Proxy client won't work in this case?

I have not made my client side work for a single device yet but I want to make sure I'm spending time on the right track:
My code will be running on the application server side, there will be hundreds of devices each has its own IP address, but runs exactly the same web services, and my code will need to talk to each of the selected ones using its IP address. I see that some generated proxy class contain the particular IP of the device where the web service is running on? If so, then it won't be possible for me to use a client proxy class to talk to all the devices.
I'm currently trying something like this:
URL url = new URL("http://172.21.159.52:80/cmi?wsdl");
QName serviceName = new QName("urn:Genus", "genusService");
Service service = Service.create(url, serviceName);
//These are for testing purpose
System.out.println ("List of QNames of service endpoints:");
Iterator it = service.getPorts();
*while (it.hasNext ()) {*
System.out.println (" " + it.next());
I can see the single service named 'genusPortType' in the output, however when the following code is reached:
QName portName = new QName("urn:Genus", "genusPortType");
Dispatch<SOAPMessage> dispatch = service.createDispatch(portName, SOAPMessage.class,Service.Mode.MESSAGE);
I got an error 'javax.xml.ws.WebServiceException: Illegal argument combination [type=javax.xml.soap.SOAPMessage,mode=MESSAGE]'. I'm confused because I see the same usage in lots of sample codes. Is it because the 'portName' part decides that these two parameters are not fit based upon the definition of 'genusPortType'? If the operation is defined like this in the wsdl file, wht values should I use?
*<wsdl:binding name="genusBinding" type="cmi:genusPortType">*
*     <wsdlsoap:binding style="rpc" transport="http://schemas.xmlsoap.org/soap/http" />*
*     <!-- login operation -->*
*     <wsdl:operation name="login" >*
*          <wsdlsoap:operation soapAction="com.controlmod.terminal.services.web.services.Login#Login" />*
*          <wsdl:input name="loginRequest" >*
*          <wsdlsoap:body encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" use="encoded" />*
*          </wsdl:input>*
*          <wsdl:output name="loginResponse" >*
*          <wsdlsoap:body encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" use="encoded" />*
*          </wsdl:output>*
*     </wsdl:operation>*
*.............*

Sorry for really slow response on this - are you still here/still
having the issue?
Fillattr runs two migrations - one for access rules, and another for
proxy settings. I've seen cases where one works and the other fails.
But eventually you can get it going if you find the hangup.
Craig Johnson
Novell Knowledge Partner
*** For a current patch list, tips, handy files and books on
BorderManager, go to http://www.craigjconsulting.com ***

Similar Messages

  • I've moved my Aperture library to another computer, and masters are referencing an old path name.  How can I update these references?  Reloctating masters does not work in this case :(

    I've moved my Aperture library from one computer to another using Finder.
    I merged the library with one which was already on the computer.
    Now, the photos I imported have reference to the old path name on my old computer.
    How can I update these references as "Relocate Masters" does not work in this case?

    Just one suggestion to be able to reconnect all at once:
    Create a smart album containing the images with missing masters:
    File -> New ->  Smart Album,     and add a rule: File Status is "Missing"     (or File Status is "offline")
    Then select the images in this album and go to the File menu:
    and select:   File -> Locate referenced File
    If you are lucky, Aperture will reconnect all at once, if you point the first image version to its counterpart.

  • Email client won't working

    Hi..
    I have a problem till yesterday with iphone 4 email client..it won't show me my new emails..I just see 1 new message like notification and when I go to see inbox it is not there .
    Pleas help me to fix this problem.
    All the best
    Dusica.

    Well I know recently Google has started locking accounts (well where they can check from) when they are checked on too often or from too many different places.
    My wife couldn't check her email anymore from new locations) until it got unlocked.
    https://www.google.com/accounts/DisplayUnlockCaptcha to unlock an account.
    For me I check from work computer, home computer, and my iPhone (thus AT&T ip assigned while in this home area). Never had an issue until I went on travel to another state and thus AT&T gave another IP and presto, kept getting bad user name or password messages when trying to check mail. Once doing the unlock I could check again.
    Not sure if Yahoo is doing the same.

  • Proxy Services won't work in imanager - ran fillattr success

    I upgraded to BM 3.9.2 and got the migration error. So I ran fillattr with the correct arguments and it said that the migration was successful. Yet, when I go into Imanager and try to pull up proxy services it says "Either you have selected the wrong object or the migration is not done properly (Please run the fillattr manually )".
    My Access rules transferred over fine after running fillattr. What can I do to fix this. The last few lines of my log are below:
    Got the object
    Got the object
    Got the object
    Got the object
    Success. Migration completed successfully.

    Sorry for really slow response on this - are you still here/still
    having the issue?
    Fillattr runs two migrations - one for access rules, and another for
    proxy settings. I've seen cases where one works and the other fails.
    But eventually you can get it going if you find the hangup.
    Craig Johnson
    Novell Knowledge Partner
    *** For a current patch list, tips, handy files and books on
    BorderManager, go to http://www.craigjconsulting.com ***

  • I am running a Mac G5 with OS 10.5.8. I downloaded FF4, won't work on "this architecture" -- so what "architecture" will it run on? Normally a new application lists the requirements, but FF4 says "all apps, all computers, all everything."

    I'm really angry about how this is working (or not working, actually). And every time I get a message from FireFox to "click here to do. . " it won't connect.

    Firefox 4 requires at least OS X 10.5 and an Intel Mac. There is a third party version of Firefox 4 that runs on OS X 10.4/10.5 and PPC Macs, for details see http://www.floodgap.com/software/tenfourfox
    If you prefer, you can get the latest version of Firefox 3.6 from http://www.mozilla.com/en-US/firefox/all-older.html

  • The hold button won't work and this morning i cracked the screen!!

    Hello,
    I was wondering if anyone can help me! The hold switch on my ipod nano does not work, or is very slow. It only works if I press down very hard on the hold button. This is of course very frustrating!
    However, just this morning I cracked the screen.I know that cracked screens are accidental damage and are therefore not covered by warranty, but the hold button was not working before the screen was cracked- it is just bad luck that this has happened before sending my ipod off to Apple to see if they can resolve the issue.
    Does anyone know if it is possible to get the hold-button issue fixed even though the screen is cracked? I am a bit of a disaster area with electrical goods, having cracked the screen of my laptop and of my phone in the past!
    Thanks
    iPod nano   Windows XP  

    I believe it is a hardware issue, you have to send it
    repairing.
    yeah but I won't be able to get the repair done on warranty, will I?

  • Will instanceof  work  in this case

    I am trying to simplify my code, and here is what i have
    i have the below classes :
    interface xyz
    class A implements xyz
    class B implements xyz
    i have lot of set/get methos in class A and class B. which are not in xyz.
    setfields(Obj) //this obj can be class A or class B
    Object ob=null;
    if ( obj instanceof A)
    ob=(A)obj;
    if(obj instanceof B)
    ob=(B)obj;
    //Here depending on the object i want to use set methods on that particular object. Is it possible?
    But when i do this , ob doesn't show the methods in Class A / class B.
    Am i doing the correct way.
    }

    I agree that's a good way to simplify your code, but you had a mistake
    in that...
    ob=(A)obj; // ob is a instance of Object
    because ob is just a object, so it can access only Object's method.
    you can correct code like this.
    A obj = (A)obj; // obj is a instance of Class A
    then, it can access any method in Class A.
    and I recommend make a 'int getType()' method in interface and
    use that. for example...
    interface xyx{
    public int getType();
    class A implements xyz
    public int getType(){return 0;}
    class B implements xyz{
    public int getType(){return 1;}
    setfields(xyz Obj) // this obj is must be instance of class implements xyz
    switch(Obj.getType()){
    case 0:
    A Obj2 = (A)Obj;
    break;
    case 1:
    B Obj2 = (B)Obj;
    break;
    you can make static int variable or enumeration class for more elegant code.
    I'm sorry about I can't explain well so that english isn't my mother tongue. -_-;;                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       

  • Sign out dowsn't work in this case .. any solution

    Hi
    I have a module called "main screen" that shows another modules in the system in <af:inlineFrame>. it has a login page and
    sign out commandNavigationItem. when user selects a mdoule from a tree in the main page the inlinframe loads the specified module.
    the problem occurs when after I load the module the Signout doesn't work ... it reload the page only.
    if the user didn't select a module and the inlineframe is empty sign out works!!!.
    I tried:
    . session.invalidate();
    . /adfAuthentication?logout=true&end_url=
    but works only if the user didn't load any modules.
    Any Ideas?!!!!

    Hi
    At last, I have found solution for this problem. I don't got it fully!.. but seems that session.invalidate() doesn't clear all user session data when called.
    SO, on weblogic there is a method I have found that make a full clear to user session data and go back to the login page.
    [Programming WebLogic HTTP Servlets|http://e-docs.bea.com/wls/docs81/servlet/progtasks.html].
    the method is called
    weblogic.servlet.security.ServletAuthentication.invalidateAll()
    I hope this helps other

  • Software update 10.0.2 says that Logitech Device detection 1.23.05 won't work with this new update and will be disabled so I need Help.

    I have a Logitech Y-BH52 keyboard so I don't want to have it disabled with the new update.

    tdaignault wrote:
    I tried to change the link speed and duplex from "auto negotation" to "1 Gbps Full Duplex," but then my computer no longer even could connect.
    If your computer can't negotiate a 1Gbps connection, then that is what is limiting you to 94Mbps, which is about right for a "100Mbps" port.
    There could be several reasons for this.
    A bad NIC
    A bad port on the switch
    IMO, the most likely cause is a bad cable.  1Gbps requires all 4 pairs in the cable be connected.  If only 2 of the 4 pairs are connected correctly, you will get a 100Mbps connect, but will not be able to connect at 1Gbps.  Try replacing the cable with one known to work at 1Gbps.

  • Clickwheel won't work proberly with case

    Hi there,
    I've got the nano 4g and everything works fine apart from the fact that when using a skin case (soft one) the click wheel is not responsing. I have another identical nano and with that one in works just fine.
    What is it? Without the case the two ipod are completely identical in control (sensitivy of clickwheel).
    Hope anyone has an idea

    You fried it, most likely.
    Try powering it off, dry it thoroughly and let it sit for 5 or more days.
    Try it then-if it does not work you will most likely have to replace it.

  • I have a new iMac and want to use my HP 7300 All in One and to use the "Fax"feature I need a phone port on computer and of course, this computer does not have one.  Are there any aftermarket devices for USB to phone that would work in this case?

    I have a new iMac and want to use my HP 7300 "All in One and cannot use the fax feature because there is no phone port is there a solution for this?BACellli

    You don't need a phone port on the iMac. The HP 7300 DeskJet Pro has an internal modem, you connect the printer's modem connection directly to your phone line. Please re-read the install directions that came with the printer.

  • Downloaded latest version to Mac OS X 10.4.11, deleted previous version, now flag tells me 4.0 won't work on this Mac -- how can I download a previous version?

    Received a prompt today to download the latest Firefox version, it proceeded as a normal download, asked if it should replace previous version, I selected "Yes", and...a flag telling me, a little tartly, that this version is incompatible with my operating system. Help!

    Firefox 4 requires at least OS X 10.5 and an Intel Mac.
    * http://www.mozilla.com/firefox/4.0/system-requirements/
    For Mac OS X 10.4.11 or Mac OS X 10.5.8 you can look at:
    * http://www.floodgap.com/software/tenfourfox/
    Firefox 3.6.x can be found here:
    * http://www.mozilla.com/en-US/firefox/all-older.html

  • I have a pile of images that I need to add dimensions to. (show height, width, depth with arrows). I think data merge won't work for this because the images are all different. Is there any way to automate this?

    I've been doing this manually for hundreds of images, and it seems like i could create an action that would reference the folder of images and also reference a text file and merge the two. Is this possible?

    basically, i have 300 images that are silhouetted and on a white background. i need to draw in arrows and then place the dimensions on the image, much like this:
    But you're right - i will check on the scripting forum. thanks for pointing me in the right direction!

  • Unary Operator - how it is working in this case??

    Hi,
    I have got a piece of code...some one plz clearify the output of this code.
    int a = 10;
    a = a++; //1
    a++; //2
    System.out.println(a);
    now the problem is that if i comment line //2 then the output is 10. If I comment the line at //1 and removing the comment at line at //2 then the output is 11.
    Can some one plz help me out to know what is happening behind the scene.
    Thanks
    GShankar

    a = a++; //1In the above expression the right hand side of the assignement is first evaluted (to 10 in your example). Then a is incremented to 11. Finally a is assigned the result of the right hand side evaluation so it becomes 10 again.
    So the total effect of the expression is that nothing happens.

  • How to add black bars to SIDES of image? Using 'Crop' won't work for this. I explain why.

    I have some clips that are off axis and I applied a bit of tilt to correct them. However, I now need to crop the sides giving me a 1.66:1 effect inside the 1.85:1 frame. But how/ Using CROP in Premiere Pro just crops the image off axis. Here is a screen grab...
    How can I get a nice, perfect side crop? I've searched, folks, so would appreciate some guidance. Thanks!

    4 point garbage matte to crop.
    Or overlay a title with black pillars.

Maybe you are looking for