Need RCOMPTE to the group by the name of the team

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• activation [plan of the team question]

  I bought the

  Plan of cloud creative photography (one year)

  but can not understand how to assign to the person who needs it in the team.

  It does not appear in the team page as periodic contributions of CS complete - so how can I give it?

  Team license links that can help

  - https://creative.adobe.com/plans?plan=team team plans

  -http://www.adobe.com/creativecloud/buy/business.html

  -https://helpx.adobe.com/contact/creative-cloud-teams.html using the team

  -manage your account http://forums.adobe.com/thread/1460939?tstart=0 team

• Need to call the PL/SQL Package in the group dynamics of approval class that implements IDynamicApprovalGroup

  Hi all

  This is regarding the approval of invoice AP I need in my project where the user selects the users/people approval when creating the invoice based on logic. There may be any number of users who could be part of the approval group.

  Approval users information is stored in a table with a single Bill.

  When the user initiates invoice approval, approval of the Bill should go to all the people/users selected in the invoice.

  I created the dynamic of approval group which is the result of a java class. This java class implements IDynamicApprovalGroup. (what is the oracle Developer Guide).

  I called PL/SQL to java class package to get all the details of users of custom for the current invoice table.

  I used the DriverManager.getConnection method by hard coding (JDBC URL, Username, Pwd) connection details that works perfectly fine.

  But I don't want to hard-code the connection details because it will change in bodies SIT/UAT/PROD.

  Please ask you to guide me if there is no solution for the information of connection without hard coding it.

  Following code used to get the connection details

  try {}

  Class.forName ("oracle.jdbc.driver.OracleDriver");

  System.out.println ("= class loaded");

  }

  catch (ClassNotFoundException ex) {}

  System.out.println ("= class loading error");

  ex.printStackTrace ();

  }

  Fitting out = null;

  CallableStatement cstmt = null;

  Try

  {

  connection = DriverManager.getConnection ("jdbc:oracle:thin:@xxhostname:1521/SID", "FUSION", "*");

  Connection = GetConnection ();

  System.out.println ("= connection =" + connection);

  prepare the call

  cstmt = connection.prepareCall ("{call XX_AP_INV_DYN_APPROVAL_PKG. {GET_GROUP_PARTICIPANTS (?,?)} ») ;

  cstmt.setString (1, '12345');   ID on invoice

  Exit of the register

  cstmt.registerOutParameter (2, java.sql.Types.VARCHAR);

  Call the stored procedure

  cstmt. Execute();

  System.out.println ("= procedure being executed");

  approvalGroupMembers = cstmt.getString (2).split("#");

  System.out.println ("= output:" + cstmt.getString (2));

  }

  catch (Exception ex) {}

  ex.printStackTrace ();

  }

  Finally

  {

  Try

  {

  If (cstmt! = null)

  close the callable statement

  {

  cstmt. Close();

  cstmt = null;

  }

  System.out.println ("= stmt closed");

  }

  catch (SQLException ex)

  {

  System.out.println ("= stmt close err");

  ex.printStackTrace ();

  }

  Try

  {

  If (connection! = null)

  close the connection

  {

  Connection.Close;

  connection = null;

  }

  System.out.println ("= closed conn");

  }

  catch (SQLException ex)

  {

  System.out.println ("= Close conn err");

  ex.printStackTrace ();

  }

  }

  Thank you

  Dipak.

  HI Dembélé - you should be able to call the provider database from the java class. Documentation is here. However, you need to specify the name of the data provider that is the same in all environments. Do you know what it is? If it's not I'll take a look and get back to you.

  __

  Peter Maurer

  Relationship with the developers of Applications in fusion

  https://blogs.Oracle.com/fadevrel/

• Need to find the version number referring to the last name change for each emplid.

  emplID	Name age	Version
  ---	-----   ---	--------
  1	ABC	25	2
  1	ABC	26	5
  1	def	27	9
  1	def	28	15
  1	def	29	18
  2	JKL	15	5
  2	MNO	16	8
  2	MNO	17	11
  2	PQR	18	20

  Need to find the version number referring to the last name change for each emplid.

  As indicated by the version number change when there is change in age as well, but I need a query that gives the version number referring to the last name change for each emplid.

  with

  DATA_TABLE (EmplId, Name, Age, version) as

  (select 1, 'abc', 25, 2 double Union all

  Select 1, 'abc', 26, 5 Union double all the

  Select 1, 'def', 27, 9 double Union all

  Select 1, 'def', 28, 15 double Union all

  Select 1, 'def', 29, 18 double Union all

  Select 2, 'jkl', 15 5 Union double all the

  Select 2, 'mno', 16, 8 double Union all

  Select 2, 'mno', 17, 11 double Union all

  Select 2, 'pqr', 18, 20 double

  )

  Select emplid,

  Max (Name) name of Dungeon (dense_rank last order by version).

  Max (Age) age of Dungeon (dense_rank last order by version).

  version Max (version)

  from (select emplid, name, version,)

  -case where name! = lag(name,1,name) on (emplid version order partition)

  then "renamed".

  end change

  of data_table

  )

  where the change is not null

  Emplid group

  EMPLID	NAME	AGE	VERSION
  1	def	27	9
  2	PQR	18	20

  Concerning

  Etbin

• Need help in the optimization of the query with the Group and joins by clause

  I'm having the problem by running the following query... It takes a lot of time. To simplify, I added the two tables FILE_STATUS = stores the file load details and COMM table Board table job showing records treated successfully and which was communicated to the other system real. Records with status = T is trasnmitted to another system and traansactions with P is waiting.

  CREATE TABLE FILE_STATUS
  (FILE_ID VARCHAR2(14),
  FILE_NAME VARCHAR2(20),
  CARR_CD VARCHAR2(5),
  TOT_REC NUMBER,
  TOT_SUCC NUMBER);
  
  CREATE TABLE COMM
  (SRC_FILE_ID VARCHAR2(14),
  REC_ID NUMBER,
  STATUS CHAR(1));
  
  INSERT INTO FILE_STATUS VALUES ('12345678', 'CM_LIBM.TXT', 'LIBM', 5, 4);
  INSERT INTO FILE_STATUS VALUES ('12345679', 'CM_HIPNT.TXT', 'HIPNT', 4, 0);
  
  INSERT INTO COMM VALUES ('12345678', 1, 'T');
  INSERT INTO COMM VALUES ('12345678', 3, 'T');
  INSERT INTO COMM VALUES ('12345678', 4, 'P');
  INSERT INTO COMM VALUES ('12345678', 5, 'P');
  COMMIT;

  Here's the query I wrote to give me the details of the file that has been loaded into the system. He reads the table of State and the commission files to display the name of the file, total records loaded, total at the table of the commission and the number of records which has finally been passed successfully loaded (Status = T) with other systems.
  

  SELECT 
      FS.CARR_CD 
      ,FS.FILE_NAME 
      ,FS.FILE_ID
      ,FS.TOT_REC
      ,FS.TOT_SUCC
      ,NVL(C.TOT_TRANS, 0) TOT_TRANS
  FROM FILE_STATUS FS
  LEFT JOIN
  (
      SELECT SRC_FILE_ID, COUNT(*) TOT_TRANS
      FROM COMM
      WHERE STATUS = 'T'
      GROUP BY SRC_FILE_ID
  ) C ON C.SRC_FILE_ID = FS.FILE_ID
  WHERE FILE_ID = '12345678';

  In production, this request has several joins and takes a long time to deal with... the main culprit for me is the join on the COMM table to count the number of number of transactions sent. Please can you give me tips to optimize this query to get results faster? What I need to delete the Group and use the partition or something else. Help, please!

  Don't know if it will be faster based on the information provided, but analytical functions offer an alternative approach;

  select carr_cd, file_name, file_id, tot_rec, tot_succ, tot_trans
    from (select fs.carr_cd,
                 fs.file_name,
                 fs.file_id,
                 fs.tot_rec,
                 fs.tot_succ,
                 count(case
                          when c.status = 'T' then
                           1
                          else
                           null
                        end) over(partition by c.src_file_id) tot_trans,
                 row_number() over(partition by c.src_file_id order by null) rn
            from file_status fs
            left join comm c
              on c.src_file_id = fs.file_id
           where file_id = '12345678')
   where rn = 1;
  
  CARR_CD FILE_NAME            FILE_ID           TOT_REC   TOT_SUCC  TOT_TRANS
  ------- -------------------- -------------- ---------- ---------- ----------
  LIBM    CM_LIBM.TXT          12345678                5          4          2
  

• How do I change my name that appears at the beginning of an email? I need to change my e-mail account name?

  My wife and I use the same email. I need to change the name that appears at the beginning of an email, a common name, rather than just my name. We use our email AOL. I need to change the name on the email account AOL to make this change? or is it possible to change this in Firefox? Thanks for help.

  The replacement must be made within the parameters of AOL. See the next page for instructions.

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• I created a custom and registered with a specific name date format. However, when I open a new worksheet, my saved format does not display in the menu drop-down? Am hoping that I don't need to create the same format for each new sheet?

  I created a custom and registered with a specific name date format. However, when I open a new worksheet, my saved format does not appear in the menu drop-down? Am hoping that I don't need to create the same format for each new sheet?

  You must save the spreadsheet containing the new date as a model format and use this custom template for each new spreadsheet where you want that this date format personalized at your disposal.

• Contact - removal of the entire group and its name.

  In Vista, I can delete individual members of a group, but how to remove the group bearing his name. Tried all I know. Help!  Thanks adavance if you can spare the time to calm down a user of profanity.

  It is clear. That's also what I thought that you said in your original post. If I gave you this...
  Assuming you are using Windows Mail...
  Open Windows Mail > toolbar, click Contacts > highlight the group in question, right click > delete see screenshot for illustration:
  For the benefits of others looking for answers, please mark as answer suggestion if it solves your problem.

• Hi I have windows 7 Home Edition premium, I need to put the Russian folder names can I do this without charge last?

  Hello

  I have windows 7 Home premium and you need the folder in Russian without the big names spend of ultimate?
  my girlfriend needs to use the laptop...
  best regards Richard

  I think you will find that you need ultimate in order to install the language packs. (ditto for win8)

• Need to change the name of the file hard

  I need to change the name of the vmdk files which will get the game of football on behalf of virtual machines. Please anyone suggest me the steps to that end, the sameange the

  Hello

  Hope you are looking steps below.

  Renaming of files in virtual machine using Cold Storage Migration

  Virtual machine files can be renamed on the fly when cold Migration is initiated from the vSphere Client, only documents of the virtual machine are renowned.

  To rename the files:

  Log on to the VMware vSphere Client.

  Locate the virtual machine in your inventory to the host.

  Right-click the virtual machine, and then click Rename.

  Change the name of the virtual machine as required.

  Perform a cold migration to a new virtual machine data store.

  For more information, see:

  For ESXi/ESX 4.x, see Migration cold in the Guide of Administration of vSphere Datacenter.

  For ESXi/ESX 3.5, see Migration cold in basic management Guide.

  When the migration is complete, the files of the virtual machine, the name of the virtual machine configured in step 4.

  Note: If a virtual directory already exists on the data store with the same name, then the virtual target machine is placed in a new directory with annexed _1.

• Need info regarding the behavior of Oracle UCM accounts and security groups

  Need information about the behavior of Oracle UCM accounts and security groups.
  
  Oracle UCM version: 11.1.1.5.0
  
  Steps to follow:
  1. log in with the user "weblogic" and content created with the id of "content1".
  2. apply the "@acc1 (R)" and "TestGroup1" to her quote created in step 1
  3. close the session
  4. connect as that of "acc1user1", the user is not able to see the "content1".
  5. close the session
  6 log in as "role1user1", the user is not able to see the "content1".
  
  Account information and group:
  1. the user "acc1user1" is part of the "@acc1 (R).
  2. the user "role1user1" belongs to 'role1 (R)' and is mapped to "TestGroup1" at the Complutense University of MADRID
  
  Expected:
  The 'acc1user1' and 'role1user1' should be able to see "content1" because they have at least the read permission.
  
  Please help me understand why users aren't able to see the content.

  See section 5.5.1.1 in the Manual: http://docs.oracle.com/cd/E23943_01/doc.1111/e10792/c05_security.htm#BGBGIJDJ

  If the accounts and roles are used, the result is an intersection, not a Union.

• When I try to save a document, I get the message "the file is opened by another program or read-only. I need to change the name to save him.

  When I try to save a document, I get the message "the file is opened by another program or read-only. I need to change the name to save him. If I optimize a document, I rename it to save it. If I opened a PDF file of the architect drawings and turn them into fashion landscape so that they are easier to read, I have to rename it. I searched on the parameters to look for something that I could turn off, but I found nothing.

  Make sure that the preview in Windows Explorer option is disabled.

• Need to change the name from vCenter through code

  Hi all

  I can't understand this, I hope that this is a real simple thing to do. I need to change the name of the vCenter, like what's in the screenshots below, but using the powershell code

  can someone show me how it's done?

  See you soon!

  Depending on whether you have vCenter, Update Manager, and licenses executed on the same host server, it is not a simple operation.

  See for example KB1006843.

  To emulate the rename also can be done by vSphere Client is simple.

  You must be connected to the vCenter.

  $optMgr = Get-View (Get-View ServiceInstance).Content.Setting
  $optVal = New-Object VMware.Vim.OptionValue
  $optVal.Key = "VirtualCenter.InstanceName"
  $optVal.Value = 
  $optMgr.UpdateOptions($optVal)
  

  ____________

  Blog: LucD notes

  Twitter: lucd22

• Need to change the name of the server and the IP

  Hi Lion,
  
  I need to change the server name and IP address of my Test Server, because it is the replica of the server of Production, and because the name and IP are the same because of this problem I can't access it on LAN environment, for that I have to change the name and IP address of my Test Server so that I can use it in the LAN environment. Can any body me consequence after changing, and in which files I have to change the IP address and the name, I use Oracle Enterprise Linux.
  
  Kind regards

  Salvation;

  Please check:
  Change of IP address in 11i app
  Re: What changes must be made if the IP address of the Server 12 APPS is changed!

  It may be useful

  Respect of
  HELIOS

• Need to copy the files from old XP disc in Vista by Doc

  Have the old OS XP disc I need to get the MDC in Vista64.  Drive appears in my computer on Vista and XP security will not read copy the required files to the vista system.

  Hard drive XP started under Vista64 OS, need to copy files from old drive XP in Vista by Doc security on files XP stopped, how do I solve the problem__

  Appropriate files on the old hard drive:

  Check the permissions of the file or folder the file is saved in and appropriated:
        
  1. right click on the file or folder and then click Properties.
  2. click on the Security tab.
  3. under group or user names, click your name to see the permissions you have.

  To open a file, you must have the read permission. For more information about permissions, see what are permissions?

  http://tinyurl.com/2j9vgr

  To take ownership of a folder:

  1 right click on the folder you want to take control and then click Properties.
  2. click on the Security tab, click Advanced, and then click the owner tab.
  3. click on modify. Need administrator permission if you are prompted for an administrator password or a confirmation, type the password or provide confirmation.
  4. click on the name of the person you want to give to the property.
  5. If you want this person to be the owner of the files and subfolders in this folder, select the Replace owner of subcontainers and objects to check.
  6. click on OK

  MS - MVP - Elephant Boy computers - don't panic!