bookmark if there is a duplicate.

  Open the HTML files and then CTRL + SHIFT + B (or show/organize all bookmarks) then open the folder you want to left click on the mouse for the first bookmark , then shift + down arrow on the keyboard to catch everything then in Organize menu choose transfer (or move) , select the folder you want and click OK.

  Exit Firefox and restart.

  Thank you

  Please check 'Resolved' the answer really solve the problem, to help others with a similar problem.