Getting 401 URL

Hi,
when i am creating the java proxy from the .NET web service using the third party tool apche SOAP Axis tool, and i used the command for this is
java org.apache.axis.wsdl.WSDL2Java <web service> then
its giving the error as
java.io.IOException: Server returned HTTP response code: 401 for URL: http://10.113.10.78:8080/OracleSPS/OracleSPS.asmx?WS
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:709)
at java.net.URL.openStream(URL.java:960)
at org.apache.crimson.parser.InputEntity.init(InputEntity.java:209)
at org.apache.crimson.parser.Parser2.parseInternal(Parser2.java:471)
at org.apache.crimson.parser.Parser2.parse(Parser2.java:305)
at org.apache.crimson.parser.XMLReaderImpl.parse(XMLReaderImpl.java:442)
at org.apache.crimson.jaxp.DocumentBuilderImpl.parse(DocumentBuilderImpl.java:185)
at org.apache.axis.utils.XMLUtils.newDocument(XMLUtils.java:369)
at org.apache.axis.utils.XMLUtils.newDocument(XMLUtils.java:420)
at org.apache.axis.wsdl.symbolTable.SymbolTable.populate(SymbolTable.java:482)
at org.apache.axis.wsdl.gen.Parser$WSDLRunnable.run(Parser.java:361)
at java.lang.Thread.run(Thread.java:536)
i can view the web service with my username and password. and also i set the all jars in classpath. i am not under standing why should getting this error,
so please help me, its very urgent to my progress...
Thanks

Sorry, I have no answer for this, but I am getting a similar problem and would really like an answer. It seems that when communicating between servlets, there needs to be some sort of security setting that will allow one to communicate via http to the other.
Surely someone has encountered this and knows what needs to be done?

Similar Messages

Getting around URL aliasing or forwarding in Automator?

Hi,
I am new to automator but have a simple task that seems not straightforward to implement. I have a page on Google Sites (unfortunately not open to the public) and I upload and download ruby code to it. I'd like to automate downloading all the files that are have certain file extension and are linked from the main page. Those urls, when I copy them as links from the Safari browser, have a format like;
http://sites.google.com/a/some_school.edu/my-cool-proj/Home/builder.c?attredirec ts=0
My workflow is four safari actions;
Get Specified URLs (the url is the top page of the google site for the project)
Get Link URLs from Webpages
Filter URLs
Download URLs
What I notice when I download those files manually from the Google Site page, using Safari, is that there is a second url that the link I started downloading gets forwarded to, apparently, and I can't seem to capture that manually to see what it is (it flashes across the browser's address bar and then dissappears pretty quickly.)
Is there some simple way to follow these forwarded URLs, that is relatively foolproof, and that could be automated? I could take it on faith that the implementation of those forwarded links is not going to change, though that seems iffy. I could use or build some other site to post and pull code from, but that seems risky, and honestly the Automator script wouldn't save me more time than I would lose in such an effort.
Any suggestions would be appreciated -
C

Hello Prashanth,
It may not be as simple as adding an http header through the Oracle HTTP Server (OHS) in the stack. The web page is returned by both the Parallel Page Engine in OC4J_Portal and the Web Cache, not by the OHS. Since OHS doesn't play a role in the response, it is not possible to tag the headers.
You could consider to install a reverse proxy server (e.g. Apache 2) in front of your web cache and add the header tag in this configuration. That should work.
Thanks,
EJ

How to get the URL of the request.

I am trying to get the URL from where the request is coming to the servlet. If http://www.xx.com/ calls my servlet i want to know that the request is coming from http://www.xx.com/ URL. If i use request.getRequestURL() which is displaying my servlet URL. Please help me which is very urgent.

You actually want the referrer? You can obtain it from the request headers as follows:String referrer = request.getHeader("referer"); // Yes, with the legendaric misspelling in there.Having said that, be aware that the client has full control over what he sends with the request headers. So the client can change/spoof/remove the referrer entry to his taste anytime. Your code should not rely that much on it, it should rather be used for statistics only.

How to get the URL of an iView using AbstractPortalComponent

Hi All,
I need to get the URL of an iView using AbstractPortalComponent using the following code.
But i am unable to open the URL from the Browser
This is the following code snippet
public class AbsClass extends AbstractPortalComponent
public void doContent(IPortalComponentRequest request, IPortalComponentResponse response)
IPortalComponentURI componentURI = request.createPortalComponentURI();
componentURI.setContextName("pcd:portal_content/LOG_Viewer/LogViewer");
String URI=componentURI.toString();
response.write(URI);
The output i got from the browser is:
/irj/servlet/prt/portal/prtroot/pcd!3aportal_content!2fLOG_Viewer!2fLogViewer
Actual Path: pcd:portal_content/LOG_Viewer/LogViewer
Output Path: /irj/servlet/prt/portal/prtroot/pcd!3aportal_content!2fLOG_Viewer!2fLogViewer
Why some extra numerics are embeded in the output path when compare with the Actual Path.
Can any one send me the modified code and your suggestions, how to achieve this problem.
Regards
Phani

Hi,
Now i need to get the URL of this component.
IPortalComponentRequest request = (IPortalComponentRequest) this.getRequest();
IPortalComponentURI componentURI = request.createPortalComponentURi();
componentURI.setContextName("com.sap.portal.appintegrator.sap.bwc.Transaction");
String URI=componentURI.toString();
response.write(URI);
I export this component into Portal and i tested with an iView based on this par file, i got the URL in the Preview.
But when i tried to run this URL by adding the http://<IPADDRESS>:<PORTNO>+URL
I am facing one AccessDenied Exception for this Object. ( USER/GUEST)
Is there any security zone, or other Permissions Problem, if so can you plz guide me what are the settings i have to change inorder to resolve this exception
Regards
Phani

How do I get a URL to embed with FormsCentral add-on to Acrobat XI Pro?

I have been going around in circles for the last few HOURS trying to find out why my FormsCentral (signed up and signed in, free account) does not give me the options shown in all of the help areas...
When I open my uploaded form (I made a fillable PDF form using Acrobat XI Pro), the drop down menus from the top are Imported Form, Options, Distribute, View Responses and Summary Report... there is no option to Test and no option in the Distribute area to get the URL to embed...
It says pretty much everywhere that a PDF made in Acrobat can be imported to FormsCentral for distribution, however it won't give me the option to distribute online at all...
Going insane in the maze of forums/ tutorials etc...

Forms Central forum http://forums.adobe.com/community/formscentral

I am unable to reply to your e-mail convermation--how can I get the URL to past into my browser?

Received a confirmation e-mail requestion me to reply to it and I cannot get the URL to paste to my browser--how can I do this???????/

Hi June,
Sorry for the trouble you've had.
I've verified your account on our end and you should be able to access the ExportPDF service now.
Please let us know if you have any questions!
-David

How to get the url of a document in content management?

Hello,
I have used API 'SKWF_PHIO_CONTENT_URL_GET to get the url of a file in CRM content management. But when I tried to open this url in explorer got error message "SSF error: Decoding not possible". Could any one please help me to resolve this problem.
Thanks
Nisha

Thanks for replying - unfortunately, there is not much
support on these forums.
I wish there was some way to emulate the mouse from
Javascript to click the play, pause, and slider buttons... but one
would't even know, programmatically, the location on screen of the
play, pause, and slider buttons.
quote:
Originally posted by:
davexnet
You'll notice that questions like this don't get answered on
this
forum. The Flash player has a philosophy that skews the power
over almost everything about it to the content providers,
and precious little control, if at all, to the client player.
It's a crappy situation - I use Flashblock in Firefox and
only view the
content *I* want to see.
It works for me, particularly since I'm using an older
(slower )
computer that really gets bogged down by aggressive Flash
content.

This is showing up on videos I try to view. I have latest update. It is a box that displays over video that says, get video URL, get embed code, etc. how do I correct it?

A Box has just started showing up that has the following.
get video url
get video URL at current time
Get embed code
report playback issue
get debug info
stats for nerds
about the HTML.5 player
this is on every video on face book. How to fix?

I had the same issue.
I had a youtube downloader app that had a plug-in installed on my Home Safari. I deleted that app. and the plugin. Not sure if that was the culprit, but I figured I'd try.Thought maybe some bug crossed over when syncing the two safaris.
Anyway, I ended up unlinking my Mobile Safari to my Home Safari (iPad; iCloud/Safari/OFF)
A prompt will come up asking what you want to do with previously synced data: Delete from My iPad (don't worry, you'll get it back)
Your bookmarks should disappear.
Reset HOME Safari : Safari/Reset Safari (everything checkmarked)
Turn ON the Mobile Safari again (if you want to relink to the Home version) or keep off (and rebuild it from scratch on iPad alone)
Personally, I turned it on, got my bookmarks back then turned it off and pressed Keep on my iPad.
Now the two aren't linked and I have the bulk of my bookmarks back.
There might be variations of this solution so play around with the process and be sure to refresh your Mobile Safari browser on a video clip to see if it takes.

Webkit get the URL of a link

I am trying to add policy functionality to an app that I have written so that, for example, instead of opening a link to a page within my app itself it might open the link in the default browser or open a window to handle the functionality internally rather than use the functionality provided online.
At this stage I am only trying to log what the URL is so that I can see what the format is. Once I've worked that out, I'll put together the policy code. The code that I have written so far is this:
<pre>
- (void)webView:(WebView *)sender decidePolicyForNavigationAction:(NSDictionary *)actionInformation
request:(NSURLRequest *)request
frame:(WebFrame *)frame
decisionListener:(id<WebPolicyDecisionListener>)listener
// get the url from the information dictionary:
NSURL *url = [\[actionInformation objectForKey:@"WebActionElementKey"] objectForKey:@"WebElementLinkURL"];
// log the url's scheme
NSLog(@"absolute string is \[%@]", [url absoluteString]);
// NSLog(@"relative string is \[%@]", [url relativeString]);
// NSLog(@"fragment is \[%@]", [url fragment]);
// NSLog(@"host is \[%@]", [url host]);
// NSLog(@"parameter string is \[%@]", [url parameterString]);
// NSLog(@"user is \[%@]", [url user]);
// NSLog(@"password is [%@]", \[url password]);
// NSLog(@"path is \[%@]", [url path]);
// NSLog(@"relative path is \[%@]", [url relativePath]);
// NSLog(@"port is \[%@]", [url port]);
// NSLog(@"query is \[%@]", [url query]);
// NSLog(@"resource specifier is \[%@]", [url resourceSpecifier]);
// NSLog(@"scheme is \[%@]", [url scheme]);
}</pre>
As you can see, I've tried every component of the WebElementLinkURL - but without luck. The results mostly came back null - only absoluteString seemed useful (hence it's the only one not commented out). But even that didn't so far enough.
For example, when clicking a link, absoluteString returns
22/04/2009 09:58:00 Test[646] absolute string is [https://test.internal.com/#]
Hovering over the same link in Safari, or clicking on it, returned https://test.internal.com/client/help
The question, therefore, is how can I get my code to return the URL in the same manner that Safari does. The website I'm trying to view does contain a fair bit of Javascript - I don't know if this is relevant though.
Methinks I am looking in the wrong place!

Hi Jun,
Before creating a customised framework page, copy the default framework
page provided my SAP to your own namespace and make changes.
Just open the copied page, set masthead to invisible ,this can be done
for any other component that you want to hide on the framepage.Then
make only the inner page visible, if that's what you want.
Then create a url iview that points to this page and pass room details as
parameters.This can be assigned to any role that you wish.
<i>Pls reward points if helpful</i>
Regards
Vineeth

Help needed in getting file url

Hi,
I work in a project where i need to get the url of the specified file. The code for that is as,
URL fileURL = Thread.currentThread().getContextClassLoader().getResource();
The required resource is available in the concerned path. But the file url returns null
Can you please help in solving the problem.
Thanks and Regards,
Mclaren

Where (text area, label, button) do you want it to do it.
What do you want it do (browse within the local text area, open the default browser...)
Text area, use above, JButton (or alike) and open default browser, use ActionListener and JDIC.

How to get the url of the document in quick parts?

I have document library in which there is one content type with the document template uploaded. In the footer of the document, I want to show the URL of the document which is opened in word application.
Is there any way to get the URL? I am not able to see the field in Quick Parts while configuring footer of document template.

Alexdu_,
You can create a single line text site column in that content type to hold URL information.
Create a SPD workflow to populate URL field using update field of current list.
This custom column will be available as Quick part in word.
This could be an alternate approach as URL is not available as quick part field.
Please 'propose as answer' if it helped you, also 'vote helpful' if you like this reply.

"Get Started" URL is rejected and XE database does not appear to start

Hi,
Brand new to Oracle, so I apologize for my ignorance. Have spent 3 days trying to launch the Get Started URL that Oracle puts on my desktop, or connect with Rapid SQL. I've spent all day today trolling these forums and am unable to resolve this issue.
OS: Windows 7 Enterprise SP1
System: i7 @ 2.7GHz, 8GB RAM, 64-bit
DB: OracleXE 11g
Error message from Chrome when I try to connect to http://127.0.0.1:8080/apex/f?p=4950: Error 102 (net::ERR_CONNECTION_REFUSED): The server refused the connection.
Here is the result of my TNS Ping Utility for 32-bit Windows: Version 11.2.0.2.0 - Production on 16-NOV-2
012 14:42:28
Used parameter files:
C:\oraclexe\app\oracle\product\11.2.0\server\network\admin\sqlnet.ora
Used TNSNAMES adapter to resolve the alias
Attempting to contact (DESCRIPTION = (ADDRESS = (PROTOCOL = TCP)(HOST = CONNECTL
AP28.awi.state.fl.us)(PORT = 1521)) (CONNECT_DATA = (SERVER = DEDICATED) (SERVIC
E_NAME = XE)))
OK (0 msec)
Here is the result of my LSNRCTL> status
Connecting to (DESCRIPTION=(ADDRESS=(PROTOCOL=IPC)(KEY=EXTPROC1)))
STATUS of the LISTENER
Alias LISTENER
Version TNSLSNR for 32-bit Windows: Version 11.2.0.2.0 - Produ
ction
Start Date 16-NOV-2012 13:41:57
Uptime 0 days 1 hr. 2 min. 50 sec
Trace Level off
Security ON: Local OS Authentication
SNMP OFF
Default Service XE
Listener Parameter File C:\oraclexe\app\oracle\product\11.2.0\server\network\a
dmin\listener.ora
Listener Log File C:\oraclexe\app\oracle\diag\tnslsnr\CONNECTLAP28\liste
ner\alert\log.xml
Listening Endpoints Summary...
(DESCRIPTION=(ADDRESS=(PROTOCOL=ipc)(PIPENAME=\\.\pipe\EXTPROC1ipc)))
(DESCRIPTION=(ADDRESS=(PROTOCOL=tcp)(HOST=CONNECTLAP28.awi.state.fl.us)(PORT=1
521)))
Services Summary...
Service "CLRExtProc" has 1 instance(s).
Instance "CLRExtProc", status UNKNOWN, has 1 handler(s) for this service...
Service "PLSExtProc" has 1 instance(s).
Instance "PLSExtProc", status UNKNOWN, has 1 handler(s) for this service...
The command completed successfully
Here is the result of my sqlplus / as sysdba
SQL*Plus: Release 11.2.0.2.0 Production on Fri Nov 16 14:47:53 2012
Copyright (c) 1982, 2010, Oracle. All rights reserved.
ERROR:
ORA-28056: Writing audit records to Windows Event Log failed
OSD-00512: Message 512 not found; product=RDBMS; facility=SOSD
O/S-Error: (OS 1722) The RPC server is unavailable.
ORA-28056: Writing audit records to Windows Event Log failed
OSD-00512: Message 512 not found; product=RDBMS; facility=SOSD
O/S-Error: (OS 1722) The RPC server is unavailable.
Enter user-name: system
Enter password:
ERROR:
ORA-01034: ORACLE not available
ORA-27101: shared memory realm does not exist
Process ID: 0
Session ID: 0 Serial number: 0
I have uninstalled and reinstalled the 64-bit version, the 32-bit version, repaired, stopped and started services, changed KEY_XE.reg, and the properties of the Get Started URL. No luck so far. Please let me know what I can do.
Thanks in advance.

Thank you for getting back to me so quickly!
When I signed in to Event Viewer, I got an error message that my event services weren't running, so I downloaded this fix from Microsoft: http://support.microsoft.com/kb/2478117
That started my Event Services and allowed me to sign into Event Viewer and clear my logs. They are all clear now.
Then I repaired my version of 11G and rebooted, but am still unable to connect to the database or the Get Started URL.
Here are my statuses:
LSNRCTL> status
Connecting to (DESCRIPTION=(ADDRESS=(PROTOCOL=IPC)(KEY=EXTPROC1)))
STATUS of the LISTENER
Alias LISTENER
Version TNSLSNR for 32-bit Windows: Version 11.2.0.2.0 - Produ
ction
Start Date 16-NOV-2012 16:31:35
Uptime 0 days 0 hr. 2 min. 30 sec
Trace Level off
Security ON: Local OS Authentication
SNMP OFF
Default Service XE
Listener Parameter File C:\oraclexe\app\oracle\product\11.2.0\server\network\a
dmin\listener.ora
Listener Log File C:\oraclexe\app\oracle\diag\tnslsnr\CONNECTLAP28\liste
ner\alert\log.xml
Listening Endpoints Summary...
(DESCRIPTION=(ADDRESS=(PROTOCOL=ipc)(PIPENAME=\\.\pipe\EXTPROC1ipc)))
(DESCRIPTION=(ADDRESS=(PROTOCOL=tcp)(HOST=CONNECTLAP28.awi.state.fl.us)(PORT=1
521)))
Services Summary...
Service "CLRExtProc" has 1 instance(s).
Instance "CLRExtProc", status UNKNOWN, has 1 handler(s) for this service...
Service "PLSExtProc" has 1 instance(s).
Instance "PLSExtProc", status UNKNOWN, has 1 handler(s) for this service...
The command completed successfully
LSNRCTL> exit
sqlplus / as sysdba
SQL*Plus: Release 11.2.0.2.0 Production on Fri Nov 16 16:34:42 2012
Copyright (c) 1982, 2010, Oracle. All rights reserved.
Connected to an idle instance.
SQL>
At least it's a different error message now! Any ideas?
Edited by: 971714 on Nov 16, 2012 1:40 PM
Edited by: 971714 on Nov 16, 2012 1:41 PM
Edited by: 971714 on Nov 16, 2012 1:41 PM

How to get full url using UCM RIDC api

Hi,
I am using RIDC api to get the full url of the document (including protocol, host, port info). I tried using DOC_INFO
and GET_SEARCH_RESULTS api. The GET_SEARCH_RESULTS api returns the path but no protocol, host, port info.
DOC_INFO does not contain any uri. Is there a way to get full url? I only have access to the idc socket connection
configuration to connect to the server.
E.g. I know the idc connection url is idc://hostname:4444
The user supplies document name as "myDoc.txt" and dId = 2000.
I used GET_SEARCH_RESULTS for myDoc.txt and DOC_INFO for dId=2000.
I need to return absolute url to the user that user can copy paste in the browser and download the doc.
Thanks,
--dhiraj

You can ge the docUrl using doc info services DOC_INFO or DOC_INFO_BY_NAME.
dataBinder.putLocal("IdcService", "DOC_INFO");
dataBinder.putLocal("dID", docId);
ServiceResponse response = idcClient.sendRequest (idcContext, dataBinder);
DataBinder respBinder = response.getResponseAsBinder();
System.out.println("DocUrl"+ respBinder.getLocal("DocUrl"));

Get original URL for Link to a Document content type

Hello,
I'm having a SharePoint doc library with ItemAdded event receiver attached to it and "Link to a Document" content type enabled. My general question is:
How to get URL that the link points to?
I'll show you how I'm trying to achieve that:
SPWeb parentWeb = properties.Web;
if (properties.ListItem.ContentType.Name == LinkContentTypeName)
object urlFieldValue = properties.AfterProperties["URL"];
if (urlFieldValue != null && urlFieldValue.ToString() != String.Empty)
// continue
But I always get null so I can't even get inside IF statement. I've tried to install ItemAdded event receiver as synchronous, but it didn't make a difference. I'd appreciate any ideas.
Regards
Tomasz

This is not very convenient solution, because ItemUpdated can be executed more than once. However this certainly will work, so I'll accept this as a workaround. Actually I've found another one. I'm parsing link aspx file and I'm getting the URL
from mso:URL tag:
<%@ Assembly Name='Microsoft.SharePoint, Version=14.0.0.0, Culture=neutral, PublicKeyToken=71e9bce111e9429c' %><%@ Register TagPrefix='SharePoint' Namespace='Microsoft.SharePoint.WebControls' Assembly='Microsoft.SharePoint' %><%@ Import Namespace='System.IO' %><%@ Import Namespace='Microsoft.SharePoint' %><%@ Import Namespace='Microsoft.SharePoint.Utilities' %><%@ Import Namespace='Microsoft.SharePoint.WebControls' %><html xmlns:mso="urn:schemas-microsoft-com:office:office" xmlns:msdt="uuid:C2F41010-65B3-11d1-A29F-00AA00C14882"><Head> <META Name='progid' Content='SharePoint.Link'></head><body><form id='Form1' runat='server'><SharePoint:UrlRedirector id='Redirector1' runat='server' /></form></body></html>
Regards
Tomasz

How do I get the URL of a jsp page?

I have a jsp page which has a link to a servlet page, in the servlet page, I want to get the URL of the jsp page.
I tried to use request.getHeader("referer"), it seems like if it is a link in the jsp directly lead to the servlet, it works, but if the link calls a javascript function which opens up a window runing the servlet, it is not working. I have to do the second way, so it is not working by doing request.getHeader("referer")
I tried to use javascript "location.href", I get the whole url by alerting the url in the jsp page:
http://localhost:8080/pos-webapp/serv/qt_old?currentpage=portal&bcd=ABC123&pnm=P1001830&prnm=1&psd=236863.0&ln=01&sln=NYPPA&stp=2.1&stt=A&_POSID=1F8464122ADAF4EEC9B1A94C286EB252
then I pass this url when calling with my servlet, in the servlet page, when I do request.getParameter("url"), the url is actually cut off like this:
http://localhost:8080/pos-webapp/serv/qt_old?currentpage=portal
anything after &pnm is cut off, anyone has any idea?
Thanks in advance.

Thank you.
but how do I put encoding in the javascript code? do I need to decode in my servlet page?
here is my code in the jsp page:
linkset[3]+='<div class="menuitems">Email Diagnostics to AutoOne</div>'<%
function emailDiagnostics()
win = window.open("/pos-webapp/serv/email_diagnostics?url=" + location.href + "&browser=" + navigator.appName, "emailDiagnostics",
"width=350, height=140, resizable=no, scrollbars=no, toolbar=no, location=yes, directories=no, status=no, menubar=no, copyhistory=no, left=450, top=300");
win.focus();
}

Getting 401 URL

Similar Messages

Maybe you are looking for