Number of nodes in the Group Rep

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• Odd number of nodes in the Cluster controller NSX

  Why odd number of knots are required in the controller NSX Cluster?

  The election of the master for each role requires a vote by a majority of all active and inactive nodes in the cluster. This is the main reason why a Cluster controller must always be deployed based on an odd number of nodes.

  See Page 10-11 of this guide for more information: https://www.vmware.com/files/pdf/products/nsx/vmw-nsx-network-virtualization-design-guide.pdf

• How to limit the small number of users in the group to update the fields in the table of the adv

  I have integrated a custom in customer support specialist area. Custom zone advanced table with some columns editable.
  
  My cusotmization works fine, but only some users need access to
  addmorerows - button in SD in my rn custom table
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  These users are in the same group of resources than others, but some people need more energy than others...
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  Thank you

  Hello

  Amol, I could filter the users and rendered all property, but they would add more users in the future, and this change would require the amendment of the code and therefore recompile

  You can create a list of choices and store valid usernames here.

  And in the application process first enter the username.

  User String = pageContext.getUsername ();

  Now check whether or not this user is in this search

  Create a prepared statemnt as select count (*) in the fnd_lookups where lookup_type = 'test' and lookup_code =: 1

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  In this way, you have no need to recompile any code... just add new values in search of...

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• The number of nodes of BPEL table + xpath

  Hello
  
  Try to get the size or the number of the group to the node, but get the error of selection, an idea?
  
  below variableentree date node < ns1:error > is table and I use the xpath query,
  County (bpws:getVariableData('outputVariable','payload','/ns1:transaction/ns1:exceptions[1]/ns1:error[1]')) to count the number of nodes of the error, but get selection error
  
  < status = taskId 'dia' = '1' txnPriority = '1' >
  < ns1:txnIdentification >
  DG of < ns1:txnId > < / ns1:txnId >
  DG of < ns1:instanceId > < / ns1:instanceId >
  DG of < ns1:processName > < / ns1:processName >
  DG of < ns1:branchCode > < / ns1:branchCode >
  Hg < ns1:moduleCode > < / ns1:moduleCode >
  godu < ns1:currentUser > < / ns1:currentUser >
  Hg < ns1:txnComment > < / ns1:txnComment >
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  Hg < ns1:operation > < / ns1:operation >
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  < ns1:etype > 1 < / ns1:etype >
  < ns1:edesc > 1 < / ns1:edesc >
  < / ns1:error >
  < ns1:error >
  < ns1:ecode > 1 < / ns1:ecode >
  < ns1:etype > 1 < / ns1:etype >
  < ns1:edesc > 1 < / ns1:edesc >
  < / ns1:error >
  < / ns1:exceptions >
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  Thank you
  < ns1:prevRemarks > fg < / ns1:prevRemarks >
  < ns1:currRemarks > fg < / ns1:currRemarks >

  Ora:countNodes('outputVariable','payload','/ns1:transaction/ns1:exceptions/ns1:error)

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  N

• Performance issues with large number of nodes

  I am creating an application to display graphics (large), for example:

  

  But I ran into some performance issues, even for a relatively small number of nodes in the scene graph (+-2000 in the picture above). The graph is built, step by step, adding circles and paths to a StackPane. Circles and paths can be semi-transparant. For a small number of nodes, I get a solid 60 FPS, but this decreases over time to about 5 frames per second. As soon as I stop adding new nodes, the framerate shoot again up to 60 images per second. The framerate drop even when all the nodes are outside the viewport.

  My questions are:

  * Is Platform.runLater () call to 2000 times a minute too?

  * This might just be a problem with my graphics card? (I have an Intel HD Graphics 3000)
  * JavaFX pulse recorder says such things, are there meaningful information in that I'm missing?

  PULSE: 1287 [163MS:321MS]

  T14 (0 + 0ms): col of CSS

  T14 (0 + 5ms): layout pass

  T14 (6 + 152ms): waiting for the minutes of the previous

  T14 (158 + 0ms): copy the State for graphic rendering

  T12 (159 + 0ms): dirty opts calculated

  T12: Path of the slow form for null

  T12 (159 + 160ms): painted

  T12 (319 + 2ms): Presentable.present

  T12 (321 + 0ms): completed the presentation of painter

  Counters:

  Background image of the region used cached: 14

  NGRegion renderBackgroundShape slow path: 1

  Nodes displayed: 1839

  Nodes visited during rendering: 1840

  Kind regards

  Yuri
  

  Basically, try some optimization of performance ranging from the simple to the complex.   Each of the changes below may provide you with an increase in performance.  Some will probably increase performance much more that others (depending on where is the real bottleneck).  I would probably start by replace the paths to the lines and reduce the number of Platform.runLater calls (as adding nodes that fall within the viewport can be difficult only).

  > The framerate drop even when all the nodes are outside the viewport.

  Place the nodes in the graph, which fall inside the viewport.

  > Is Platform.runLater () call to 2000 times a minute too?

  Yes, there is no reason to call it more than 60 times a minute when the framerate of JavaFX is capped at 60 frames per second by default.

  > This might just be a problem with my graphics card? (I have an Intel HD Graphics 3000)

  Yes, it's a graphics system relatively low-end.  But, look at the comment of developer below - your CPU and choice of graphic primitives can also affect rendering speed.

  > Adding circles and the paths to a StackPane

  You don't need a StackPane to this, a group is a container easier and probably better.

  > may be semi-transparant

  Remove transparency * may * cause acceleration.

  ----

  You run may be:

  https://JavaFX-JIRA.Kenai.com/browse/RT-20405 : improve the rendering of path performance

  Maybe if you use lines rather than the paths, performance might improve.

  A comment by a developer on this performance tweak application is:

  «It is quite normal for applications that use arbitrary paths (if the node path objects, SVGPath, polyline, or polygon) because these paths are rendered in software.» As card circle, Ellipse, line and Rectangle very primitive forms easily to the operations that can be performed entirely on the GPU, which makes them essentially cheap. There is no need to compare the rendering of a complicated shape for rendering of simple primitives for this reason. »

  ----

  Setting the cache indicators can help, but probably only if you animate nodes.

  ----

  Present level of detail of your graph functionality so a chart with zoom out is not make as many nodes as a part of zoomed in.

  ----

  You run any code important calculation on the JavaFX application thread that could stall it?

  Can make you available a ftom so that others can reproduce your problems?

• Assignment of students to group learning and the creation of the Group of the learner.

  Hi all
  
  Are there standard ways to create groups of R12.1.3. I tried looking for the standard API Oracle but do not have to find them. Is there another way of backend, we can create a group of learners. I know to create from request, but we need to loose several groups of learners. I wanted to back-end.
  
  We will also need to assign the number of learners in the Group of the learner. Can you please explain the possible ways to make these assignments.
  
  All entries are the most expected.
  
  Thank you
  Abhishek.

  Hi Abhishek,

  You can use "Compulsory registration" If you're working with an online offer. It will automatically register all learners in a learning group.

  Best regards Anders Northeved

• How to count the number of nodes under each parent in any given xml

  How to count the number of nodes under each parent in any given xml. for example the xml below was
  books has 3 childern, library [1] has 4, [2] library has 6 and bookshop [3] has 2. is it possible to get the number of tags in a childnode duplicate IE library [1]
  Book1 Tagus repeated twice... vice versa. do we need to make plsql lie we can achieve through sql
  
  < book >
  < library >
  ABC < book1 > < / book1 >
  BCA < book2 > < / book2 >
  ACR < book1 > < / book1 >
  Lac < Book4 > < / Book4 >
  < / book >
  < library >
  ABC < book1 > < / book1 >
  BCA < book2 > < / book2 >
  ACR < book3 > < / book3 >
  ACR < book3 > < / book3 >
  tray of < bookn_1 > < / bookn_1 >
  adjusted cost base < bookn > < / bookn >
  < / book >
  < library >
  ABC < book1 > < / book1 >
  BCA < book2 > < / book2 >
  < / book >
  < / books >
  
  
  I tried this... query.
  
  Select
  XMLQUERY ('count($doc/Books/Bookstore[1]/descendant::*)' in the way of xmltype ("< books >
  < library >
  ABC < book1 > < / book1 >
  ACR < book1 > < / book1 >
  Lac < Book4 > < / Book4 >
  < / book >
  < library >
  ABC < book1 > < / book1 >
  BCA < book2 > < / book2 >
  < / book >
  (< / books > ')
  as 'doc' of happy return) .getNumberVal () as node_count
  of the double
  
  Select
  XMLQUERY ('count($doc/Books/descendant::*)' in the way of xmltype ("< books >
  < library >
  ABC < book1 > < / book1 >
  ACR < book1 > < / book1 >
  Lac < Book4 > < / Book4 >
  < / book >
  < library >
  ABC < book1 > < / book1 >
  BCA < book2 > < / book2 >
  < / book >
  (< / books > ')
  as 'doc' of happy return) .getNumberVal () as node_count
  of the double
  
  How can I get the counts for each parent in a single query
  
  Published by: user7955917 on August 24, 2012 07:26

  Generic function how you want the query to be.

  If the structure is known in advance, as in your example a 'books' and then a 'library' root element, it's as simple as:

  SQL> select x1.parent_id, x2.child_name, count(*)
    2  from tmp_xml t
    3     , xmltable(
    4         '/books/bookstore'
    5         passing t.object_value
    6         columns parent_id   for ordinality
    7               , child_list  xmltype path '*'
    8       ) x1
    9     , xmltable(
   10         '/*'
   11         passing x1.child_list
   12         columns child_name varchar2(30) path 'name()'
   13       ) x2
   14  group by x1.parent_id, x2.child_name
   15  order by x1.parent_id, x2.child_name
   16  ;
  
   PARENT_ID CHILD_NAME                       COUNT(*)
  ---------- ------------------------------ ----------
           1 book1                                   2
           1 book2                                   1
           1 book4                                   1
           2 bookn                                   1
           2 bookn_1                                 1
           2 book1                                   1
           2 book2                                   1
           2 book3                                   2
           3 book1                                   1
           3 book2                                   1
  
  10 rows selected
   
  

  If you want a generic solution that works without knowledge of the structure, you will need a recursive approach, and most importantly, you should know which nodes in distinct, I suppose that the leafs?

  Also, please help if you can provide the link of reference on the function name() years what are the other expression, that I can use after the path

  You can start reading the documentation: http://docs.oracle.com/cd/E11882_01/appdev.112/e23094/xdb_xquery.htm#CBAGCBGJ

  The clause PATH expects a XQuery expression however before 11.2, we can put only simple XPath expressions.

• Error FPGA - object internally in pipeline not connected to the sufficient number of nodes comments

  Hi all

  I'm currently implementing LMS on module FPGA of myRIO 1900. I couldn't fully understand the following error.

  "Reading memory: internal pipeline object not connected to the sufficient number of nodes feedback"

  Details:

  "The selected object has a built-in shift register that makes the output on a particular loop iteration correspond to the entries in the previous iteration.

  Connect the outputs of the object directly to a minimum number of nodes of Feedback or uninitialized shift registers. You cannot connect the outputs to another object. "

  I just tried to access the values of two briefs and add them. The address for two memory values are the same, but the memories are different. I tried to implement this on SCTL, it doesn't but on normal everything in a loop, it works.

  Help me understand the error.

  Thank you.

  Post an excerpt of your code would help us to be more confident, but I think that you run in the error described in the second note of this help documentation.

  http://zone.NI.com/reference/en-XX/help/371599L-01/lvfpgaconcepts/fpga_memory_items/

  Using the parameters of memory default point that you have to attach a feedback with a x 2 node latency on the release of reading to use within a single cycle timed loop.

• Calculate the number of nodes in a version

  Hi gurus,

  Need a little help. Is it possible to discover a total number of nodes (all hierarchies combined) in a version?

  Kind regards

  Olivier M.

  Hi Olivier,.

  Run these queries on the main tables of DRM.

  All nodes in a version (inclusive orphans)

  SELECT COUNT (1)

  OF DRM_Schema.RM_NODE

  WHERE I_VERSION_ID = XXX

  NOTE- Where DRM_Schema represents the name of the repository of DRM and XXX is the version id that you can get if you do a right click on the version and check its properties.

  
  

  Total number of orphans in the Version-

  SELECT COUNT (1)

  OF DRM_Schema.RM_NODE

  WHERE I_VERSION_ID = XXX

  AND I_NODE_ID NOT IN (SELECT I_CHILD_ID

  OF DRM_Schema.RM_RELATIONSHIP

  WHERE I_VERSION_ID = XXX)

  Thank you

  Denzz

• Validation of DRM to restrict the number of nodes under a Parent

  Hi, I am creating a real-time validation, to limit the number of nodes under a parent to 1000. All new Member beyond 1000 should be erroneous outside. In order to achieve this, I created a property:

  VerParMemCount: String, node, derived

  Parameters: LessThanOrEqual (PropValue (Core.LeafChildren), 1000)

  This property was then used in the VerEssChldMemCountRT of Validation:

  Class: Property value is equivalent to the

  Level: node

  To check value: false

  Run this Validation: real-time batch and two

  When I test manually, it works. However in real time when I insert 1001 nodes or more from a parent, there is no notifications/errors. Any guidance how to do this job? DRM version: 11.1.2.1.107. Thanks in advance!

  Hello

  Please can you confirm - you have been assigned the validation to the level of hierarchy as the validation in real-time?

  Thank you

  Best regards

  Durai Gandhi

• Why jump nodes in my group originally from the root of the scene?

  My project code is too big to post, so I'll try to explain the problem that I am facing. I'm creating a tool of graphic layout as part of the largest visualization tool which we build.

  The problem I have is a group that I've positioned using layoutX, we jump to the origin whenever we do the graphic layout.

  Here is the basic flow of the program:

  1. We create a graph of JUNG's Circle (vertices) and Lines (edge)
  2. JUNG layout classes allows us to generate a layout for the chart. The layout generates an x, position y for each vertex in the graph in a particular area. Say a box of 400 by 400.
  3. We use the positions generated to define the properties centerX, centerY of the circles. The lines are tied to the centers of the nodes they're going between, so they update automatically.

  All this is done to present the original graphic layout. The user can select nodes and choose them to be prepared in a new way. This creates a sublayout on the Web. When the circles are selected move us to a group position in the exact spot where they were selected. The groups are made adjustable and can be moved across the canvas. Update the lines and circles do not move compared to the benchmark of their parent group. Only their container is moved inside sceneRoot.

  The scengraph looks like:

  VBox

  Group: sceneRoot

  Circles and lines that make up the original graph

  Group: sublayout

  Circles and selected lines

  The problem comes when we try to make the layout of the selected nodes. We generate a subgraph of JUNG. Then we make the graphical layout with the subgraph to the circles that it contains new positions. We ask the page layout to generate a layout, the size of the sublayout group. This method works. JUNG generates new values for the positions of the circles. When we created the centers of the circles, the group goes to the top left corner - origin of the sceneRoot. In my view, the Group should not move at all. Is layoutX, there have been established and the circles are positioned relative to origin of sublayout. I can't understand why the group jumps to the origin. It should stay in place.

  I thought about it!

  The problem was by using a group as the root of the Web. Limits of the group are the sum of the limits of his childhood. This means that the limits of the group will change according to what is in it. When the VBox kicked out, it would move because the new bounding box. The fix was to use a component. A pane is resized according to the rules of pref/min/max and does not fear for the sum of its children. This is one containing more "solid".

• How to display the current page and the total number of pages for a group

  Hi all
  I have a requirement to indicate the current page number and the total number of pages for a group. For example:
  
  < A >
  < 1 >
  Row1
  row2
  ....
  Clotilde
  < / 1 >
  < 2 >
  Row1
  row2
  ....
  Clotilde
  < / 2 >
  < 3 >
  Row1
  row2
  ....
  Clotilde
  < / 3 >
  < /A >
  
  Now my requirement is Group 1 can have lines going into several pages say x, so I need to show with in the Group 1 x and on the next page page 1 page 2 x etc.
  Although Group 2 spans several pages, I need to see page 1 of y... etc.
  
  Can someone help me please how to do this?
  
  
  Thank you
  Sunny

  Try if@section

• How to get the Total number of nodes XML?

  Hi all

  I have a Flash program that I do in Actionscript 3, using CS6.

  I use the XMLSocket class to read XML data. I'll write some examples of XML that is sent to the Flash

  program below...

  I know with this line here (below) I can access the 4th 'element or node' of XML data.

  To access the XML nodes/elements:

  * I created a Variable XML, called xml, and "e.data" contains ALL the XML data
  var XML = XML (e.data);

  Access to the 4th item of the data:

  . xml MESSAGE[3].@VAR;          -> 'loggedOutUsers '.

  . xml MESSAGE[3].@TEXT;         --->     "15"

  EXAMPLE OF XML DATA:

  
  

  < FRAME >

  0 < VAR MESSAGE = "screen2Display" TEXT = "FRAME_1" / >

  1 < VAR MESSAGE = "numUsers" TEXT = "27" / >

  2 < VAR MESSAGE = "loggedInUsers" TEXT = "12" / >

  3 < VAR MESSAGE = "loggedOutUsers" TEXT = "15" / >

  4 < VAR MESSAGE = "admins" TEXT = "2" / >

  < / FRAMEWORK >

  I'm new to Flash and Actionscript but I am very familiar with other languages and how paintings and other work, and I know for

  example, in a Shell Script to get the total number of items in an array called 'myArray' I wanted to write something like

  This-> ${#myArray [@]}. And since the data processing XML looks a lot like a table I thought that there was perhaps

  a way to access the total number of "nodes or elements" in the XML data...?

  Any thoughts would be much appreciated!

  Thanks in advance,

  Matt

  even if you don't need to run to the loop to know the length

  var levelList: XMLList = xml.children ();

  levelList.length (); This will give you the length of all children in your case of length MESSAGE of nodes;

  Pls click useful if my answer is helpful.

  Thank you

  Bala

• Residential group number! I can't change the password or leave the group. Convenience store can not solve the prob

  I did a clean install of OS win7 ultimate. If I try to change the PW, it says that windows cannot configure a homegroup on this computer... but still, he said that this computer belongs to a homegroup. If I try to leave the homegroup, it says windows could not remove the computer from the group. Troubleshooting, declare that the Peer Name Resolution Protocol service or peer network identity Manager is not open. If I go in services, it shows that the NIBPM is automatic and the PNRP protocol is manual. If I try to start the PNRP protocol, it is said that windows didn't start the service because of the error code # 0 x 80630801. Peer Networking Grouping also gets error code # 1068.

  Hello

  On the Start Menu, open the control panel and open Folder Options.  Click on the view tab, check the radio button titled "Show hidden files, folders and drives" and out the box "Hide extensions of known file types".  Click on apply to apply these changes.

  Then go to C\Windows\ServiceProfiles\LocalService\AppData\Roaming to Local disk and open the PeerNetworking folder.  Inside, you should see a file called idstore.sst.  Delete this file.  Close this window and return to Folder Options then re - hide hidden files folders and drives and extensions of known file types.

  Restart the PC and let Windows load completely.

  Now, check if the following services appear as shown below, and if they are not, if you can set them up now.

  Peer Networking Grouping this should see the service type as Manual and the service is expected to show that started.

  HomeGroup Provider should show as Manual startup type, and the service should show as started.

  Peer name resolution should see commissioning type as Manual and the service is expected to show that started.

  Peer Networking Identity Manager should display type as Manual startup and service should show as started.

  If they are all able to start, or have already started, see if Homegroup now works properly.

  Kind regards

  DP - K

• Who decides the surviving node during the detection of split brain?

  How oracle decides which node be deported upon detection of split brain when all nodes have written in the voting record, but unable to check the heart rate through interconnection.

  Ajay

  Hi Ajay,

  the complexity of this depends on the amount of nodes in the cluster and how is isolated (and how), but generally speaking "the cluster creates an invisible group (cohorts) between nodes, which could potentially cause corruption and should be resolved. » Oracle Clusterware manages the "split brain" scenario in putting an end to all nodes in the smallest cohort. "If the cohorts are the same size, the cohort with the smallest number node inside survives." (Cited in Apress - Expert Oracle RAC diagnostic Performance and Tuning by Murali vandenberghe)

  Best regards

  Stefan Koehler

  Researcher and consultant independent of Oracle performance

  Twitter: @OracleSK