Need to extract only file name from path.........

Hi All,
I have a parameter.This calls the function
"CALL FUNCTION 'F4_FILENAME' to get the file from C drive.
After selecting the file the path is displayed in the Parameter field.
My problem is I need to extract only file name from the path.Please advice.
Example : Prameter id C:\folder\file.xls
I shd extract only file.xls from the path.Please advice.

Hi,
Use the below logic:
data: begin of itab,
val type char20,
end of itab.
SPLIT l_f_path AT '\' INTO TABLE itab.
The last record of the internal table holds the file name.
describe table itab lines l_f_lines.
read itab index l_f_lines.
l_f_filaname = itab-val.
Hope this helps u.

Similar Messages

File name from path

We are converting an application that has a table with a full path text column and want to extract only the file name and wondered how we can accomplish this. The table is named BatchImport and the column is named FilePath varchar(40). An example of a
row would contain G:\Everyone\Billfiles\AATAX.XLS and we want to extract just AATAX.XLS out of it. Thanks.

REVERSE, CHARINDEX and RIGHT will be your friends here:
DECLARE @table TABLE (value NVARCHAR(MAX))
INSERT INTO @table (value) VALUES ('G:\Everyone\Billfiles\AATAX.XLS')
SELECT *, RIGHT(value,CHARINDEX('\',REVERSE(value))-1)
FROM @table
Don't forget to mark helpful posts, and answers. It helps others to find relevant posts to the same question.

Separating file name from path...

HI,
I have a jsp page in which the user selects a file by using
<INPUT type=file name =fname value="browse">
I am passing this to the javascript in the same page as:
var file = document.f1.fname.value;
What I want to do is strip of the file name and add another filename to it:
i.e if var file = C:\path\to\folder\myfile.txt
then I want another variable to have the value:
var x = C:\path\to\folder\ .
Which means that I want to remove everything until the first " \ " (from the end)
is there any way to do this (and that too in javascript)
Thanks,
-G.

why are you using java script? I think all this can be done without java script. If you take the file name which is selected by the user in a String. Then you can very easily change the file name once you have the name of the file which the user selected.

Need to extract faces and names from a multi-portrait high school year book scanned page

     I am responsible for generating individual high school student portraits (5" x 7") from scanned 600 dpi pages of portraits from my high school beacon (year book). There are 16 - 20 black and white portraits per page and there are between 800 and 850 students in my graduating class.
     The end product is taking an estimated 850 portraits from multi- portrait pages and convert them into individual 5" x 7" B&W photos.
     Attached is one of the pages of portraits to experiment with. Other than selecting each individual portrait and pasting it into a new image from the clipboard, I don't know of a faster way to do it. This will nearly take forever.
     Might there be a way to create a grid of selections template that applies to each page and extracts the portraits and names separating them into individual images?
I couldn't download a sample for some reason. The portraits are arranged 4 wide and 5 down like below table but without the header. The names are below the faces. If one of you has an idea, I could email the full size image to you.
Header 1
Header 2
Header 3
Header 4

Nothing shows up.
1. Go to www.pixentral.com
2. Put your cursor in the blank provided for the upload (where it says your image)
3. Click on the Browse button
4. When the "Choose File" box pops up, navigate to where your photo is stored, click on your photo file, then at the bottom of the box, click the Open button
5. It will come back to Pixentral with the path to your photo shown in the blank
6. Now on the Pixentral screen, click on the Send button
7. Once your photo is loaded (you'll see it on the Pixentral screen) scroll down to the bottom of the photo where you'll see two lines of text, one that says HTML and one that says Forum
8. You need to copy all of the text in the box after "Forum"
9. An easy way to select it is to triple click on the text OR drag your mouse all the way from the first character to the last character of the text
10. Once it is selected, press Ctrl-C to copy it
11. Then go to your message on the forum here and paste your link by pressing Ctrl-V

Extract actual file name from the filepath

I have a String something like
String str ="C:\Documents and Settings\abc\Desktop\file_Aug_2007.xls";
how can i extract the substring file_Aug_2007.xls with or without using io apis
Regards,
ivin

ivinjacob wrote:
I have a String something like
String str ="C:\Documents and Settings\abc\Desktop\file_Aug_2007.xls";
how can i extract the substring file_Aug_2007.xls with or without using io apis
Regards,
ivinHi Ivin,
First, you will need to use double backslashes in your file path; \ is an escape character so for literal strings you have to double them up.
String str ="C:\\Documents and Settings\\abc\\Desktop\\file_Aug_2007.xls";For the next part, you don't need anything terribly fancy; the backslashes in your file should be plenty to find the filename.
From the String class, use the substring(int beginIndex, int endIndex), lastIndexOf(String str), and length():
String sub;
sub = str.substring(str.lastIndexOf("\\") + 1, str.length()); I hope that helps.
Thok

Truncate file name from path

hi friends..
i have file path
/abc/xyz/file.pqr
i want to truncate file.pqr
/ may be any numbers .
plz help

Hi,
Search the string for '/' in loop
eg:-
lv_last_index type i.
lv_last_index = 0.
do
SEARCH file_path FOR '/' starting at lv_last_index.
if sy-subrc <> 0.
exit.
else.
lv_last_index = SY-FDPOS.
endif.
lv_last_index -
will contain last ocurrence of '/'.....
no ur filepath will be
file_path = file_path+0( lv_last_index ). "..specifies the within the bracket
check if the last char is '/'.. in that case cate lv_last_index = lv_last_index - 1.
Mark helpfull answers & close the thread.
Regards
Manoj

Separating file name and path

Hi,
I need to separate the file name and path of the file which I am uploading from my BSP appl.I am using SO_SPLIT_FILNAME_PATH ,but it is not working because this fm identifies '\' only from GUI front end.
How do I separate the file name and path in BSP ?
Regards,
Ananya

Assuming that you are using File Upload tag for file upload:
DATA: fileUpload TYPE REF TO CL_HTMLB_FILEUPLOAD.
fileUpload ?= CL_HTMLB_MANAGER=>GET_DATA(
                       request = request
                       id      = 'myUpload'
                       name    = 'fileUpload' ).
file_name      = fileUpload->file_name.
file_mime_type = fileUpload->file_content_type.
file_length    = fileUpload->file_length.
file_content   = fileUpload->file_content.
Data : filepath type string,
       filename type string,
       t1 type string.
filepath = file_name .
while filepath CA ''.
split filepath at '' into t1 filename.
endwhile.
In my sample code above, filename attribute will contain the name of the file uploaded.
Hope this solves the problem.
Regards,
Ravikiran.

Get the File Name from DB

I have values stored in the database with relative ads
"/partners/41394/Ads/offers_41394_1by2.jpg"
I only need to extract the file name itself any time this
kind of record is encountered. I am not interested in the relative
path of the record.
eg extract only "offers_41394_1by2.jpg"
Thanks

List functions are very helpful, be sure to check out the
full range in
the documentation.
listLast("/partners/41394/Ads/offers_41394_1by2.jpg","/")

How to get the target file name from an URL?

Hi there,
I am trying to download data from an URL and save the content in a file that have the same name as the file on the server. In some way, what I want to do is pretty similar to what you can do when you do a right click on a link in Internet Explorer (or any other web browser) and choose "save target as".
If the URL is a direct link to the file (for example: http://java.sun.com/images/e8_java_logo_red.jpg ), I do not have any problem:
URL url = new URL("http://java.sun.com/images/e8_java_logo_red.jpg");
System.out.println("Opening connection to " + url + "...");
// Copy resource to local file
InputStream is = url.openStream();
FileOutputStream fos=null;
String fileName = null;
StringTokenizer st=new StringTokenizer(url.getFile(), "/");
while (st.hasMoreTokens())
                fileName=st.nextToken();
System.out.println("The file name will be: " + fileName);
File localFile= new File(System.getProperty("user.dir"), fileName);
fos = new FileOutputStream(localFile);
try {
    byte[] buf = new byte[1024];
    int i = 0;
    while ((i = is.read(buf)) != -1) {
        fos.write(buf, 0, i);
} catch (Throwable e) {
    e.printStackTrace();
} finally {
    if (is != null)
        is.close();
    if (fos != null)
        fos.close();
}Everything is fine, the file name I get is "e8_java_logo_red.jpg", which is what I expect to get.
However, if the URL is an indirect link to the file (for example: http://javadl.sun.com/webapps/download/AutoDL?BundleId=37719 , which link to a file named JavaSetup6u18-rv.exe ), the similar code return AutoDL?BundleId=37719 as file name, when I would like to have JavaSetup6u18-rv.exe .
URL url = new URL("http://javadl.sun.com/webapps/download/AutoDL?BundleId=37719");
System.out.println("Opening connection to " + url + "...");
// Copy resource to local file
InputStream is = url.openStream();
FileOutputStream fos=null;
String fileName = null;
StringTokenizer st=new StringTokenizer(url.getFile(), "/");
while (st.hasMoreTokens())
                fileName=st.nextToken();
System.out.println("The file name will be: " + fileName);
File localFile= new File(System.getProperty("user.dir"), fileName);
fos = new FileOutputStream(localFile);
try {
    byte[] buf = new byte[1024];
    int i = 0;
    while ((i = is.read(buf)) != -1) {
        fos.write(buf, 0, i);
} catch (Throwable e) {
    e.printStackTrace();
} finally {
    if (is != null)
        is.close();
    if (fos != null)
        fos.close();
}Do you know how I can do that.
Thanks for your help
// JB
Edited by: jb-from-sydney on Feb 9, 2010 10:37 PM

Thanks for your answer.
By following your idea, I found out that one of the header ( content-disposition ) can contain the name to be used if the file is downloaded. Here is the full code that allow you to download locally a file on the Internet:
      * Download locally a file from a given URL.
      * @param url - the url.
      * @param destinationFolder - The destination folder.
      * @return the file
      * @throws IOException Signals that an I/O exception has occurred.
     public static final File downloadFile(URL url, File destinationFolder) throws IOException {
          URLConnection urlC = url.openConnection();
          InputStream is = urlC.getInputStream();
          FileOutputStream fos = null;
          String fileName = getFileName(urlC);
          destinationFolder.mkdirs();
          File localFile = new File(destinationFolder, fileName);
          fos = new FileOutputStream(localFile);
          try {
               byte[] buf = new byte[1024];
               int i = 0;
               while ((i = is.read(buf)) != -1) {
                    fos.write(buf, 0, i);
          } finally {
               if (is != null)
                    is.close();
               if (fos != null)
                    fos.close();
          return localFile;
      * Returns the file name associated to an url connection.<br />
      * The result is not a path but just a file name.
      * @param urlC - the url connection
      * @return the file name
      * @throws IOException Signals that an I/O exception has occurred.
     private static final String getFileName(URLConnection urlC) throws IOException {
          String fileName = null;
          String contentDisposition = urlC.getHeaderField("content-disposition");
          if (contentDisposition != null) {
               fileName = extractFileNameFromContentDisposition(contentDisposition);
          // if the file name cannot be extracted from the content-disposition
          // header, using the url.getFilename() method
          if (fileName == null) {
               StringTokenizer st = new StringTokenizer(urlC.getURL().getFile(), "/");
               while (st.hasMoreTokens())
                    fileName = st.nextToken();
          return fileName;
      * Extract the file name from the content disposition header.
      * <p>
      * See <a
      * href="http://www.w3.org/Protocols/rfc2616/rfc2616-sec19.html">http:
      * //www.w3.org/Protocols/rfc2616/rfc2616-sec19.html</a> for detailled
      * information regarding the headers in HTML.
      * @param contentDisposition - the content-disposition header. Cannot be
      *            <code>null>/code>.
      * @return the file name, or <code>null</code> if the content-disposition
      *         header does not contain the filename attribute.
     private static final String extractFileNameFromContentDisposition(
               String contentDisposition) {
          String[] attributes = contentDisposition.split(";");
          for (String a : attributes) {
               if (a.toLowerCase().contains("filename")) {
                    // The attribute is the file name. The filename is between
                    // quotes.
                    return a.substring(a.indexOf('\"') + 1, a.lastIndexOf('\"'));
          // not found
          return null;
     }

File name from an attribute of an element of the payload

Dear Experts,
Could you suggest the syntax for specifying the file name from a variable in the file receiver communication channel? I need to grab the file name from an ATTRIBUTE of an element of the payload.
Thanks a lot.
Example: Need to pass the attribute "descReceipt" of the element "promotionEvent" to the variable in the following payload:
<?xml version="1.0" encoding="UTF-8"?>
<promotionExport>
<promotionEvent eventIdentifier="30" status="A" descReceipt="FILE1">

The scenario is IDOC to FILE. I need to get the value of the receiving partner.
The message header has the node "DynamicConfiguration". Please see the xml below. I need to grab the key RCVPRN (0000002104) from this and pass as the file name in the receiver file adapter.
How do I setup my receiver file adapter for this? I tried changing the values for the variable, and also enabling the adapter specific message attributes. No solution yet.
<SAP:DynamicConfiguration xmlns:SAP="http://sap.com/xi/XI/Message/30" xmlns:SOAP="http://schemas.xmlsoap.org/soap/envelope/" SOAP:mustUnderstand="1">
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="RCVPOR">LXDCLNT250</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="SNDPRT">LS</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="RCVPRN">0000002104</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="SNDPOR">SAPLCD</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="SNDPRN">LCDCLNT220</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="CIMTYP" />
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="IDOCTYP">WPDBBY01</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="MESTYP">WPDBBY</SAP:Record>
<SAP:Record namespace="http://sap.com/xi/XI/System/IDoc" name="RCVPRT">KU</SAP:Record>
</SAP:DynamicConfiguration>

How do I put the file name and path into either the header or the footer?

I need to keep track of pdf documents as their names or locations sometimes change. I need to show both file name and path, preferably in the header. I would also like it to refresh automatically when I change, say, the file name, but that's not as essential as the basic function of showing the file name and path.

You will need to do this with a text field and a Javascript that executes on open.

How can I extract a file name / path from a local path variable in TestStand?

I have local TestStand string variable, call it 'locals.path', that contains a path to a file including the file name (c:\inputs\input.txt). I need to be able to split up the path (c:\input) and the file name (input.txt) and save them into 2 local variables. What is the best way to do this?
After reading through some of the other forums, it looks like there are some built-in functions that can accomplish this, but I am unable to find how to use them anywhere on the NI web site. One forum said to use the File I/O>Strip Path.file function. How is this called? Is this function in a DLL?
I know that there are a number of DLLs that are installed with TestStand into the c:\windows\system32 directory. One forum made note of CVI_OpenFile / CVI_ReadFIle functions in the cvirt.dll that I used to solve a problem that I had in the past. The problem is that I had no idea that that these functions even existed and would have never known unless a similar question had been posted previously. Is there some place that these DLL function interfaces are defined or documented? Is there a function that can extract the file name out of a string for me?
Thanks,
Mike

Hi,
There sound like functions in say LabVIEW or CVI.
I have attached a small example which may help. (I have not allowed for any error trapping, if say you dont find the file and cancel)
Regards
Ray Farmer
Message Edited by Ray Farmer on 10-16-2006 10:04 PM
Regards
Ray Farmer
Attachments:
Sequence File1.seq ‏33 KB

Import From Folder: How to Extract the File Name in a Custom Column.

Hello All
Here´s what we´re trying to do:
We have a folder with csv files named like this:
Sales_2013-02-05.csv
Sales_2013-02-04.csv
Sales_2013-02-03.csv
Sales_2013-02-02.csv
Sales_2013-02-01.csv
And in the csv files there are the sales columns but not the date column.
So we want to extract the date from the file name.
I´ve tried entering = Source[Name] in a custom column, but it adds a "LIST" link, and on a click on expand, it adds ALL file names from the folder in each row, instead of just the needed one.
If we could get the proper file name in each row (from where they got extracted), we could split the column and get the date from there. But I don´t know how put the filename there properly.
Can you help?

This isn't entirely straightforward, but it's definitely possible. What you need to do is to apply all of your transforms to each individual file instead of the combined files. I do that as follows:
1) Use Folder.Files as generated by the GUI to look at the list of my files.
2) Pick one file and do all the transformations to it that I want to apply to all of the files. Sometimes, this just amounts to letting the autodetection figure out the column names and types.
3) Go into the advanced editor and edit my code so that the transformations from step 2 are applied to all files. This involves creating a new function and then applying that function to the content in each row.
4) Expand the tables created in step 3.
As an example, I have some files with names that match the ones you suggested. After steps 1 + 2, my query looks like the following:
let
    Source = Folder.Files("d:\testdata\files"),
    #"d:\testdata\files\_Sales_2013-02-01 csv" = Source{[#"Folder Path"="d:\testdata\files\",Name="Sales_2013-02-01.csv"]}[Content],
    #"Imported CSV" = Csv.Document(#"d:\testdata\files\_Sales_2013-02-01 csv",null,",",null,1252),
    #"First Row as Header" = Table.PromoteHeaders(#"Imported CSV"),
    #"Changed Type" = Table.TransformColumnTypes(#"First Row as Header",{{"One", Int64.Type}, {"Two", type text}, {"Three", type text}})
in
    #"Changed Type"
For step 3, I need to take steps 3-5 of my query and convert them into a function. As a check, I can apply that function to the same file that I chose in step 2. The result looks like this:
let
    Source = Folder.Files("d:\testdata\files"),
    Loader = (file) =>
        let
            #"Imported CSV" = Csv.Document(file,null,",",null,1252),
            #"First Row as Header" = Table.PromoteHeaders(#"Imported CSV"),
            #"Changed Type" = Table.TransformColumnTypes(#"First Row as Header",{{"One", Int64.Type}, {"Two", type text}, {"Three", type text}})
        in
            #"Changed Type",
    #"d:\testdata\files\_Sales_2013-02-01 csv" = Source{[#"Folder Path"="d:\testdata\files\",Name="Sales_2013-02-01.csv"]}[Content],
    Loaded = Loader(#"d:\testdata\files\_Sales_2013-02-01 csv")
in
    Loaded
Now I apply the same function to all of the rows, transforming the existing "Content" column into a new value:
let
    Source = Folder.Files("d:\testdata\files"),
    Loader = (file) =>
        let
            #"Imported CSV" = Csv.Document(file,null,",",null,1252),
            #"First Row as Header" = Table.PromoteHeaders(#"Imported CSV"),
            #"Changed Type" = Table.TransformColumnTypes(#"First Row as Header",{{"One", Int64.Type}, {"Two", type text}, {"Three", type text}})
        in
            #"Changed Type",
    Transformed = Table.TransformColumns(Source, {"Content", Loader})
in
    Transformed
Finally, I need to expand out the columns in the table, which I can do by clicking on the expand icon next to the Content column header. The resulting query looks like this:
let
    Source = Folder.Files("d:\testdata\files"),
    Loader = (file) =>
        let
            #"Imported CSV" = Csv.Document(file,null,",",null,1252),
            #"First Row as Header" = Table.PromoteHeaders(#"Imported CSV"),
            #"Changed Type" = Table.TransformColumnTypes(#"First Row as Header",{{"One", Int64.Type}, {"Two", type text}, {"Three", type text}})
        in
            #"Changed Type",
    Transformed = Table.TransformColumns(Source, {"Content", Loader}),
    #"Expand Content" = Table.ExpandTableColumn(Transformed, "Content", {"One", "Two", "Three"}, {"Content.One", "Content.Two", "Content.Three"})
in
    #"Expand Content"
From here, you should be able to get to what you want.

Get file name and path from adf inputFile

Hi,
I use adf's inputFile component. I need to get the file name and filePath. Does anyone knows how to do that?
Thanks in advance

You may bind the value to an UploadedFile object and get the name from this object.
You may use a valueChangeListener backin bean method or a managed bean.
Here the code i just write for something similar. Note that i'm not able to test it because of a bug in 10.1.3.1 with inputFile and web.xml parameters. I still waiting for the patch.
public void uploadedFile(ValueChangeEvent valueChangeEvent) throws IOException,
Exception {
final int BUFFER = 2048;
byte data[] = new byte[BUFFER];
int currentByte;
String fileName;
UploadedFile uploadedFile =
(UploadedFile)valueChangeEvent.getNewValue();
if (uploadedFile != null) {
String mimeType = uploadedFile.getContentType();
if (mimeType == "application/x-zip-compressed") {
// get the uploaded file as a zip file
ZipFile zipFile = new ZipFile(uploadedFile.getFilename());
// verify the zip archive contains only one entry
if (zipFile.size() != 1) {
FacesContext context =
FacesContext.getCurrentInstance();
ResourceBundle errorMessage =
ResourceBundle.getBundle(context.getApplication().getMessageBundle());
Exception ZipFileContentException =
new Exception(errorMessage.getString("error.fileUpload.zipFileContent.moreThanOneEntry").replace("{0}",
String.valueOf(zipFile.size())));
throw ZipFileContentException;
// get the entries in the zip file even it is only one
Enumeration zipFileEntries = zipFile.entries();
// Process each entry
while (zipFileEntries.hasMoreElements()) {
// grab a zip file entry
ZipEntry entry =
(ZipEntry)zipFileEntries.nextElement();
// check the entry is not a directory
if (entry.isDirectory()) {
Exception ZipFileContentException =
new Exception(JSFUtils.getStringFromBundle("error.fileUpload.zipFileContent.isDirectoryInsteadFile"));
throw ZipFileContentException;
File destFile = new File(entry.getName());
Magic magic = new Magic();
// getMagicMatch accepts Files or byte[],
// which is nice if you want to test streams
MagicMatch match = magic.getMagicMatch(destFile, true);
if (match.getMimeType() != "application/xml") {
Exception ZipFileContentException =
new Exception(JSFUtils.getStringFromBundle("error.fileUpload.isNotXMLFile").replace("{0}",
match.getMimeType()));
throw ZipFileContentException;
//TODO get the repository directory from classification-param.xml
BufferedInputStream is =
new BufferedInputStream(zipFile.getInputStream(entry));
FileOutputStream fos = new FileOutputStream(destFile);
BufferedOutputStream dest =
new BufferedOutputStream(fos, BUFFER);
// read and write until last byte is encountered
while ((currentByte = is.read(data, 0, BUFFER)) !=
-1) {
dest.write(data, 0, currentByte);
dest.flush();
dest.close();
is.close();
zipFile.close();
} else if (mimeType == "application/xml") {
String currentEntry = uploadedFile.getFilename();
File destFile = new File(currentEntry);
BufferedInputStream is =
new BufferedInputStream(uploadedFile.getInputStream());
FileOutputStream fos = new FileOutputStream(destFile);
BufferedOutputStream dest =
new BufferedOutputStream(fos, BUFFER);
// read and write until last byte is encountered
while ((currentByte = is.read(data, 0, BUFFER)) != -1) {
dest.write(data, 0, currentByte);
dest.flush();
dest.close();
is.close();
} else {
Exception ZipFileContentException =
new Exception(JSFUtils.getStringFromBundle("error.fileUpload.isNotXMLFile").replace("{0}",
mimeType));
throw ZipFileContentException;
}

Seperating file name from the path

Hi,
im trying to validate the file path and trying to seperate file name from the path.
for eg.
if the path is f:\sapfilepath\doc\ext.txt
need to get f:\sapfilepath\doc im looking for a dynamic way of doing this.
your help will be appreciated.
Regards,
ravi.

Hi,
Use the function module
SO_SPLIT_FILE_AND_PATH
Thanks,
Naren

Need to extract only file name from path.........

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