Authentication failed. (Microsoft.AnalysisServices.AdomdClient). The target principal name is incorrect (Microsoft.AnalysisServices.AdomdClient)

Hi Experts,
We had a task to  migrate SQL Server all the components to another server, the migration went well and had no issues at all. but We can login to SSAS service locally wihtout any issues. when we are connecting the analysis services from the other machines(servers)
it is givng the below error.
Authentication failed. (Microsoft.AnalysisServices.AdomdClient)
The target principal name is incorrect (Microsoft.AnalysisServices.AdomdClient)
1) it is a stand alone server
2) it is connecting to default instance but not to a named instance
3) SPN's were set correctly. Double checked with the tool(MS Kerberos configuration Tool).
4) The SQL server analysis start account has domain admin privileges.
5) we can connect to Database services from the other machine remotely.
6) none of the analysis services are  connecting.
Thank you in advance.

Hi Ramu,
According to your description, you migrated SQL Server to another server, everything works fine except that cannot connect to SSAS remotely with the error, right?
Authentication failed. (Microsoft.AnalysisServices.AdomdClient)
The target principal name is incorrect (Microsoft.AnalysisServices.AdomdClient)
Based on my research, this issue is caused by that the SPN for account that run the Analysis Services is corrupt. You said that the SPN were set correctly, however the error message indicate that the problem is related to SPN. So in your scenario, you can
delete the SPN under the service account, and register SPN for Analysis Services instance. Please refer to the link below to see the details.
http://msdn.microsoft.com/en-IN/library/dn194200.aspx
Besides, here is a blog which describe the similar issue.
http://www.wolfsoftwaresystems.com/code/sql/the-target-principal-name-is-incorrect-microsoft-analysisservices-adomdclient/
Regards,
Charlie Liao
TechNet Community Support

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                      I want to know ,how to find out the target component and target view in WEB UI,when i click on a field which shows as a hyper link in WEB UI .At GUI level ,i know the how it will work.Any way to find out the component name and view name which is show the details of that field at GUI level .IF you go to transaction CRMD_ORDER then open the service contract id .then goto the item level value .there is 1 service data tab is available at item level.there is two button is available.first one is available time and second one is response time .if i click on any button then one popup is open which is shows the details of that button.I dont know how to find out the component name and view name from GUI level.
    Thanks in Advance....
    Vishwas Sahu

    Hi Vishwa,
                 The control would be something like this for navigation in Get_p_xxx method u mention as link and u mention a event name which gets triggered on the click of this hyperlink. So your GET_P_XXX method would have the following code:
    CASE iv_property.
        WHEN if_bsp_wd_model_setter_getter=>fp_fieldtype.
          rv_value = cl_bsp_dlc_view_descriptor=>field_type_event_link.
        WHEN if_bsp_wd_model_setter_getter=>fp_onclick.
          rv_value = 'EXAMPLE'.
    Now you have to create a method as EH_ONEXAMPLE at your IMPL class and within which you would give an outbound plug method. Within the outbound plug the target data would be filled in the collection and window outbound plug would be triggered.
    This is a huge topic and  i have just mentioned you basic things.
    Regards,
    Bharathy.

  • Error SOURCE IS the same as the target Task name Clear Cube Data

    Hi,
    When I am trying to use the Clear Package I am getting an error -
    "Source is the same as the Target"
    Any help with how to resolve this ?
    thx

    Hi,
    Please check if the Clear DM package is linked with the Process chain for Clearing the Data.
    We usually get the error you are mentioning in Copy Function when we have same source and Destination members.
    Hope this helps,
    Regards,
    G.Vijaya Kumar

  • How to typecast an object, if the target class name is stored in a str var

    Hi,
    I have a method, in which i will receive an object as an attribute. I need to typecast it to another class, and that class name is stored in a string variable.
    public void printContent(Object obj)
    /* Here i have a variable strClassName - in which a class name is stored.
    I need to create a variable for that class and type cast the object obj into that class. */
    Please kindly help me to find out a way to do this.
    Thanks in advance

    Typecasting does one thing: it enables you to store an object in a reference variable of the target type (if it is of a type which is allowed to be in that variable). It doesn't alter the object in any way.
    Therefore it simply isn't meaningful to talk about typecasting to a type name known only at run time. The whole point of typecasting is to tell the compiler to treat a reference as being of a specified type.
    I don't know what you're trying to do but I think you should probably be looking at the reflection api, getting field an method information from the object's Class object.

  • Webapp authentication failed when using chinese characters as login name

    Hello,
    I have tried webapp authentication on tomcat and oc4j, via BASIC and FORM auth-method. All failed when the login name contains non-English characters. It seems an encoding issue, therefore, I also tried to change the page encoding of the login form to utf-8. None of the above is successful. Is there any solution? I really appreciate any help!
    Thanks in advance!!

    Enterprise support:
    Call enterprise support  (866) 752-7753  to create  a case ID number
    Get an account at
    http://developer.apple.com/  then submit a bug report to http://bugreporter.apple.com/
    Once on the bugreporter page,
       -- click on New icon
       -- See if you need to attach a log file or log files, clicking on Show instructions for gathering logs.  Scroll down to find the area or application that matches the problem.
       -- etc.
    Developers:
    "Submitting Bugs and Feedback
    Your feedback goes a long way towards making our products even better. With Apple Bug Reporter, you can submit bug reports or request enhancements to APIs and developer tools."
    https://developer.apple.com/bug-reporting/

  • How to get the target file name from an URL?

    Hi there,
    I am trying to download data from an URL and save the content in a file that have the same name as the file on the server. In some way, what I want to do is pretty similar to what you can do when you do a right click on a link in Internet Explorer (or any other web browser) and choose "save target as".
    If the URL is a direct link to the file (for example: http://java.sun.com/images/e8_java_logo_red.jpg ), I do not have any problem:
    URL url = new URL("http://java.sun.com/images/e8_java_logo_red.jpg");
    System.out.println("Opening connection to " + url + "...");
    // Copy resource to local file                   
    InputStream is = url.openStream();
    FileOutputStream fos=null;
    String fileName = null;
    StringTokenizer st=new StringTokenizer(url.getFile(), "/");
    while (st.hasMoreTokens())
                    fileName=st.nextToken();
    System.out.println("The file name will be: " + fileName);
    File localFile= new File(System.getProperty("user.dir"), fileName);
    fos = new FileOutputStream(localFile);
    try {
        byte[] buf = new byte[1024];
        int i = 0;
        while ((i = is.read(buf)) != -1) {
            fos.write(buf, 0, i);
    } catch (Throwable e) {
        e.printStackTrace();
    } finally {
        if (is != null)
            is.close();
        if (fos != null)
            fos.close();
    }Everything is fine, the file name I get is "e8_java_logo_red.jpg", which is what I expect to get.
    However, if the URL is an indirect link to the file (for example: http://javadl.sun.com/webapps/download/AutoDL?BundleId=37719 , which link to a file named JavaSetup6u18-rv.exe ), the similar code return AutoDL?BundleId=37719 as file name, when I would like to have JavaSetup6u18-rv.exe .
    URL url = new URL("http://javadl.sun.com/webapps/download/AutoDL?BundleId=37719");
    System.out.println("Opening connection to " + url + "...");
    // Copy resource to local file                   
    InputStream is = url.openStream();
    FileOutputStream fos=null;
    String fileName = null;
    StringTokenizer st=new StringTokenizer(url.getFile(), "/");
    while (st.hasMoreTokens())
                    fileName=st.nextToken();
    System.out.println("The file name will be: " + fileName);
    File localFile= new File(System.getProperty("user.dir"), fileName);
    fos = new FileOutputStream(localFile);
    try {
        byte[] buf = new byte[1024];
        int i = 0;
        while ((i = is.read(buf)) != -1) {
            fos.write(buf, 0, i);
    } catch (Throwable e) {
        e.printStackTrace();
    } finally {
        if (is != null)
            is.close();
        if (fos != null)
            fos.close();
    }Do you know how I can do that.
    Thanks for your help
    // JB
    Edited by: jb-from-sydney on Feb 9, 2010 10:37 PM

    Thanks for your answer.
    By following your idea, I found out that one of the header ( content-disposition ) can contain the name to be used if the file is downloaded. Here is the full code that allow you to download locally a file on the Internet:
          * Download locally a file from a given URL.
          * @param url - the url.
          * @param destinationFolder - The destination folder.
          * @return the file
          * @throws IOException Signals that an I/O exception has occurred.
         public static final File downloadFile(URL url, File destinationFolder) throws IOException {
              URLConnection urlC = url.openConnection();
              InputStream is = urlC.getInputStream();
              FileOutputStream fos = null;
              String fileName = getFileName(urlC);
              destinationFolder.mkdirs();
              File localFile = new File(destinationFolder, fileName);
              fos = new FileOutputStream(localFile);
              try {
                   byte[] buf = new byte[1024];
                   int i = 0;
                   while ((i = is.read(buf)) != -1) {
                        fos.write(buf, 0, i);
              } finally {
                   if (is != null)
                        is.close();
                   if (fos != null)
                        fos.close();
              return localFile;
          * Returns the file name associated to an url connection.<br />
          * The result is not a path but just a file name.
          * @param urlC - the url connection
          * @return the file name
          * @throws IOException Signals that an I/O exception has occurred.
         private static final String getFileName(URLConnection urlC) throws IOException {
              String fileName = null;
              String contentDisposition = urlC.getHeaderField("content-disposition");
              if (contentDisposition != null) {
                   fileName = extractFileNameFromContentDisposition(contentDisposition);
              // if the file name cannot be extracted from the content-disposition
              // header, using the url.getFilename() method
              if (fileName == null) {
                   StringTokenizer st = new StringTokenizer(urlC.getURL().getFile(), "/");
                   while (st.hasMoreTokens())
                        fileName = st.nextToken();
              return fileName;
          * Extract the file name from the content disposition header.
          * <p>
          * See <a
          * href="http://www.w3.org/Protocols/rfc2616/rfc2616-sec19.html">http:
          * //www.w3.org/Protocols/rfc2616/rfc2616-sec19.html</a> for detailled
          * information regarding the headers in HTML.
          * @param contentDisposition - the content-disposition header. Cannot be
          *            <code>null>/code>.
          * @return the file name, or <code>null</code> if the content-disposition
          *         header does not contain the filename attribute.
         private static final String extractFileNameFromContentDisposition(
                   String contentDisposition) {
              String[] attributes = contentDisposition.split(";");
              for (String a : attributes) {
                   if (a.toLowerCase().contains("filename")) {
                        // The attribute is the file name. The filename is between
                        // quotes.
                        return a.substring(a.indexOf('\"') + 1, a.lastIndexOf('\"'));
              // not found
              return null;
         }

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