Get the current file name of uploaded file

I'm looking for a way (an API?) to use files in content areas in
portlets, dynamic pages and html. ie <img
src='http://he130r.us.oracle.com/pls/portal30/docs/FOLDER/CUSTOM/
IMAGES/app1.gif"> This syntax works until I delete the file
and/or replace with a new one.... At that time the file name
becomes app1_0.gif..... Is there a way to find the proper URL
based on the display name? I've looked at the wwsbr_all_items
view but can't figure out where to find the path... If this
view is the right place do you have some sample queries?
thx

While some of the info is in that table I can not create a
complete picture of how to find the file. I need to recreate a
vaild url based upon the display name of the file.
wwdoc_document$ contains a column called filename which holds in
my case 'CONTACTS.GIF' while the file is displayed
as 'Contacts.gif'.... The name column holds the
value '755.GIF'... I would really like an API that I can pass
the value 'Contacts.gif' to and it
returns 'http://he130r.oracleads.com/pls/portal30/doc/FOLDER/CUST
OM/IMAGE/Contacts_1.gif' or whatever.... Short of that I need
to create a procedure that I can call that will build this url
based on the display name I pass it..... HELP! Any ideas?

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How to get the current schema name

Hi,
Can anybody please tell me how to get the current schema name, there is some inbuilt function for this,but i am not getting that. Please help me.
Thanks
Jogesh

ok folks, I found the answer at Tom's as usual.
http://asktom.oracle.com/tkyte/who_called_me/index.html
I rewrote it into a function for kicks. just pass the results of DBMS_UTILITY.FORMAT_CALL_STACK to this function and you will get back the owner of the code making the call as well some extra goodies like the name of the code and the type of code depending on the parameter. This ignores the AUTHID CURRENT_USER issues which muddles the schemaid. Quick question, does the average user always have access to DBMS_UTILITY.FORMAT_CALL_STACK or does this get locked down on some systems?
cheers,
paul
create or replace
FUNCTION SELF_EXAM (
   p_call_stack VARCHAR2,
   p_type VARCHAR2 DEFAULT 'SCHEMA'
) RETURN VARCHAR2
AS
   str_stack   VARCHAR2(4000);
   int_n       PLS_INTEGER;
   str_line    VARCHAR2(255);
   found_stack BOOLEAN DEFAULT FALSE;
   int_cnt     PLS_INTEGER := 0;
   str_caller VARCHAR2(30);
   str_name    VARCHAR2(30);
   str_owner   VARCHAR2(30);
   str_type    VARCHAR2(30);
BEGIN
   str_stack := p_call_stack;
   -- Loop through each line of the call stack
   LOOP
     int_n := INSTR( str_stack, chr(10) );
     EXIT WHEN int_cnt = 3 OR int_n IS NULL OR int_n = 0;
     -- get the line
     str_line := SUBSTR( str_stack, 1, int_n - 1 );
     -- remove the line from the stack str
     str_stack := substr( str_stack, int_n + 1 );
     IF NOT found_stack
     THEN
        IF str_line like '%handle%number%name%'
        THEN
           found_stack := TRUE;
        END IF;
     ELSE
        int_cnt := int_cnt + 1;
         -- cnt = 1 is ME
         -- cnt = 2 is MY Caller
         -- cnt = 3 is Their Caller
         IF int_cnt = 1
         THEN
            str_line := SUBSTR( str_line, 22 );
            dbms_output.put_line('->' || str_line);
            IF str_line LIKE 'pr%'
            THEN
               int_n := LENGTH('procedure ');
            ELSIF str_line LIKE 'fun%'
            THEN
               int_n := LENGTH('function ');
            ELSIF str_line LIKE 'package body%'
            THEN
               int_n := LENGTH('package body ');
            ELSIF str_line LIKE 'pack%'
            THEN
               int_n := LENGTH('package ');
            ELSIF str_line LIKE 'anonymous%'
            THEN
               int_n := LENGTH('anonymous block ');
            ELSE
               int_n := null;
            END IF;
            IF int_n IS NOT NULL
            THEN
               str_type := LTRIM(RTRIM(UPPER(SUBSTR( str_line, 1, int_n - 1 ))));
             ELSE
               str_type := 'TRIGGER';
             END IF;
             str_line := SUBSTR( str_line, NVL(int_n,1) );
             int_n     := INSTR( str_line, '.' );
             str_owner := LTRIM(RTRIM(SUBSTR( str_line, 1, int_n - 1 )));
             str_name := LTRIM(RTRIM(SUBSTR( str_line, int_n + 1 )));
          END IF;
       END IF;
   END LOOP;
   IF UPPER(p_type) = 'NAME'
   THEN
      RETURN str_name;
   ELSIF UPPER(p_type) = 'SCHEMA.NAME'
   OR    UPPER(p_type) = 'OWNER.NAME'
   THEN
      RETURN str_owner || '.' || str_name;
   ELSIF UPPER(p_type) = 'TYPE'
   THEN
      RETURN str_type;
   ELSE
      RETURN str_owner;
   END IF;
END SELF_EXAM;

How to get the current function name in java

How to get the current function name in java.
In c it is done as
printf("%s",__func__);
Thanx in advance.

j0o wrote:
System.out.println("Class Name: " + new Exception().getStackTrace()[0].getClassName() +
"/n Method Name : " + new Exception().getStackTrace()[0].getMethodName() +
"/n Line number : " + new Exception().getStackTrace()[0].getLineNumber());
I pointed the OP at this approach yesterday in one of his multi-posts. I still have not been given my Dukes!

How to get the current class name in static method?

Hi,
I'd like to get the current class name in a static method. This class is intended to be extended. So I would expect that the subclass need not to override this method and at the runtime, the method can get the subclass name.
getClass() doesn't work, because it is not a static method.
I would suggest Java to make getClass() static. It makes sense.
But in the mean time, does anybody give an idea to work around it?
Thank you,
Bill

Why not create an instance in a static method and use getClass() of the instance?
public class Test {
public static Class getClassName() {
return new Test().getClass();

Get the current space name

hi,
i am working on webcenter space application.i need to pass current space name to bean.how can i get the current space name??
Regards
Vinay Kumar

No, that will not compile. You probably need to do:
SpacesContext sc = SpacesContext.getCurrentInstance();
String spaceName = sc.getCurrentSpaceName();
Beware that the doc says "Return the SpacesContext instance for the request that is being processed by the current thread, if any."
So the SpacesContext might be null.
Jaap

Uix frameset - how to get the current frame name?

I have a frameset with 3 different frames in it. a top, center and bottom.
There is a menu in the top frame that changes the contents of the center or bottom frame.
How can I get the value of the frame name for the current frame into a bound value like a httpsession or page?
for example if I change the center frame to "listadmin.uix" how can I load a bound value with the frames new source?
Is there a way to pass javascript to a boundvalue? I could get the frames new properties from javascript

In Jdev 9.0.3 you could do something like this:<frameBorderLayout>
<start>
   <frame name="sideFrame">
    <boundAttribute name="source">
     <ctrl:pageURL name="sideFramePage">
      <ctrl:properties>
       <ctrl:property key="currentFrame"
                      value="sideFrame"/>
      </
     </
    </
   </
</
</then in startFramePage.uix you can access the current frame name by doing:<styledText data:text="currentFrame@ctrl:page"/>

How to get the current user name in Provider hosted app using appOnlyAccessToken

Hi,
Please help me, how to get the HostWeb UserName in Provider Hosted App
i have Provider hosted App, and Anonymous Authentication is enabled on AppWeb, using appOnlyAccessToken
Below code does not return current user who Log in in hostweb, it is returning
SharePoint App (app@sharepoint)
Web web = clientContext.Web;
clientContext.Load(web);
clientContext.ExecuteQuery();
clientContext.Load(web.CurrentUser);
clientContext.ExecuteQuery();
clientContext.Web.CurrentUser.LoginName;
Below code gives a blank name when Anonymous Authentication is enabled, if Anonymous Authentication is disabled
app prompts for credentials
HttpContext.Current.User.Identity.Name
Thanks
Ram

Hi,
Since you are using a provider Hosted app if you want to get the current logged in name than do not use AppOnlyAccessToken else use AccessToken which is App + user Context AccessToken.
then
Web web = clientContext.Web;
clientContext.Load(web);
clientContext.ExecuteQuery();
clientContext.Load(web.CurrentUser);
clientContext.ExecuteQuery();
clientContext.Web.CurrentUser.LoginName;will return proper user Name.
HttpContext.Current.User.Identity.Name will never return the user as this object is related to IIS server of your App Server not sharepoint.you should set this as Anonymous in case of provider hosted app.you can download the below sample which uses the AccessToken which has user name in it.https://code.msdn.microsoft.com/Working-provider-hosted-8fdf2d95
Whenever you see a reply and if you think is helpful,Vote As Helpful! And whenever you see a reply being an answer to the question of the thread, click Mark As Answer

How to get the current user name of the host who is occupying a specific VM?

I'm developing a winform app with c# code to manage Hyper-V. I need to remind someone if he/she would take a VM which has already been occupied by others.
Is there any powershell cmd or WMI interface to get the current user of a specific VM?
Thanks!

Hiya,
from cmd there are quser(Query user) and qwinsta(Query Session)
which should give you that. Don't know how you can incoorperate that in C#, but that should give you something to work with :)
https://technet.microsoft.com/en-us/library/cc785434.aspx
https://technet.microsoft.com/en-us/library/cc788125.aspx

How to get the complete path name when uploading a file from servlet

Hi,
I write a servlet to upload a file from html page
<intput type=file name=fileupload>I am using
import org.apache.commons.fileupload.to upload file. i want to get the all fields in the form and file name and content of the file also.
It give the file name only
String filename= fileItem.getName();
o/p krish.jpgBut i want complete path naem like
d:/krishna/images/funny/krish.jpgI serach the API org.apache.commons.fileupload.*
But i did nt find the method to get it.
plz help me , which api or method to use here..

Krishna_Rao_chintu wrote:
But i need path and have to do some calculations on it.No, you don't. If you have requirements which say you do then the requirements are wrong. You couldn't do anything useful with the path on the client system even if you could get it.
is there any alternatives in java
I need path and have to calculate MD5, Presumably you need to calculate MD5 on the contents of the file and not on the name of the file.
and convert the file to binary format.... etc oprations on itSorry, "convert a file to binary format" is basically meaningless.
but we can get the content of the file using
byte [] get()/ getString() methods
If i get content is there any performance degrades?
ie if the content is lengthy is it take more time?Take more time than what? Degrading performance from what? It's certainly true that it would be quicker to not upload the file, but that's a pointless comparison. If you have some other process to compare with, let us know what it would be.

How to get the pull path name from a file upload window

Hello everyone!
I have encountered the following problem with the following JSP code:
<form method="post" action="filename.jsp">
Upload JAVA program:
<input type=file size=20 name="fname" accept="java">
<input type=submit value="go">
</form>
<%
String s = "";
if (request.getParameter("fname") != null)
s = request.getParameter("fname")
%>
The value of s is alway the filename. However I want to get the full path in addition to the filename, so that I can read the file. Does anyone know how to get the pull name of the file?
thanks a lot in advance,

Dear Sir,
thanks a lot for your reply. Please let me explain what I intended to do: I want to upload a file from the local machine and then read the content of the file. Therefore I need to the fullpath of the filename like /var/local/file.java instead of file.java. The latter is what I got.
The problem I have with your code is that the function like "request.getServerScheme()" is not recognized. Maybe is it because I didn't install servelet package? I only installed javax package btw. Also my application runns on Tomcat server if this could give you some information. The error message I had is as follows:
org.apache.jasper.JasperException: Unable to compile class for JSP
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
[javac] Compiling 1 source file
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:133: cannot resolve symbol
symbol : method getServerScheme ()
location: interface javax.servlet.http.HttpServletRequest
url = request.getServerScheme()
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:136: cannot resolve symbol
symbol : method getServerScheme ()
location: interface javax.servlet.http.HttpServletRequest
+ ((("http".equals(request.getServerScheme()) && request.getServerPort() != 80)
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:137: cannot resolve symbol
symbol : method getServerScheme ()
location: interface javax.servlet.http.HttpServletRequest
||("https".equals(request.getServerScheme()) && request.getServerPort() != 443))
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:139: cannot resolve symbol
symbol : method getServletConfig ()
location: interface javax.servlet.http.HttpServletRequest
+ "/" + request.getServletConfig().getServletName()
^
An error occurred at line: 36 in the jsp file: /addExercise.jsp
Generated servlet error:
/usr/local/jakarta-tomcat-5.0.12/work/Catalina/localhost/tutor/org/apache/jsp/addExercise_jsp.java:140: cannot resolve symbol
symbol : variable path
location: class org.apache.jsp.addExercise_jsp
+ "/" + path
^
5 errors
     org.apache.jasper.compiler.DefaultErrorHandler.javacError(DefaultErrorHandler.java:128)
     org.apache.jasper.compiler.ErrorDispatcher.javacError(ErrorDispatcher.java:351)
     org.apache.jasper.compiler.Compiler.generateClass(Compiler.java:413)
     org.apache.jasper.compiler.Compiler.compile(Compiler.java:453)
     org.apache.jasper.compiler.Compiler.compile(Compiler.java:437)
     org.apache.jasper.JspCompilationContext.compile(JspCompilationContext.java:555)
     org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:291)
     org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:301)
     org.apache.jasper.servlet.JspServlet.service(JspServlet.java:248)
     javax.servlet.http.HttpServlet.service(HttpServlet.java:856)

How to get the full path name of a file

Hey everybody, I'm new here and in Java.
so I will explain my question by giving an example:
I want to send file from the Desktop by the "Send to" on the popup menu to my program.
I want to know how can I read the full path name and the file name, and show it on my text box.
Thanx alot

If this is a client side application then I'd say look into drag-n-drop tutorials.
i.e. drag file to your application, action listener fires, create File object giving you full name and path of users action.
If this is a web application then look into the "multipart/form-data" content type specifications on how to upload files.
i.e. user specifies file from <input type='file' ... /> type, submits, servlet receives data and recreates file locally on application server side.
If you are thinking that all you need to send a file to a program is the full path and name in a textbox its a little bit more complicated then that.
Good luck, hope that helps!

Read / Get the link destination name in PDF file by JavaScript

Is there is any way to read or retrieve the link destination name by the JavaScript?
In my document there is more then 100 of link with destination name of the named destination, now I want read all the link and retrieve the destination name in the link.
Here I was used getLinks(,) method to get the links in the document. And also I am able to get the properties of getLinks such as borderColor, borderWidth, highlightMode, rect and so on.
But I want to get the destination name of the link.
Is possible?
Thanks & Regard
Thirusanguraja Venkatesan

You may need to use a plug-in.
One of the experts on this forum, Gilad D, has a tool called Batch Edit URL Links here:
http://try67.blogspot.com.au/2010/03/acrobat-batch-edit-links.html
There is also one called AutoBookmark that also handles links:
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Check CSV file name while uploading file?

Hi All,
I am uploading data from CSV file.. That part is working fine.
Now i want to check CSV file name. Ex : CSV file name = W1234H4321AB.
I want to check 6th letter, if 6th letter is eq 'H', i want to upload data otherwise should display some message.
I tried by reading this attribute, but i declared this type as XSTRING, i checked value after using get_Attribute in lv_datasouce. It contains some XSTRING value. How to get CSV file name??
Cheers,
Venkys.

>
Thomas Jung wrote:
> >I want to check 6th letter, if 6th letter is eq 'H', i want to upload data otherwise should display some message.
>
> Everything that you have been told about using the filename property of the FileUpload is correct. However you should note that the filename will not be available until after the file has already been upload to the server. So you can't stop the fileUpload based upon a check of the filename - its too late and the upload has already occured.
yeah that is true. but is there anyway to check it otherwise.
or we need to wait for something in coming versions 7.3 or 7.4.

Indesign CS3: How to get the current document name?

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Thanks,
Alois Blaimer

...Plese Ignore the last post...<br /><br />docRef is the UIDRef of the document. Plese consider the following code<br /><br />IDocument *docFrom = Utils<ILayoutUIUtils>()->GetFrontDocument(); <br />if (docFrom == NULL ) <br /> break; <br />UIDRef docRef = GetUIDRef(docFrom )); <br /><br />InterfacePtr<IDocument> iDocument (docRef, IID_IDOCUMENT); <br />if(iDocument == nil) <br /> break; <br /><br />PMString docName; <br />iDocument->GetName(docName);

Re: Getting the current server name

If you are running inside the server as a startup class or a server side
component, you can use weblogic.management.Admin.getServerName().
-Sabha
Anand Shukla wrote:
Hi,
How can i get the running weblogic server name.
Anand

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Get the current file name of uploaded file

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