Not able to read a part of the document but chrome and foxit Reader pdf reader

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  Per-Erik Pedersen

  Hi Peders,

  Thanks for posting on the Adobe forums.

  This camera is supported by Camera Raw 8.2, which is available in Photoshop CS 6. Please check the following KB Articles:

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  Applications of camera Raw compatible Adobe

  Kind regards

  Sandeep

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  I myself had this new cool device and everything is working fine beside it is not able to read files audio wma.
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• Qosmio G20 is not able to read a cd after burning

  Hello

  I need help please.

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  Thank you

  Read your ad I didn't understand what the service does exactly, but it's really odd if this occurs after the installation of clean operating system with recovery DVDs.

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  Hello

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  Hello

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  But let's try some common tips.
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  3. expand Control and class.
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• DVD-ROM not able to read any Dvd or cd...

  Hi, my dvd rom is not able to read any disc, I can see the drive in my computer, when I insert a disc, it runs for the first time, the light shows and thn it stops completely, and when I click on the dvd icon in my computers it drive ejects, when I tried to google my problem I got to know tht, this may be due the updates for vista and the solution was to remove the upper filter or lower in the registry, I tried tht so but thn too the problem persists, try wht should I do?

  try to use the DVD rom in another IDE or SCSI port in your computer.

• UTL_FILE.get_line not able to read large files...

  Dear all,

  UTL_FILE.get_line is not able to read data from file. can you suggest me to fix this problem...

  Here is my code;

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  l_input_line varchar2 (6000);

  Start

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  end;

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  Thank you.

  Kind regards

  Ravi.

  Your call to FOPEN() function specifies a size of 32K maximum line. When you call GET_LINE() without specifying the number of characters to get, Oracle will therefore assume you want to get 32K characters. However, your stamp, L_INPUT_LINE, is defined as VARCHAR2 (6000). That is, it is much too small.

  The solution is to call GET_LINE() with the third parameter set to 6000.

  BTW, you can also set a condition of release on the loop.

• I have a copy of photoshop cs2 (version 9) and I am not able to record by phone or the Internet.  Is there a number I can contact for this?

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• someone is using my email address to send emais. I tried to reset my password but I am not able to. He says that the answer to my secret question does not.

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  You'd better post in the Windows Live Solution Center-Hotmail like that is dedicated to Windows Live Hotmail.

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  Good luck!

• Not able to install updates to get the message: 'updates were not able to configure correctly, returning from changes'...

  Original title: I have Vista with Service Pack 2, but were unable to come down and configure updates since 07/12/11. Need help, thank you.

  Do not get the error codes. Just when I reboot, I get the message "updates were not able to configure correctly, returning from the changes".

  Hello

   
  You are able to start in safe mode, then you can perform the clean boot in safe mode and then try to boot into normal mode.
  Put your boot system helps determine if third-party applications or startup items are causing the problem.
   
  See the following article in the Microsoft KB for more information on how to solve a problem by performing a clean boot in Windows Vista or Windows 7:
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  http://support.Microsoft.com/kb/929135
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• All-in-one printer, HP Photosmart 7510 does not recognize my computer as part of the network.

  Range of all-in-one printers, HP Photosmart 7510

  The function scan of my 7510 new does not recognize my computer as part of the network. (He said that the scanner can not find the computer)   It prints fine from the computer, however.  HP ran diagnostics and cannot find the problem in their system.  Suggested that I try to find a solution to the millisecond.  I was unable to find a solution yet.  Can anyone help?

  Hello

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• Not able to read a text file or binary

  I am reading in a file text or binary, as shown in the Blackberry "how - adds an application of plain text or binary files"

  http://www.BlackBerry.com/knowledgecenterpublic/livelink.exe/fetch/2000/348583/800332/800620/How_To _...

  I use Eclipse 3.4.2 I implemented a project of Blackberry with the following structure:

  Test1 (project name)

  -src (folder)

  |   + com.name.xxx

  |   + com.name.yyy

  |            Test1.Java

  + NET_RIM_BLACKBERRY

  |

  -test

  I'm reading in the test file is at the level of the main project folder.  For example, the Test1.jdp file is located in the folder of the main project and the test file (text file or binary).  I am trying to access the file using the line of code below:

  Class classs = Class.forName(this.getClass().getName());
  
  InputStream is = classs.getResourceAsStream("/test");
  

  The getResourceAsStream indicates that the file will be in the same location as the location of the Test1.jdp file if the bar oblique is in front of the name of the file, or that it will look in the source for the main class file.  When I have the slash I get error

  CAFETERIA: could not find the test file

  I tried several different attempt to get to the CAFETERIA to find the file, but I'm running out of ideas and I would appreciate someone else's ideas on what to try.

  Thank you

  Thank you klyubin and stef

  I found a solution that works for me with Eclipse plug-ins, JDE.

  First, the example of Blackberry in the how-to (see original thread number) with test as file is named simply 'test '.  This does not work for me at all in Eclipse.  I gave up to place the file in the folder of the project Eclipse (main folder) and placed in the src folder, which of course gives a compile error because is not a Java file, and of course when using file name in my java code I changed the code correctly as in the first line that follows :

  
  // this will look in the source folder of the class // that myClass refers tois = myClass.getResourceAsStream("test.txt");
  
  // this is supposed to refer to a file called 'test' // in the source folder of the class that myClass refers to // but apparently for Eclipse it must have a .txt extension// as in the above exampleis = myClass.getResourceAsStream("test");
  
  // this is supposed to refer to a file called 'test' // in the main project folder but not apparenlty for Eclipseis = myClass.getResourceAsStream("/test");
  
  // this is supposed to refer to a file called 'test.txt' // in the main project folder but not apparenlty for Eclipseis = myClass.getResourceAsStream("/test.txt");             } 
  

  Note: I was also getting the error message when I place the file named "test" in the src below directory.  I managed to get rid of this error message by configuring the path building (similar to deleting the file of the JDP file including klyubin suggested above for Blackberry JDE ordinary rather than the plugin Eclipse) to remove the file 'test' for compiling so I would not get the compile error.

  However, this was not my main problem - if you name the file test.txt and place it in the src directory you do not get this error I believe.

  Errors during construction

  Errors running builder "Blackberry resources Builder" on the project...

  Could not compile files

  java.io.FileNotFoundException: C:\path to the file I had added in the directory\.rrh of the source (or some truncation of the file name .rrh - for example te.rrh)

  Finally, I have checked the jar (as klyubin suggested) file that is created when I place the file 'test' or 'test.txt' in my Eclipse project main folder.  The jar file did not contain the name file and so when I run in the Simulator, the file is not found.  Even when I put the file in the jar (main directory) myself the Simulator still could not find the file.   When I placed the test.txt file in my directory src the jar resulting file contains the file named and the Simulator can find the file ok.

  So, it seems to me that for the plug-in Eclipse JDE contextual explanation of the method getResourceAsStream does not apply as described at this time.

  getResourceAsStream

  public InputStream getResourceAsStream(String name)
  

  Looking for a resource with a given name. This method returns null if no resource with this name is found. The rules looking for resources associated with a given class are specific profile. RIM implementation notes: this method searches the resource from the directory in which the current JDP project file is located. By default, it is assumed that this JDP file resides in the parent directory of the package of the current project space. Thus, the package space tree (for example, com/rim/PackageName) is added to name so that the search begins from this space package, that is, in the source directory of the current project. However, if the current JDP project file does not lie in its presumed location, the behavior above prevent the resource located. In this case, the user must add a slash (/) in the name of the resource before calling getResourceAsStream. This treats the name as an absolute path, which is compared to a tree whose root is the location of the JDP file. Parameters: name - name of the desired resource Returns: a java.io.InputStream object. Since: JDK1.1

  If Alain klyubin (or someone else) can enlighten us more on why the file should be named *.txt to find the file in the source directory, which can be useful to me and others in the future.  My real goal is to read in an image file named *.jpeg.  We'll see how it goes.

  And thank you klyubin for the suggested code change which I had also found in another thread...

  // this was more than what was necessary//Class myClass = Class.forName(this.getClass().getName());
  
  // this is simpler  Class myClass = this.getClass();
  

  And so many other Eclipse users who try to follow the Blackberry How To - add a text or a binary file in an application, do not forget to consult the file name in your code java as in the examples that I gave above.

• Not able to connect to MWAGUI. Telnet works, but MWAGUI in error "the connection failed". Help, please...

  Not able to connect to MWAGUI

  Telnet works, but MWAGUI giving the error "connection failed". Help, please...

  We bounced back the application and services many times MWA. Also, we have disabled enabled several port but no solution yet.

  There are 3 work of the plates for MWA GUI on the Windows Machine.

  Patch # 12574324 for 11i

  Patch # 12780257 for 11i

  Patch # 20983924 for R12

  All these patches work on Windows 7 and 8, with all versions of Java.

  The first two patches for 11i work on R12 also but with a minor problem. When you press tab even though visually the cursor moves to the next field, but in reality, it remains in the previous filed and what you enter is taken as a stream in the previous field. The replacement of the tab in the first two tasks is .

  To solve this problem the 3rd patch was created, which is the latest patch to work for R12 too but this patch after a few ORS and Patch requests (other modules) does not work anymore and gives the error "unable to connect". The first two patches can be used s a solution until a new patch coming to replace 20983924.

  Please note that CMD/batch of all three patches files is performed differently, so follow the guideline strictly for the respective patch.

  Kind regards

  Pierre Paul

• Tried everything: 1 address, but not able to connect via HTTPS to the server:

  Hello

  I am trying to connect to the third of the OSB business service web service.

  objective WS is protected with SHA1 base 64 encoded password.

  I am able to connect to the service target of SOAP UI. I am also able to Telnet to my server for dev to the URL of the WS.

  But when connecting from OSB BS I'm tried them all: 1 address, but not able to connect via HTTPS to the server: error.

  Can you please help me solve the problem.

  I tried different policies, but still does not work.

  Kind regards.

  Problem solved. We need to use the proxy server.