create link db with timeout parameter

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• How to create the event with the parameter schematically

  Hello

  I created the button schematically, I want to pass the parameter for this event.

  And I must capture this setting in LICS pls help someone.

  OASB OASubmitButtonBean = (OASubmitButtonBean) pageContext.getWebBeanFactory () .createWebBean (pageContext, "BUTTON_SUBMIT");

  oasb.setID ("ZoomBtn");

  oasb.setUINodeName ("ZoomBtn");

  oasb.setEvent ("Zoomeve");

  oasb.setText ("zoom");

  webBean.addIndexedChild (oasb);

  OASubmitButtonBean zoomBt = (OAButtonBean) webBean.findChildRecursive ("ZoomBtn");

  If (zoomBt! = null) {}

  Hashtable paramsWithBinds = new Hashtable (2);

  paramsWithBinds.put ("PoHeaderId", new OADataBoundValueFireActionURL (webBean (OAWebBeanData),

  "{$PoHeaderId}"));

  FireAction = firePartialAction

  OAWebBeanUtils. getFireActionForSubmit (zoomBtn, "openCustRefDetForm", null, paramsWithBinds, true, true);

  zoomBtn.setAttributeValue (PRIMARY_CLIENT_ACTION_ATTR, firePartialAction);

  }

  But it gives the following error.

  • Error (50.9): method getFireActionForSubmit (oracle.apps.fnd.framework.webui.beans.nav.OAButtonBean, java.lang.String, null, com.sun.java.util.collections.Hashtable, boolean, boolean) could not be found in the oracle.apps.fnd.framework.webui.OAWebBeanUtils class

  Dilip salvation,

  Thanks for your replay instance it works fine.

  Good answer,

  Instead of

  import com.sun.java.util.collections.Hashtable;

  We have to import

  import java.util.Hashtable;

  Kind regards

  Sangu

• How to create links in a project of Muse to an anchor with a button created in Adobe Edge?

  Hi everyone, I created a button with edge animate, but when I import in my project Adobe Muse I can't link to anchors in Muse.

  What is the code?

  Thank you.

  Edge entertainment in an iFrame, which is technically a different page than you have placed the ILO tone on. To get the button to link to an anchor on the homepage, you will need to create a literal link (/ #anchor page.html) as an action on board. Make sure that the target attribute is set to _parent.

  who should do the trick but it won't be a smooth roll as Muse normally between anchor points.

  David

• Cannot create links with editing in the browser

  Sorry to post this here, but I'm out of options. Three of my clients are unable to create links on their sites using IBE. I tried to reach out to the IBE team on it for more than a month. I called Adobe Support the other day, that escalating the issue to the BIE team, who sent an automatic e-mail response that I was here in 24 hours, which was a week ago.

  These issues in IBE makes null and void, and I'll have to do all updates for customers of myself while apologizing constantly telling them that they would be able to make changes to their sites as themselves.

  Is there anyone on this forum who can attract the attention of the IBE team? Please note - I have tried all the normal channels.

  Thank you.

  This is a known bug and there is a current fix. Basically what is happening, it's there are two changes taking place simultaneously: 1) addition of text in a text frame and 2.) adding a link. The solution for all those who faced this problem is to add your text, click outside the text frame, click inside and add your link to the text (to break in two separate actions).

  Update (October 11, 2016): this issue has been resolved by the release of today. Adding hyperlinks to the new text should now work the first time.

• I created an account with Hightail (and received and responded to the verification email) but when I click on the down arrow next to Lacrosse "join".

  I created an account with Hightail (and received responded to the verification email), but when I click on the arrow next to the button "join" the 'link' button does not appear.

  Have you added hightail in the menu (alt + T) tools > options > Annexes > out.

• Urgent: how to create a shortcut with a relative path?

  I put a set of files and folders nested on a server on a network location.  (Use of office machines Vista; I don't know that the server is running.  The files are a text file and two Office 2007 files).  Users will copy this set-up files on their computer (there is a program of VBA macro in a Word document, they need).  They need tomorrow (Friday) and the shortcut path gives me fits!

  The hierarchy looks like:

  MainDir
  File.txt
  SubDir1
  Shortcut to the Word file
  SubDir2
  Word file
  Excel file

  I don't think even the paths associated with the shortcut until I looked a little deeper.  Because I put all the files and folders together on my computer, when I did a drag - move in the folder on the server, the target properties and start in the shortcut always pointed at my computer!  So now, I'm afraid that a user copy this on their machine will have a shortcut that points to ~ my ~ computer, instead of the file on their computer.

  I deleted the shortcut located on the server and creates a new one from the Word file in the directory on the server.  Better - I think - but it still points to the location of the server.  And it is even more likely that, when the user opens the shortcut, it will try to open the file on the server, vice the file on their computer.

  How can I create a shortcut to this Word file that will pick up the relative path from around where to reside?  I can just click > properties and change the target and start in paths?  Or do I need a VBScript or VBA?

  I have to get this done fast!  A kick - in the right direction is greatly appreciated.

  Ed

  Hi Ed,

  It is not possible to create a shortcut with a relative path.

  You can try the following allows to bypass and check if it helps:

  Create a shortcut to cmd.exe running the start command to open the current directory. Then set the MDC in the

  The shortcut properties.

  a. make a new shortcut to: cmd.exe /c start.

  b. now right-click on it and get the properties of the shortcut (shortcut tab).

  c. the shortened full path must have been set to this:

  %windir%\system32\CMD.exe /c start.

  d. set the directory of "start in" (i.e. DLG) to be a relative path:

  (Example for a shortcut on the desktop)

  ...\SendTo

  e. click on "change icon...". "and set" look for icons in this file: "to: %SystemRoot%\system32\SHELL32.dll and press ENTER.

  Standard folder icon will probably be in the lower right.

  f. now you can copy this file and simply change the "start in:" to make new related shortcuts directory.

  NOTE: The above steps are just an example, so try this at your own risk. And check if it works.

  You can also read the following article and check:

  Folder redirection fails when you specify the mapped network drive or the Environment Variable

  http://support.Microsoft.com/kb/321805

  You can check out the following link and also post your request in the Technet forum for further assistance:

  http://social.technet.Microsoft.com/forums/en-us/winserverPN/threads

  Hope this information is useful.

• tried to create an account with xbox (software on my computer) and it won't let me download anything from there. Is there a setting I need to change or take it back to the store?

  I just bought a new laptop about 4 days ago. It came with 10 Windows. The problem I have is, it does not download anything. I tried to share a youtube video, and it wouldn't let me. I tried to create an account with xbox (software on my computer), and it won't let me download anything from here. Is there a setting I need to change or take it back to the store?

  Thank you

  Michelle

  Hi Michelle,

  Thanks for posting your query on the Microsoft Community.

  According to the description, I understand that you are unable to download anything.

  I would like to know some information.

  1. You are connected to the internet?

  2. You get no matter what error code or error message?

  Please check the date and time system and check if it is up-to-date.

  If it is not set correctly, please follow the link to change the date and time:

  http://Windows.Microsoft.com/en-us/Windows-8/clock-out-of-sync

  I suggest you follow the steps to run the internet connection Troubleshooter:

  Follow the steps below:

  (a) affects Windows and x together and select Control Panel.
  (b) click on troubleshooting.
  (c) click show all in the left side panel.
  (d) click on Internet connections.
  (e) follow the on-screen instructions to resolve this problem.

  Hope this information helps. Please let us know if you need any other help with Windows in the future. We will be happy to help you.

• Create link DB in a procedure

  Hello

  I am trying to create the DB link in a procedure

  CREATE or replace PROCEDURE cre_db_lnk AS

  BEGIN

  dbms_output.put_line ('before exce');

  EXECUTE IMMEDIATE "CREATE DATABASE LINK remote_link_3"

  ||' CONNECT to c1upgrade03cm IDENTIFIED BY c1upgrade03 '

  ||' USING "sidev11g" ';

  dbms_output.put_line ('afetr exce');

  END cre_db_lnk;

  This procedure compiled successfully, but when I check in the browser schema that i cannot find the DB link with that name.

  And it is not even showing the DBMS_OUTPUT.

  But when I tried to create using simple SQL, then it was created

  CREATE DATABASE LINK remote_link_2

  CONNECT to c1upgrade03cm IDENTIFIED BY c1upgrade03

  With the HELP of "punvm-oracle11g/sidev11g";

  I can't understand why it does not create in the process what Miss me here!

  My version of DB is: -.

  Oracle Database 11 g Enterprise Edition Release 11.2.0.3.0 - 64 bit Production

  Hello

  you mention that your procedure COMPILES successfully, but you can also RUN it? When you run it, you see the 'before' as well as 'after' messages (you do SQL > SET SERVEROUTPUT ON before calling the procedure)?

  If so, the database link must be created.

  Otherwise, you should see an error message (and not the message 'after'); for example: ORA-01031: insufficient privileges

  It may be that the owner of the procedure user can create links from database only through a ROLE that you have given, but the roles are NOT active when you define a ((Eh bien, j'ai juste parler de cas standard) procedure). In this case: you can create a DB link with the immediate command in SQL * more, but not through the procedure.

  Then you will have to directly grant the privilege: LINK of DATABASE CREATE and GRANT to scott;

  Best regards

  Bruno Vroman.

• Create a rectangle with an image inside

  Hi all

  Issue of great starter... how to create a rectangle with an image inside.

  My project:

  -Create a dialog box where to put the name of the image.

  -create the link between the image to the active document (the photo is in the 'Links' folder)

  -make a rectangle in my document with an image inside.

  Think it all is ok except the last step (a start, I was spirit "which is the easier"... but no).

  Thank you very much and sorry for my English.

  to add a rectangle:

  var rect = doc.rectangles.add ();

  to place an image inside:

  Rect.place (file (image. Path));

• Failed to get the results of a report from beep with a SQL with like ' %: Parameter.

  Hello

  I am trying to create a beep with sql like query report below:

  Example:

  Select usr_key from usr where usr_login like ' %: Parameter.

  In above query parameter before the user.

  The above query returns not all the results in the report.

  If I hard top query below it returns the result-code:

  Select usr_key from usr where usr_login like '% Test % '.

  Please let me how can know we use % with a parameter in the NOSE?

  Thank you

  Hi I had the solution how to do this at RANDOM.

  CONCAT function should be used for the same

• get list from a SQL database and create links to each lines in AS3

  I'm doing an AIR application with these conditions:

  I have a SQL database. In my table, there is a column 'categories' (with different categories (computer, books... etc.)).

  

  In my AS3, I managed to recover "theDescription" when the user select a category. With the URLMethod and a php file.

     // create SQL $sql = "SELECT * FROM annonces where categorie = '$categorie'"; 
  $sql_result = mysql_query($sql, $connection) or die ("Couldn't execute query."); 
  $num = mysql_numrows($sql_result); $phptheDescription = ""; 
  $counter = 0; 
  while ($row = mysql_fetch_array($sql_result)) { 
  $theDescription = $row["theDescription"]; 
  $phptheDescription = $theDescription; 
  } 
  echo  "phptheDescription=" . $theDescription;
  

  So my AS3 code retrieves the $phptheDescription since php and displays in a output_txt .

  Problem: in my output_txt , a 'theDescription"appears. But I have TWO Articles in the category 'Computers' (and I have 100 points in the same categories).

  How do I show all the "theDescription' that are in the same categories?

  Ex: If I choose "Computer", it should display "a Surface Pro 3" and "An IMAC". But it only shows the last element "IMAC".

  And, after that, it is possible to create "links" for each item posted?

  Here are 2 videos of short films (20 sec) explaining my problems:

  https://vid.me/DS2r

  http://sendvid.com/6iesrygk

  THX

  Hi-I've never used PHP so I could be wrong here - but it seems to me you're a loop in the records, each time as reset $theDescription to all that is in the current (rather than accumulate output) record. If you are effctively echo only the last record in the query.

  I think you should be either running the command echo inside your loop, or alternatively the concatenation of a string inside the loop and then echoing the end of the loop?

  I would be likely to generate XML data in PHP (just by concatenating strings) and then use it in my AS3 code.

• ORA-20002: you can't create links from database

  HI Experts,

  Creating database link, I get the error below. Anyone of you could please have some light on this error. I googled for this error and did not get any update.

  Oracle version: 11.2.0.3

  The user with whom I created this link of database DBA privilege.

  SQL > create database link temp_LINK connect to account identified by * using "(DESCRIPTION = (ADRESSE = (COMMUNUITY=tcp.world) (PROTOCOL = TCP)(Host=myself.us.lev.com) (Port = 1525)) (CONNECT_DATA = (SID = LNDB1DB)))'; "
  create database link temp_LINK connect to account identified by * using "(DESCRIPTION = (ADRESSE = (COMMUNUITY=tcp.world) (PROTOCOL = TCP)(Host=myself.us.lev.com) (Port = 1525)) (CONNECT_DATA = (SID = LNDB1DB)))'; "
  *
  ERROR at line 1:
  ORA-00604: error occurred at the SQL level 1 recursive
  ORA-20002: you can't create links from database
  ORA-06512: at line 4

  
  

  Thank you..
  

  Well Yes, this request confirmed suspicion of SB.

  The datadictionary is your friend

• Creating web forms with Acrobat

  Is it possible to create a form with Acrobat and host it on my own server and not on Adobe?

  You need an embed code when you use a web form created in FormsCentral, but because FormsCentral doesn't host your PDF forms as it does your web forms, you do not use an embed code. Just place somewhere on your site and link to it as you would for any other type of content on your site.

• Create links for bookmarks

  
  I created a file in robohelp with all the information I need. Now, I starts the process of exporting the file in HTML format so that it can sit in a report Center and be referenced from there. I would like to be able to create links between each page bookmark placed at various locations on the report Center. When I click on the bookmark I want and create a shortcut, the link that gives me takes me to the appropriate bookmark but it does not display the table of contents on the side. Is there a way to make this link produce the bookmark, I want to be pulled upward and the table of contents with it? I don't want to use the feature where you have to click "Show" to display the table of contents. Thank you.

  Hello

  I don't know understand what you do. Especially when you say you "created a file in robohelp with information" and "begin the process of exporting the file in HTML format. This means you use RoboHelp for Word? If you use RoboHelp HTML, you are already 'in' HTML. That's what you work in.

  When you say you want to create a page to bookmark links (don't know what a bookmarked the page here). Are you want to bind RoboHelp topics to your application in a way so that if a user displays help, they click on a link and your application open? Or are you meaning to the contrary? You are eager to links in the application to open the help?

  In regards to opening a page with the complete set of frames showing, you simply link that you see in the address bar after clicking the "show" link. Then click on a link "view, then look at the address bar of your browser and note the path.

  See you soon... Rick

• How to create a StackedAreaChart with a desabscisses (xAxis) axis?

  Hi all
  
  While it is not a problem to create an AreaChart with an axis desabscisses I wonder why I can't create a StackedAreaChart with a desabscisses axis.
  
  When executing this Code I am getting and java.lang.ClassCastException (java.lang.Double cannot be cast to java.lang.String) in the last line "primaryStage.show ()";
  
  I would appreciate your suggestions.
  

  public class JavaFXCharts extends Application {
  
      final ObservableList<String> categories2 = FXCollections.observableArrayList("1", "3","6", "9","12","15","18","21","24","27","30","31");
      final CategoryAxis xAxis = new CategoryAxis(categories2);
      final NumberAxis yAxis = new NumberAxis();
      final StackedAreaChart<String, Number> sac =
              new StackedAreaChart<>(xAxis, yAxis);    
      
      
      @Override
      public void start(Stage primaryStage) {
  
          primaryStage.setTitle("Area Chart Sample");
          sac.setTitle("Temperature Monitoring (in Degrees C)");
          XYChart.Series<String, Number> seriesApril =
                  new XYChart.Series<>();
          seriesApril.setName("April");
          seriesApril.getData().add(new XYChart.Data("1", 4));
          seriesApril.getData().add(new XYChart.Data("3", 10));
          seriesApril.getData().add(new XYChart.Data("6", 15));
          seriesApril.getData().add(new XYChart.Data("9", 8));
          seriesApril.getData().add(new XYChart.Data("12", 5));
          seriesApril.getData().add(new XYChart.Data("15", 18));
          seriesApril.getData().add(new XYChart.Data("18", 15));
          seriesApril.getData().add(new XYChart.Data("21", 13));
          seriesApril.getData().add(new XYChart.Data("24", 19));
          seriesApril.getData().add(new XYChart.Data("27", 21));
          seriesApril.getData().add(new XYChart.Data("30", 21));
          XYChart.Series<String, Number> seriesMay =
                  new XYChart.Series<>();
          seriesMay.setName("May");
          seriesMay.getData().add(new XYChart.Data("1", 20));
          seriesMay.getData().add(new XYChart.Data("3", 15));
          seriesMay.getData().add(new XYChart.Data("6", 13));
          seriesMay.getData().add(new XYChart.Data("9", 12));
          seriesMay.getData().add(new XYChart.Data("12", 14));
          seriesMay.getData().add(new XYChart.Data("15", 18));
          seriesMay.getData().add(new XYChart.Data("18", 25));
          seriesMay.getData().add(new XYChart.Data("21", 25));
          seriesMay.getData().add(new XYChart.Data("24", 23));
          seriesMay.getData().add(new XYChart.Data("27", 26));
          seriesMay.getData().add(new XYChart.Data("31", 26));
          Scene scene = new Scene(sac, 800, 600);
          sac.getData().addAll(seriesApril, seriesMay);
          primaryStage.setScene(scene);
          primaryStage.show();
      }
  }

  Greetings
  Thomas

  For JavaFX 2.2.3 there is a bug creating a StackedAreaChart with a desabscisses (xAxis) axis

  http://JavaFX-JIRA.Kenai.com/browse/RT-23812 "StackedAreaChart ClassCastException on axis desabscisses"

  The bug is fixed in the jdk8 branch and planned to backport and JavaFX 2.2.6 release (next year I think).

  Link to the source diff patch:
  http://Hg.OpenJDK.Java.NET/openjfx/8/controls/RT/Rev/d811a42d75ba