i was searching for a way to upload a file to server and found the following code ..
File file ;
int maxFileSize = 5000 * 1024;
int maxMemSize = 5000 * 1024;
ServletContext context = pageContext.getServletContext();
String filePath = context.getInitParameter("file-upload");
// Verify the content type
String contentType = request.getContentType();
if ((contentType.indexOf("multipart/form-data") >= 0)) {
DiskFileItemFactory factory = new DiskFileItemFactory();
// maximum size that will be stored in memory
factory.setSizeThreshold(maxMemSize);
// Location to save data that is larger than maxMemSize.
factory.setRepository(new File("/user2/tst/test"));
// Create a new file upload handler
ServletFileUpload upload = new ServletFileUpload(factory);
// maximum file size to be uploaded.
upload.setSizeMax( maxFileSize );
try{
// Parse the request to get file items.
List fileItems = upload.parseRequest(request);
// Process the uploaded file items
Iterator i = fileItems.iterator();
out.println("<html>");
out.println("<head>");
out.println("<title>JSP File upload</title>");
out.println("</head>");
out.println("<body>");
while ( i.hasNext () )
{
FileItem fi = (FileItem)i.next();
if ( !fi.isFormField () )
{
// Get the uploaded file parameters
String fieldName = fi.getFieldName();
String fileName = fi.getName();
boolean isInMemory = fi.isInMemory();
long sizeInBytes = fi.getSize();
// Write the file
if( fileName.lastIndexOf("\\") >= 0 ){
file = new File( filePath +
fileName.substring( fileName.lastIndexOf("\\"))) ;
}else{
file = new File( filePath +
fileName.substring(fileName.lastIndexOf("\\")+1)) ;
}
fi.write( file ) ;
out.println("Uploaded Filename: " + filePath +
fileName + "<br>");
}
}
out.println("</body>");
out.println("</html>");
}catch(Exception ex) {
System.out.println(ex);
}
}else{
out.println("<html>");
out.println("<head>");
out.println("<title>Servlet upload</title>");
out.println("</head>");
out.println("<body>");
out.println("<p>No file uploaded</p>");
out.println("</body>");
out.println("</html>");
}
this code is working perfectly fine .. the only problem i am facing is , the file that is updated is getting stored in apache/bin folder ..
for this it was specified to add the following code to web.xml that is available in ROOT/WEB-INF
<context-param>
<description>Location to store uploaded file</description>
<param-name>file-upload</param-name>
<param-value>
/user2/tst
</param-value>
</context-param>
even after this the file is getting stored in bin folder ..
i needed some help in this regard .. thanks in advance ..
Related
I am using org.apache.commons.net.ftp.FTPClient for retrieving files from a ftp server. It is crucial that I preserve the last modified timestamp on the file when its saved on my machine. Do anyone have a suggestion for how to solve this?
This is how I solved it:
public boolean retrieveFile(String path, String filename, long lastModified) throws IOException {
File localFile = new File(path + "/" + filename);
OutputStream outputStream = new FileOutputStream(localFile);
boolean success = client.retrieveFile(filename, outputStream);
outputStream.close();
localFile.setLastModified(lastModified);
return success;
}
I wish the Apache-team would implement this feature.
This is how you can use it:
List<FTPFile> ftpFiles = Arrays.asList(client.listFiles());
for(FTPFile file : ftpFiles) {
retrieveFile("/tmp", file.getName(), file.getTimestamp().getTime());
}
You can modify the timestamp after downloading the file.
The timestamp can be retrieved through the LIST command, or the (non standard) MDTM command.
You can see here how to do modify the time stamp: that: http://www.mkyong.com/java/how-to-change-the-file-last-modified-date-in-java/
When download list of files, like all files returned by by FTPClient.mlistDir or FTPClient.listFiles, use the timestamp returned with the listing to update timestemp of local downloaded files:
String remotePath = "/remote/path";
String localPath = "C:\\local\\path";
FTPFile[] remoteFiles = ftpClient.mlistDir(remotePath);
for (FTPFile remoteFile : remoteFiles) {
File localFile = new File(localPath + "\\" + remoteFile.getName());
OutputStream outputStream = new BufferedOutputStream(new FileOutputStream(localFile));
if (ftpClient.retrieveFile(remotePath + "/" + remoteFile.getName(), outputStream))
{
System.out.println("File " + remoteFile.getName() + " downloaded successfully.");
}
outputStream.close();
localFile.setLastModified(remoteFile.getTimestamp().getTimeInMillis());
}
When downloading a single specific file only, use FTPClient.mdtmFile to retrieve the remote file timestamp and update timestamp of the downloaded local file accordingly:
File localFile = new File("C:\\local\\path\\file.zip");
FTPFile remoteFile = ftpClient.mdtmFile("/remote/path/file.zip");
if (remoteFile != null)
{
OutputStream outputStream = new BufferedOutputStream(new FileOutputStream(localFile));
if (ftpClient.retrieveFile(remoteFile.getName(), outputStream))
{
System.out.println("File downloaded successfully.");
}
outputStream.close();
localFile.setLastModified(remoteFile.getTimestamp().getTimeInMillis());
}
I'm new to CodeIgniter and having the following issue. When I upload a file, its successfully uploaded on my local folder and the file name saved to the db. The problem is, file name has been changed after uploading.
eg:
file name "screenshot image 123.png" -> "screenshot_image_123.png"
It just convert the space into underscore;this issue occurred only when the file name having space.Other than this issue all the other files uploading successfully. Given below is my code which I used on my controller page:
public function upload_pic()
{
$image_path = realpath(APPPATH . '../uploads');
$config['upload_path'] = $image_path;
$config['allowed_types'] = "gif|jpg|jpeg|png";
$config['file_name'] = $_FILES['file']['name'];
$config['encrypt_name'] = TRUE;
$config['overwrite'] = TRUE;
$this->load->library('upload',$config);
$this->upload->initialize($config);
if(!$this->upload->do_upload('file'))
{
echo $this->upload->display_errors();
}
else
{
$finfo=$this->upload->data();
$data['uploadInfo'] = $finfo;
}
}
Can anyone help me????
Try before saving file name generate some unique
$filename = $_FILES['file']['name'];
$ext = pathinfo($filename, PATHINFO_EXTENSION);
$filename = sha1_file($filename). md5(date('Y-m-d H:i:s:u')) . '.'. $ext; //not only this you can generate any format
$config['file_name'] = $filename; //Use this file name is db too
Please help me with the below issue.
I have to read data from excel .
I have created JSPDynpage component and followd below link :
http://help.sap.com/saphelp_sm40/helpdata/en/63/9c0e41a346ef6fe10000000a1550b0/frameset.htm below is my code. I am trying to read excel file using apache poi 3.2 API
try
{
FileUpload fu = (FileUpload)
this.getComponentByName("myfileupload");
// this is the temporary file
if (fu != null) {
// Output to the console to see size and UI.
System.out.println(fu.getSize());
System.out.println(fu.getUI());
// Get file parameters and write it to the console
IFileParam fileParam = fu.getFile();
System.out.println(fileParam);
// Get the temporary file name
File f = fileParam.getFile();
String fileName = fileParam.getFileName();
// Get the selected file name and write ti to the console
ivSelectedFileName = fu.getFile().getSelectedFileName();
File fp = new File(ivSelectedFileName);
myLoc.errorT("#fp#"+fp);
try {
//
**FileInputStream file = new FileInputStream(fp);** --> error at this line
HSSFWorkbook workbook = new HSSFWorkbook(file);
myLoc.errorT("#workbook#"+workbook);
//Get first sheet from the workbook
HSSFSheet sheet = workbook.getSheetAt(0);
myLoc.errorT("#sheet#"+sheet);
//
} catch(Exception ioe) {
myLoc.errorT("#getLocalizedMessage# " + ioe.getLocalizedMessage());
}
Error :
#getLocalizedMessage# C:\Documents and Settings\10608871\Desktop\test.xls (The system cannot find the path specified)
at line FileInputStream file = new FileInputStream(fp);
I have created the PAR file and deploying it on server.
Thanks in Advance,
Aliya Khan.
i resolved the problem, i was passing the worng parameter instead of f
FileInputStream file = new FileInputStream(f);
How to take automatically backup of a log file(.txt) when it's size reached a threshold level, say 5MB. The backup file name should be like (log_file_name)_(system_date) and original log file should be cleaned(0 KB).
Please help. Thanks in advance.
Check your log file size using lenght().Then check if its bigger then 5mb call extendLogFile() func.
This is c# code u can easly convert to java
Size check:
if (size > 400 * 100 * 100)
{
extendLogFile(Path);
}
Copy old log file in archive directory and create new log file:
private static void extendLogFile(string lPath)
{
string name = lPath.Substring(0, lPath.LastIndexOf("."));
string UniquName = GenerateUniqueNameUsingDate(); // create a unique name for old log files like '12-04-2013-12-43-00'
string ArchivePath = System.IO.Path.GetDirectoryName(lPath) + "\\Archive";
if (!string.IsNullOrEmpty(ArchivePath) && !System.IO.Directory.Exists(ArchivePath))
{
System.IO.Directory.CreateDirectory(ArchivePath);
}
string newName = ArcivePath + "\\" + UniquName;
if (!File.Exists(newName))
{
File.Copy(lPath, newName + ".txt");
using (FileStream stream = new FileStream(lPath, FileMode.Create))
using (TextWriter writer = new StreamWriter(stream))
{
writer.WriteLine("");
}
}
}
I am trying to upload a file via a form and then save in in SQL as a blob.
I already have my form working fine, my database is fully able to take the blob and I have a controller that take the file, saves it in a local directory:
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult FileUpload(int id, HttpPostedFileBase uploadFile)
{
//allowed types
string typesNonFormatted = "text/plain,application/msword,application/pdf,image/jpeg,image/png,image/gif";
string[] types = typesNonFormatted.Split(',');
//
//Starting security check
//checking file size
if (uploadFile.ContentLength == 0 && uploadFile.ContentLength > 10000000)
ViewData["StatusMsg"] = "Could not upload: File too big (max size 10mb) or error while transfering the file.";
//checking file type
else if(types.Contains(uploadFile.ContentType) == false)
ViewData["StatusMsg"] = "Could not upload: Illigal file type!<br/> Allowed types: images, Ms Word documents, PDF, plain text files.";
//Passed all security checks
else
{
string filePath = Path.Combine(HttpContext.Server.MapPath("../Uploads"),
Path.GetFileName(uploadFile.FileName)); //generating path
uploadFile.SaveAs(filePath); //saving file to final destination
ViewData["StatusMsg"] = "Uploaded: " + uploadFile.FileName + " (" + Convert.ToDecimal(uploadFile.ContentLength) / 1000 + " kb)";
//saving file to database
//
//MISSING
}
return View("FileUpload", null);
}
Now all I am missing is putting the file in the database. I could not find anything on the subject... I found some way to do it in a regular website but nothing in MVC2.
Any kind of help would be welcome!
Thank you.
This could help: http://byatool.com/mvc/asp-net-mvc-upload-image-to-database-and-show-image-dynamically-using-a-view/
Since you have HttpPostedFileBase in your controllers method, all you need to do is:
int length = uploadFile.ContentLength;
byte[] tempImage = new byte[length];
myDBObject.ContentType = uploadFile.ContentType;
uploadFile.InputStream.Read(tempImage, 0, length);
myDBObject.ActualImage = tempImage ;
HttpPostedFileBase has a InputStream property
Hope this helps.
Alright thanks to kheit, I finaly got it working. Here's the final solution, it might help someone out there.
This script method takes all the file from a directory and upload them to the database:
//upload all file from a directory to the database as blob
public void UploadFilesToDB(long UniqueId)
{
//directory path
string fileUnformatedPath = "../Uploads/" + UniqueId; //setting final path with unique id
//getting all files in directory ( if any)
string[] FileList = System.IO.Directory.GetFiles(HttpContext.Server.MapPath(fileUnformatedPath));
//for each file in direcotry
foreach (var file in FileList)
{
//extracting file from directory
System.IO.FileStream CurFile = System.IO.File.Open(file, System.IO.FileMode.Open);
long fileLenght = CurFile.Length;
//converting file to a byte array (byte[])
byte[] tempFile = new byte[fileLenght];
CurFile.Read(tempFile, 0, Convert.ToInt32(fileLenght));
//creating new attachment
IW_Attachment CurAttachment = new IW_Attachment();
CurAttachment.attachment_blob = tempFile; //setting actual file
string[] filedirlist = CurFile.Name.Split('\\');//setting file name
CurAttachment.attachment_name = filedirlist.ElementAt(filedirlist.Count() - 1);//setting file name
//uploadind attachment to database
SubmissionRepository.CreateAttachment(CurAttachment);
//deleting current file fromd directory
CurFile.Flush();
System.IO.File.Delete(file);
CurFile.Close();
}
//deleting directory , it should be empty by now
System.IO.Directory.Delete(HttpContext.Server.MapPath(fileUnformatedPath));
}
(By the way IW_Attachment is the name of one of my database table)