How to get full file path while uploading a file in flex Applications

How to get full file path while uploading a file in flex applications.
FileReference Object is giving file name and other details but not the actual path.
Is there any workaround to to get the file path?.
Thanks

Why not ask in the Flex forum; it is more likely that someone over there knows.

Similar Messages

How to give the unix server path while uploading a file

Hi All,
Am very new to this file operations in Oracle.
here i have a requirement like we need to upload table to unix server using oracle. I have read about File operations in Oracle but in that i did not understand one point like how and where to give the unix server path .. like username@servername and Path where that file needs to be stored.
I have googled for file operations in oracle.. in that there are asking me to create one object like as follows
Create or replace directory rtvms2 as 'MYDIR:'
The MYDIR: i need to use while uploading the file .
UTL_FILE.FOPEN('MYDIR:','Report.txt','W');
My Question is how and where should i give my unix server information .?
and how and in which step oracle connect to unix server and upload the file.?
Can any one please help me.
Thanks
Sree

you're close...
(this example shows it for a windows server, but you just replace the windows directory path with a unix server path)
CREATE OR REPLACE DIRECTORY mydir AS 'c:\myfiles';Note: This does not create the directory on the file system. You have to do that yourself and ensure that oracle has permission to read/write to that file system directory.
Then, grant permission to the users who require access e.g....
GRANT READ,WRITE ON DIRECTORY mydir TO myuser;Then use that directory object inside your FOPEN statement e.g.
fh := UTL_FILE.FOPEN('MYDIR', 'myfile.txt', 'r');Note: You MUST specify the directory object name in quotes and in UPPER case for this to work as it is a string that is referring to a database object name which will have been stored in uppercase by default.
For security reasons, Oracle requires a directory object and it is that object that points to the actual path (it's up to you to make sure that path is valid and exists on the operating system). Then the object can have read and write permissions given to relevant database users as appropriate i.e. you may want one user to only be able to read files from that location whilst another user can read and write. You still need to ensure operating system level permissions exist for that path for the operating system user under which oracle was installed (on unix that's likely to be an "oracle" user).

How to get ALL validate-errors while insert xml-file into xml_schema_table

How to get all validate-errors while using insert into xml_schema when having a xml-instance with more then one error inside ?
Hi,
I can validate a xml-file by using isSchemaValid() - function to get the validate-status 0 or 1 .
To get a error-output about the reason I do validate
the xml-file against xdb-schema, by insert it into schema_table.
When more than one validate-errors inside the xml-file,
the exception shows me the first error only.
How to get all errors at one time ?
regards
Norbert
... for example like this matter:
declare
     xmldoc CLOB;
     vStatus varchar
begin
-- ... create xmldoc by using DBMS_XMLGEN ...
-- validate by using insert ( I do not need insert ;-) )
     begin
     -- there is the xml_schema in xdb with defaultTable XML_SCHEMA_DEFAULT_TABLE
     insert into XML_SCHEMA_DEFAULT_TABLE values (xmltype(xmldoc) ) ;
     vStatus := 'XML-Instance is valid ' ;
     exception
     when others then
     -- it's only the first error while parsing the xml-file :
          vStatus := 'Instance is NOT valid: '||sqlerrm ;
          dbms_output.put_line( vStatus );
     end ;
end ;

If I am not mistaken, the you probably could google this one while using "Steven Feuerstein Validation" or such. I know I have seen a very decent validation / error handling from Steven about this.

How to get the real path of the xml file

I have a java application
following is the package structure
com>>gts>>xml
having file---------> MyXML.xml
com>>gts>>java
having java program to read the file
Problem is if I use File file = new File("..\\xml\\MyXML.xml");
java.io.FileNotFoundException: E:\LEARNING_WORK_SPACE\JavaXml\..\xml\MyXml.xml (The system cannot find the path specified)
     at java.io.FileInputStream.open(Native Method)
     at java.io.FileInputStream.<init>(Unknown Source)
     at java.io.FileInputStream.<init>(Unknown Source)
     at sun.net.www.protocol.file.FileURLConnection.connect(Unknown Source)
How do I get the real path of the xml file.
Edited by: shashiwagh on Jan 29, 2010 11:46 AM

Hi,
if your XML file is inside a package you can easily get it from the classloader.
Note that your application maybe packaged inside a jar so it is not safe to use java.io.File for this purpose.
You have an xml file in :
com/gts/xml/MyXML.xml
in a class of the same module (that will packaged in the same jar) for example com.gts.java.XmlLoader :
// To get the stream :
InputStream is = this.getClass().getResourceAsStream("/com/gts/xml/MyXML.xml");
// Or the URL :
URL xml = this.getClass().getResource("/com/gts/xml/MyXML.xml");Hope it helps.

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 can i get the absolute filepath while uploading a file

Hi techies,
could you help in solving the following prob..
I am trying to upload a file path to the database...for ths i've used input type="file" tag and i can only retreive th ename of the file and could not able to retreive the entire file path starting from root directory.
I am in linux environment and using firefox as browser.
I would reapply appreciate any help in this regard.
Thank you very much.
dandu.

you don't need the path. There is no valid reason you can come up with that you need anything more then the file name. Try giving any reason you can think of and I'll explain why you don't it.

How to get full package path?

Hi all.
My program consists of 2 files:
1.) run.exe
2.) classes.zip
The second (.zip) file contains java classes. How can I gain a full path to any package in this .zip file? I tried
MyClass.class.getPackage().getName() but it returned only somethig like myPackage.mySecondPackage. But I need full path, something like C:\My program\...
Many thanks.
Miso

I tried
> MyClass.class.getPackage().getName()
but it returned only somethig like
myPackage.mySecondPackage. But I need full
path, something like C:\My program\...
Think of it a bit: your jar (or zip) file is stored in, say 'c:\MyPrograms'
as 'myjar.zip'. What should the path of your class file be then?
'c:\MyPrograms\myjar.zip\myPackage.mySecondPackage.MyClass'
wouldn't make any sense would it?
Leaving out the name of the jar/zip file itself wouldn't make any sense
either. IOW, the 'absolute path' of a class doesn't make any sense,
as long as the JVM can find your classes everything is fine.
kind regards,
Jos

SP Designer Workflow: How to Get Full URL Path to List Item and inserted same list ITem URL in another list

Hi,
I have requirement in Sp Designer 2010,Get List item URL and insert in another list as one column value.When open another list and click on same item column entry url will show the parent item information.
Here i have create work flow and insert item URL in another list but cant find appropriate item url information.I can easily make item url link through String builder in mail body with using current id and predefine link,but
when try to insert the same type of item link in another list where i cant find string builder for create custom url link,only get valur of Path,URL,Absolute URL and Relative server URL,all these links or not provide me exact
item link dispaly information.
So I opened SharePoint Designer and start creating the workflow associated to the list.
As there is some Field from source related to current item URL I start using it
Encoded Absolute URL – this one should be the one to use
Server Relative URL
URL Path
Unfortunately, none of these options were providing the correct link. Indeed, these options are providing an incorrect path:
Encoded Absolute URL
http://wfe1/Lists/bpf/1_.000
Server Relative URL
/Lists/bpf/1_.000
URL Path
/Lists/bpf/1_.000
As you can see, the item URL is composed by an ID while it should be http://wfe1/Lists/bpf/dispform.aspx?id=1
Hasan Jamal Siddiqui(MCTS,MCPD,ITIL@V3),Share Point Application Developer,TCS

Unfortunately, [%Current Item:URL%] doesn't seem to be available from a "Site Workflow" associated to a List. I'm finding less advantages to doing a "Site Workflow" when I don't necessarily need to. One problem is the workflow is initiating
twice. I'm thinking I should have just created the workflow as a a "List Workflow."
I am going to try "Association Columns" -- that may work. Anyone have other suggestions?

How to scan for the viruses while uploading a file

Hi!!,
I have an appliaction written in java wherein I take as an input .xls file.
I want to have same facility for uploading other fils also, but I need to check
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Please suggest me how can I do this??

How do you update the other fields of the database table?
I assume your database table has some fields like Changed by , Changed On, Change time etc., in its definition.
If you are making a direct table update, pass the SY-UNAME, SY-DATUM/SY-DATLO, SY-UZEIT/SY-TIMLO to the respective fields in your table along with the data modified.
Rgds
Vijay

How to get legacy customer no. while uploading Customer master and vendor

Hi All,
I want to upload Customer master and Vendor master data from legacy using LSMW. I am using SAP standard programs available for the same. I am using internal number assignment in SAP for these masters.
I case of material master these is a field in MARA named BISMT which is populated with old material no. used in legacy. Is there any corresponding field for customer and vendor master upload?. Moreover is this field automatically populated by SAP report while uploading?.I want to know which old material no. is mapped to which new internally generated number.
Your feedback is highly appreciated.

Hi,
if you are looking for a field in customer and vendor master data that should contain the old (legacy number), take a look at knb1-altkn and in case of a vendor at lfb1-altkn.
regards
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(409) Conflict Error while uploading the file into Sharepoint library

Getting the below error while uploading the file into Sharepoint library.
(409) Conflict. at System.Net.HttpWebRequest.GetResponse() at Microsoft.SharePoint.Client.SPWebRequestExecutor.Execute() at Microsoft.SharePoint.Client.File.SaveBinary(ClientContext context, String serverRelativeUrl,
Stream stream, String etag, Boolean overwriteIfExists
I have used the below code:
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Thanks in advance.

May be issue is with path.
https://server/ should be there instead of
subsite path(https://server/path/path)
in the client context
Check the below link
http://social.msdn.microsoft.com/Forums/sharepoint/en-US/fbb38b10-1127-48a6-a65f-0301edd766c4/the-remote-server-returned-an-error-409-conflict-error-while-uploading-files-to-sharepoint?forum=sharepointdevelopmentlegacy

Getting SY-SUBRC=2 while uploading excel using ALSM_EXCEL_TO_INTERNAL_TABLE

Hi Experts,
I am getting sy-subrc = 2 while uploading the file using ALSM_EXCEL_TO_INTERNAL_TABLE wiht message OLE upload error.
If i upload the same file in another system it is not giving any error.
I want to know in which situation we will get this error & who will resolve this error means either basis,technical or admin.
Please give me the best answer to resolve this issue.

Hi Naresh,
If i upload the same file in another system it is not giving any error.
if it is working fine in another system , it seems the rights issue in your system try to login with admin rights.
Regards
Parveen

How to get complete path when uploading a file in Mozzila

Hi
I was trying to upload a file using html code
<input type="file" name="cdrfile">
then I wrote
<% = request.getParameter("cdrfile")%>
trying to get the path and the file name. But Mozilla doesn't return me the complete path only the file name. Does anyone know how to get the complete path in Mozilla?
Cheers
M

Deliberate double post:
http://forum.java.sun.com/thread.jspa?threadID=5147786&messageID=9551206#9551206

How to get the content of the uploaded file.

Hi Experts,
I am using a input box to get the path of the file to be uploaded.(I am not using FILE UPLOAD UI Element).Could you please let me know how to get the content of the uploaded file.
Regards,
Arun

>
ARUN KUMAR.S wrote:
> Hi Experts,
>
> I am using a input box to get the path of the file to be uploaded.(I am not using FILE UPLOAD UI Element).Could you please let me know how to get the content of the uploaded file.
>
> Regards,
> Arun
You will not be able to use a normal InputField to upload file contents. This is not allowed by the browser security model. You must use one of the upload specific UI mechanisms - the FileUpload UI element or as of 7.01 ACFUpDownload or FlashIslands.

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How to get full file path while uploading a file in flex Applications

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