I have an application that uploads a file. I need to pass this file into another program but for that I need the file name only. Is there any simple code for that, using only Java or a servlet procedure?
while (files.hasMoreElements())
{
name = (String)files.nextElement();
type = multipartRequest.getContentType(name);
filename = multipartRequest.getFilesystemName(name);
originalFilename = multipartRequest.getOriginalFileName(name);
//extract the file extension - this can be use to reject a
//undesired file extension
extension1 = filename.substring
(filename.length() - 4, filename.length());
extension2 = originalFilename.substring
(originalFilename.length() - 4, originalFilename.length());
//return a File object for the specified uploaded file
File currentFile = multipartRequest.getFile(name);
//InputStream inputStream = new BufferedInputStream
(new FileInputStream(currentFile));
if(currentFile == null) {
out.println("There is no file selected!");
return;
}
There's a method in apache commons-io to get the file's extension. There's also guava Files class, with its getFileExtension method.
Related
In my application, I have files in FTP server one directory and I move that file source to target path. In this process, when I move selected source file that source file will not show in the source path, it will show only in target path.
I tried this below code, but I am getting error:
string sourceurl = "ftp://ftp.com/Mainfoder/Folder1/subfolder/subsubfolder/"
string Targetpat =
"ftp://ftp.com/Mainfoder/DownloadedFiles/"+subfolder+"/"+todaydatefolder+"/"+susubfolder;
Uri serverFile = new Uri(sourceurl + filename);
request = (FtpWebRequest)FtpWebRequest.Create(serverFile);
request.Method = WebRequestMethods.Ftp.Rename;
request.Credentials = new NetworkCredential(ftpUserID, ftpPassword);
request.RenameTo = Targetpat+"/"+newfilename;//folders without filename
response = (FtpWebResponse)request.GetResponse();
Stream ftpStream = response.GetResponseStream();
An unhandled exception of type 'System.Net.WebException' occurred in System.dll
Additional information: The remote server returned an error: (553) File name now allowed.
response = (FtpWebResponse)request.GetResponse(); //This line throwing the above exception
request.RenameTo = newfilename: when I set only newfilename, it renames that source same file name only.
How can I move this file to another directory within in same FTP server?
Please can anyone tell me. Thank you
As I wrote you already before:
request.RenameTo takes a path only.
So this is wrong:
string Targetpat =
"ftp://ftp.com/Mainfoder/DownloadedFiles/"+subfolder+"/"+todaydatefolder+"/"+susubfolder;
request.RenameTo = Targetpat+"/"+newfilename;
It should be:
string Targetpat =
"/Mainfoder/DownloadedFiles/"+subfolder+"/"+todaydatefolder+"/"+susubfolder;
request.RenameTo = Targetpat+"/"+newfilename;
i 'm .net developer. i want to Zip all files and make a one zip file with this technique.
ZipFile multipleFilesAsZipFile = new ZipFile();
Response.AddHeader("Content-Disposition", "attachment; filename=" + DateTime.Now.ToString("ddMMyyyy_HHmmss") + ".zip");
Response.ContentType = "application/zip";
for (int i = 0; i < filename.Length; i++)
{
string filePath = Server.MapPath("~/PostFiles/" + filename[i]);
multipleFilesAsZipFile.AddFile(filePath, string.Empty);
}
multipleFilesAsZipFile.Save(Response.OutputStream);
how ever for making this Zip i use third party library Ionic.
all files are ziped successfully but not extracted to client desktop. is there problem with my code. or this library that i'm using has been expired.
Is there free full version .net compatible library to zip all files.
Use SharpZipLib:
Nuget Package
Install-Package SharpZipLib
OR Download here
http://www.icsharpcode.net/OpenSource/SharpZipLib/
Snippet from examples:
private void CompressFolder(string path, ZipOutputStream zipStream, int folderOffset) {
string[] files = Directory.GetFiles(path);
foreach (string filename in files) {
FileInfo fi = new FileInfo(filename);
string entryName = filename.Substring(folderOffset); // Makes the name in zip based on the folder
entryName = ZipEntry.CleanName(entryName); // Removes drive from name and fixes slash direction
ZipEntry newEntry = new ZipEntry(entryName);
newEntry.DateTime = fi.LastWriteTime; // Note the zip format stores 2 second granularity
// Specifying the AESKeySize triggers AES encryption. Allowable values are 0 (off), 128 or 256.
// A password on the ZipOutputStream is required if using AES.
// newEntry.AESKeySize = 256;
// To permit the zip to be unpacked by built-in extractor in WinXP and Server2003, WinZip 8, Java, and other older code,
// you need to do one of the following: Specify UseZip64.Off, or set the Size.
// If the file may be bigger than 4GB, or you do not need WinXP built-in compatibility, you do not need either,
// but the zip will be in Zip64 format which not all utilities can understand.
// zipStream.UseZip64 = UseZip64.Off;
newEntry.Size = fi.Length;
zipStream.PutNextEntry(newEntry);
// Zip the file in buffered chunks
// the "using" will close the stream even if an exception occurs
byte[ ] buffer = new byte[4096];
using (FileStream streamReader = File.OpenRead(filename)) {
StreamUtils.Copy(streamReader, zipStream, buffer);
}
zipStream.CloseEntry();
}
string[ ] folders = Directory.GetDirectories(path);
foreach (string folder in folders) {
CompressFolder(folder, zipStream, folderOffset);
}
}
Taken from : https://github.com/icsharpcode/SharpZipLib/wiki/Zip-Samples
Works awesome!
I have built a Server that you can upload files to and download, using Eclipse, servlet and jsp, it's all very new to me. (more info).
Currently the upload system works with the file's name. I want to programmatically assign each file a random key. And with that key the user can download the file. That means saving the data in a config file or something like : test.txt(file) fdjrke432(filekey). And when the user inputs the filekey the servlet will pass the file for download.
I have tried using a random string generator and renameTo(), for this. But it doesn't work the first time, only when I upload the same file again does it work. And this system is flawed, the user will receive the file "fdjrke432" instead of test.txt, their content is the same but you can see the problem.
Any thoughts, suggestions or solutions for my problem?
Well Sebek, I'm glad you asked!! This is quite an interesting one, there is no MAGIC way to do this. The answer is indeed to rename the file you uploaded. But I suggest adding the random string before the name of the file; like : fdjrke432test.txt.
Try this:
filekey= RenameRandom();
File renamedUploadFile = new File(uploadFolder + File.separator+ filekey+ fileName);
item.write(renamedUploadFile);
//remember to give the user the filekey
with
public String RenameRandom()
{
final int LENGTH = 8;
StringBuffer sb = new StringBuffer();
for (int x = 0; x < LENGTH; x++)
{
sb.append((char)((int)(Math.random()*26)+97));
}
System.out.println(sb.toString());
return sb.toString();
}
To delete or download the file from the server you will need to locate it, the user will input the key, you just need to search the upload folder for a file that begins with that key:
filekey= request.getParameter("filekey");
File f = new File(getServletContext().getRealPath("") + File.separator+"data");
File[] matchingFiles = f.listFiles(new FilenameFilter() {
public boolean accept(File dir, String name) {
return name.startsWith(filekey);
}
});
String newfilename = matchingFiles[0].getName();
// now delete or download newfilename
I'm developing an eclipse plug-in and I need to traverse a directory and whole content of the directory. I found the method which reads a file in plug-in (bundleresource) as InputStream.
InputStream stream = Activator.class.getResourceAsStream("/dir1/dir2/file.ext");
this method works for files only. I need a way to read directories, list subdirectories and files like File.io.
Thanks.
Do you want to read a resource directory of your plugin? Otherwise you have to traverse a directory and open one stream per file:
String path = "c:\\temp\\";
File directory = new File(path);
if (directory.isDirectory()) {
String[] list = directory.list();
for (String entry : list) {
String absolutePath = path + entry;
System.out.println("processing " + absolutePath);
File file = new File(absolutePath);
if (file.isFile()) {
FileInputStream stream = new FileInputStream(file);
// use stream
stream.close();
}
}
}
If you want to traverse subdirectories as well you should wrap this into a recursive method, check if file is a directory and call the recursive method in this case.
I'm trying to create a new file in an eclipse plugin. It's not necessarily a Java file, it can be an HTML file for example.
Right now I'm doing this:
IProject project = ...;
IFile file = project.getFile("/somepath/somefilename"); // such as file.exists() == false
String contents = "Whatever";
InputStream source = new ByteArrayInputStream(contents.getBytes());
file.create(source, false, null);
The file gets created, but the problem is that it doesn't get recognized as any type; I can't open it in any internal editor. That's until I restart Eclipse (refresh or close then open the project doesn't help). After a restart, the file is perfectly usable and opens in the correct default editor for its type.
Is there any method I need to call to get the file outside of that "limbo" state?
That thread does mention the createFile call, but also refers to a FileEditorInput to open it:
Instead of java.io.File, you should use IFile.create(..) or IFile.createLink(..). You will need to get an IFile handle from the project using IProject.getFile(..) first, then create the file using that handle.
Once the file is created you can create FileEditorInput from it and use IWorkbenchPage.openEditor(..) to open the file in an editor.
Now, would that kind of method (from this AbstractExampleInstallerWizard) be of any help in this case?
protected void openEditor(IFile file, String editorID) throws PartInitException
{
IEditorRegistry editorRegistry = getWorkbench().getEditorRegistry();
if (editorID == null || editorRegistry.findEditor(editorID) == null)
{
editorID = getWorkbench().getEditorRegistry().getDefaultEditor(file.getFullPath().toString()).getId();
}
IWorkbenchPage page = getWorkbench().getActiveWorkbenchWindow().getActivePage();
page.openEditor(new FileEditorInput(file), editorID, true, IWorkbenchPage.MATCH_ID);
}
See also this SDOModelWizard opening an editor on a new IFile:
// Open an editor on the new file.
//
try
{
page.openEditor
(new FileEditorInput(modelFile),
workbench.getEditorRegistry().getDefaultEditor(modelFile.getFullPath().toString()).getId());
}
catch (PartInitException exception)
{
MessageDialog.openError(workbenchWindow.getShell(), SDOEditorPlugin.INSTANCE.getString("_UI_OpenEditorError_label"), exception.getMessage());
return false;
}