How do I get the file name?

I'm kind of new to javascript -- working with a colleague who has Adobe Acrobat Professional -- I've got an *old* version of Adobe Acrobat and most of the functions don't work -- but how do we get the file name -- we're trying to develop an application -- so if the user opens c:\something.pdf, we'd like to know the file name (something.pdf).
If we get a fully qualified path, that's fine -- I think I can parse it.
Thank you in advance.

The 'paty' property of the 'doc' object has been available since version 3.01.
Using the following in the JavaScript console when the Acobat JS API Refenece is open:
this.path;
returns:
this.path;
/C/Program Files/Adobe/Acrobat 8.0/Help/ENU/JS_api_reference.pdf

Similar Messages

How do I get the file names on a customized proof sheet?

How do I get the file names on a customized proof sheet?

Actually, I really don't know what you say.
But when you see the properties of a file in Windows, you can see there are two file size. One is called "SIze" and the other one is called "Size on disk".
Therefore, I would like to know the size on disk but not just the size.

How do I get the file name to print up closer to a horizontal photo?

Kind of a nit-picking question, but here goes anyway:
I use Lightroom's Print module to create contact sheets (three rows by three columns). I check the "Photo Info" check box and select "Filename" to have the file names printed with each photo. When the photo is shot vertically, the file name prints at the lower edge of the photo, as you would expect. However, when the photo is shot horizontally, the file name still prints down in the same place where it would print if the photo were vertical. That can be confusing because if the next photo down on the sheet is vertical, a file name can end up being much closer to the vertical photo below it--to which it is completely unrelated--than to the horizontal photo above it, to which it actually refers.
Is there any way to get Lightroom to print file names closer to horizontal photos?

Thanks to both of you for your speedy responses. Making all the shots horizontal or vertical (i. e., unchecking the "Rotate to Fit" box on the right) is not what I want. Increasing the vertical spacing is a possibility, but that'll make the vertical shots smaller. There's a program called "ImageBuddy" which does exactly what I want LIghtroom to do in this instance: print the filenames of only the horizontal shots higher and therefore closer to the photos to which they refer. It's quite cheap, but I'd rather not use Lightroom for everything else and ImageBuddy just for contact sheets, so if anybody knows of a way to do what I want to do within Lightroom, please do clue me in.

How is it posible to get the File name, size and type from a File out the H

How is it posible to get the File name, size and type from a File out the HttpServletRequest. I want to upload a File from a client and save it on a server with the client name. I want to conrole before saving the name, type and size of the file.How is it posible to get the File name, size and type from a File out the HttpServletRequest.
form JSP
<form name="form" method="post" action="procesuploading.jsp" ENCTYPE="multipart/form-data">
File: <input type="file" name="filename"/
Path: <input type="text" readonly="" name="path" value="c:"/
Saveas: <input type="text" name="saveas"/>
<input name="submit" type="submit" value="Upload" />
</form>
proces JSP
<%@ page language="java" %>
<%@ page import="java.util.*" %>
<%@ page import="FileUploadBean" %>
<jsp:useBean id="TheBean" scope="page" class="FileUploadBean" />
<%
TheBean.doUpload(request);
%>
BEAN
import java.io.*;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.ServletInputStream;
public class FileUploadBean {
public void doUpload(HttpServletRequest request) throws IOException
          String melding = "";
          String filename = request.getParameter("saveas");
          String path = request.getParameter("path");
          PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("test.java")));
          ServletInputStream in = request.getInputStream();
          int i = in.read();
          System.out.println("filename:"+filename);
          System.out.println("path:"+path);
          while (i != -1)
               pw.print((char) i);
               i = in.read();
          pw.close();
}

Thanks it works great.
Here an excample from my code
import org.apache.commons.fileupload.*;
public class FileUploadBean extends Object implements java.io.Serializable{
String foutmelding = "geen";
String path;
String filename;
public boolean doUpload(HttpServletRequest request) throws IOException
     try
     // Create a new file upload handler
     FileUpload upload = new FileUpload();
     // Set upload parameters
     upload.setSizeMax(100000);
     upload.setSizeThreshold(100000000);
     upload.setRepositoryPath("/");
     // Parse the request
     List items = upload.parseRequest(request);
     // Process the uploaded fields
     Iterator iter = items.iterator();
     while (iter.hasNext())
     FileItem item = (FileItem) iter.next();
          if (item.isFormField())
               String stringitem = item.getString();
     else
          String filename = "";
               int temp = item.getName().lastIndexOf("\\");
               filename = item.getName().substring(temp,item.getName().length());
               File bestand = new File(path+filename);
               if(item.getSize() > SizeMax && SizeMax != -1){foutmelding = "bestand is te groot.";return false;}
               if(bestand.exists()){foutmelding ="bestand bestaat al";return false;}
               FileOutputStream fOut = new FileOutputStream(bestand);
               BufferedOutputStream bOut = new BufferedOutputStream(fOut);
               int bytesRead =0;
               byte[] data = item.get();
               bOut.write(data, 0 , data.length);
               bOut.close();
     catch(Exception e)
          System.out.println("er is een foutontstaan bij het opslaan de een bestand "+e);
          foutmelding = "Bestand opsturen is fout gegaan";
     return true;
     }

How to get the file name in one of the Xml node field?

Hi Experts,
i have one requirement with file name in xml file.
i need to get <b>the file name of file</b> in one of the node in xml content of that file.
how can i achieve that req?
treat this as Very Urgent.

Hi Bheem,
Explain your scenario in detail..
>>i need to get the file name of file in one of the node in xml content of that file.
1. Create an node <b>Filename</b> in the target datatype.
2. Check the Option Filename in the Adapter Specific Message Attributes in the sender file CC.
3. Use the following UDF and map it to the Filename field in the target message type.
public String getFileName(Container container)
DynamicConfiguration conf = (DynamicConfiguration) container.getTransformationParameters().get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey key = DynamicConfigurationKey.create("http://sap.com/xi/XI/System/File","FileName");
String ourSourceFileName = conf.get(key);
return ourSourceFileName;
Regards
San

C Programming: How can we get the filesystem name for a given file-path?

C Programming: How can we get the filesystem name for a given file-path?
Say I have a filepath=/mnt1/file1
Using some OS API like stat, can I get the Filesystem /mnt ?
Thanks in advance,
-V

Enter the command up to the point of entering the file path and add a space, then drag the file into the terminal window. It will fill out the path.
If you need to go further into the contents of the Application package, you can continue with /Contents...
Another way is to start typing and then hit Tab to auto-complete. It will stop where it can't determine the next letter.
So, type /App tab and it will fill in /Applications. Type a / and start with the name of the app, then tab and it should complete. Continue till you have the correct path.
Spaces will be replaced with \<space>, so, App Store would end up as /Applications/App\ Store.app

How to get the file name from Oracle B2B 10g

Hi My requirement is I am getting a CSV file from Trading partner, I am using oracle 10g b2b to translate the data.
In my BPEL 10g I am using AQ adapter to get the message from IP_IN_QUEUE.
Now I want to get the file name Eg: SampleFile.dat of the CSV file in my BPEL process.
I tried using the b2b.filename property in the receive activity and it is not getting the file name.
<sequence name="main">
    <receive name="Receive_Note" partnerLink="GetB2BNote"
             portType="ns1:Dequeue_ptt" operation="Dequeue"
             variable="Receive_Note_Dequeue_InputVariable"
             createInstance="yes">
             <bpelx:property name="b2b.fileName" variable="WriteFileName"/>
    </receive>
</sequence>
Can you help me to get the file name from Oracle b2b 10g ?
Thanks,
b2b user

Hi My requirement is I am getting a CSV file from Trading partner, I am using oracle 10g b2b to translate the data.
In my BPEL 10g I am using AQ adapter to get the message from IP_IN_QUEUE.
Now I want to get the file name Eg: SampleFile.dat of the CSV file in my BPEL process.
I tried using the b2b.filename property in the receive activity and it is not getting the file name.
<sequence name="main">
    <receive name="Receive_Note" partnerLink="GetB2BNote"
             portType="ns1:Dequeue_ptt" operation="Dequeue"
             variable="Receive_Note_Dequeue_InputVariable"
             createInstance="yes">
             <bpelx:property name="b2b.fileName" variable="WriteFileName"/>
    </receive>
</sequence>
Can you help me to get the file name from Oracle b2b 10g ?
Thanks,
b2b user

How to get the file name from directory

Hi,
I have a directory called test inside i have only one .txt file. i dont know that file name.
Is it possible get the file name using PL/SQl code. ???
Using that .txt file i have to create a dynamic table.
If i have use *.txt also not working
Anyone suggest me its possible to do or not????
Cheers,
Shan

Hi Saubhik ,
Wheni execute the function i am getting the following error
Warning: compiled but with compilation errors
Errors for FUNCTION LISTDIR
LINE/COL
ERROR
7/3
PLS-00201: identifier 'DBMS_BACKUP_RESTORE.SEARCHFILES' must be declared
7/3
PL/SQL: Statement ignored
8/54
PL/SQL: ORA-00942: table or view does not exist
8/20
PL/SQL: SQL Statement ignored
10/11
PLS-00364: loop index variable 'EACH_FILE' use is invalid
10/2
PL/SQL: Statement ignored      Cheers,
Shan

How to find out the file name

Hi,
In selection screen (parameter) user will give input TXT file from presentation server to upload to SAP. I need to capture the file name only but not the path and need to concatenate with date stamp and need to download error log in XLS file to presentation server.
How to find out the file name from selection screen?
I searched SCN threads but not found relavant solution.
Thanks,
R Kumar

Hi
This code gets only filename from selection screen :
REPORT x.
PARAMETERS p_file(100).
DATA : gv_full_path LIKE ibipparms-path,
       gv_full_path_string TYPE string,
       gv_filename(100),
       gv_file_ext(3).
AT SELECTION-SCREEN ON VALUE-REQUEST FOR p_file.
Ask user to select file with a popup :
CALL FUNCTION 'F4_FILENAME'
    IMPORTING
      file_name = gv_full_path.
Get filename from path :
gv_full_path_string = gv_full_path.
CALL FUNCTION 'CH_SPLIT_FILENAME'
    EXPORTING
      complete_filename = gv_full_path_string
    IMPORTING
      extension         = gv_file_ext
      name              = gv_filename.
CONCATENATE gv_filename '.' gv_file_ext INTO gv_filename.
p_file = gv_filename.
I hope it helps.

Get the File Name received from FTP Adapter

Hi,
How to get the File name reived through the FTP Adpater. I have created a variable with the Message type from ftpAdapterinboundheader.wsdl. from there I mapped the filename attribute to a local string variable.
But I did not receive the file name. The output in the Audit trail is as follows:
<?xml version="1.0" encoding="UTF-8" ?>
- <Invoke_File_Process_FileProcess_InputVariable>
- <part xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" name="InputParameters">
- <InputParameters xmlns="http://xmlns.oracle.com/pcbpel/adapter/db/PIERS/SP_FILE_PROCESS_UPDATE/">
<AS_DIR xmlns="">I</AS_DIR>
<AS_FILE_NAME xmlns="" />
</InputParameters>
</part>
</Invoke_File_Process_FileProcess_InputVariable>
Can any one let me know how to get the recevied file name from FTP adapter.
Thanks

you have to define variable of type InboundHeader_msg. Then in receive activity click on Adapter tab and for header variable chose your newly created variable (InboundHeader_msg). Once you receive message from FTP you should see in this variable fileName.

File adapter polling and getting the file name 11g Jdeveloper

I am using JDeveloper 11g. My process is something like;
A file adapter reads a XML file
This triggers the BPM process, file content is sent to start activity as message
I have to get the file name that is read by file adapter.
How to do that? Any help is appreciated.
I have seen couple of answers like bellow in forum but did not help me much
This is a very simple task in JDeveloper 11g, file adapters are linked to the receive activity, here there is a tab called Properties which you can use to extract the file name. Please follow the steps below:
1. Create a Simple Type variable of string type, call it fileName_var or any name you wish.
2. Open the Receive activity in your bpel project.
3. Scroll down to jca.file.FileName and double click on the value column, click on the browse button (...), select Variable, then click the search icon
4. Locate your recently created variable
5. Now when your project runs it will assign the file name to this variable, you can use this same approach also to pull the directory, file size, etc.
*************************/

hi,
Like you said.. steps are simple and clearly mentioned in guide.
1. create a variable of type string
2. after configuring the file adapter , go to corresponding receive activity , go to properties tab
3. search for a property with name as jca.file.FileName . This basically keeps track of file name
4. for this property, give the value as name of variable created in step 1 above
5. Use this variable anywhere in assign / transform
HTH,
Ketan

Get the file name using XSLT mapping

Hi
How to get the file Name at receiver side using XSLT mapping.
Could any one please help me
Regards
sowmya

Sowmya
If you will use Grpahical Mesage Mapping then this can be achieved using Adapter-Specific Attribute u201CFileNameu201D
http://help.sap.com/saphelp_nw04/helpdata/en/43/03612cdecc6e76e10000000a422035/content.htm
Code Snippet -->
DynamicConfiguration conf = (DynamicConfiguration) container
.getTransformationParameters()
.get(StreamTransformationConstants.DYNAMIC_CONFIGURATION);
DynamicConfigurationKey key = DynamicConfigurationKey.create(
u201Chttp://sap.com/xi/XI/System/Fileu201D,
u201CFileNameu201D);
But in case you have to ONLY use XSLT mapping then I would suggest to use the same jave code & call it from your XSLT mapping. I never tried such thing you might use couple of jar files too.
- lalit -

How can I edit the file name of the home page?

How can I edit the file name of the home page?
Gr Ammani

I have a similar issue with regards to the filename...
When I navigate to my website, it doesnt work unless I actually type the filename at the end of it. Example: type in "www.whitelion.com" (this is my real domain name) and it doesnt work. When I type in "www.whitelion.com/index.html" then it works.
This is obviously inconvenient because my clients will never know to add the filename to the end of the site.
Is there any setting or way to get the page to appear by just typing the original domain name without adding the extension?
Thanks,
Emelia

Can you get the file name of the current executing TSQL script?

Can you get the file name of the current executing TSQL script? I wrote entries to a generic log file and would like to include the script name.

Okay, So What you can do is
1. Read get the version from your database and redirect it to a text file(SQLCMD outout can be directed to text file using -o option or windows redirection operator >)
2. Now you can read this value from the text file either inside a batch file or a powershell script and decide what operations you can do.
Satheesh
My Blog |
How to ask questions in technical forum

How do I get the directory name that a class is located in?

How do I get the directory name that a class is located in? I have a class located under /com/dhcmc and I want to be able to retrieve the "/com/dhcmc" that the class is residing under so that I can then use this directoy name to build a relative path name for the other directories supporting the application such as the images directory etc i.e.
|---com
|   \---dhcmc       (my class resides here)
|       +---images (supporting app dirs under the class dir)
|       +---config (supporting app dirs under the class dir)

You can try the methods.
Class.getClassLoader()
Class.getResourceAsStream()
Class.getResource()
You probably wont get the complete directory path, but you might
be able to load the files relative to your classes directory.

How do I get the file name?

Similar Messages

Maybe you are looking for