I'm using the Upload component in a Vaadin8 project to get a file up on the server, as shown in the source code on this page:
https://demo.vaadin.com/sampler/#ui/data-input/other/upload
After I choose the file on my pc and click upload, the window opens up just like in the sampler, and the progress bar goes all the way to the end, but the file is nowhere to be found in the project file system. Is there another step I'm supposed to be doing? How to I configure the destination folder for the uploaded files?
Taken from official documentation here Receiving upload:
The uploaded files are typically stored as files in a file system, in a database, or as temporary objects in memory. The upload component writes the received data to an java.io.OutputStream so you have plenty of freedom in how you can process the upload content.
So in your case, an uploaded file is stored as a temporary object. In V8 documentation example is cut-off, but it's presented in V7: Receiving Upload Data
public OutputStream receiveUpload(String filename,
String mimeType) {
// Create upload stream
FileOutputStream fos = null; // Stream to write to
try {
// Open the file for writing.
file = new File("/tmp/uploads/" + filename);
fos = new FileOutputStream(file);
} catch (final java.io.FileNotFoundException e) {
new Notification("Could not open file<br/>",
e.getMessage(),
Notification.Type.ERROR_MESSAGE)
.show(Page.getCurrent());
return null;
}
return fos; // Return the output stream to write to
}
public void uploadSucceeded(SucceededEvent event) {
// Show the uploaded file in the image viewer
image.setVisible(true);
image.setSource(new FileResource(file));
}
So the idea is that you could create a file yourself, once upload has succeed.
Related
I am creating a dot net core mvc app to allow my internal team to download the enterprise ios app on their phones. So I created a plist to put (serve) it in that app and allow users to download and install the mobile app through this dot net app.
Below is the Downloadapp action that takes a string with a file name and returns the plist file.
public async Task<IActionResult> Downloadapp(string filename)
{
string file_name = getFileName(filename);
string file_path = getFilePath(filename);
if (filename == null)
return Content("filename not present");
var path = Path.Combine(
Directory.GetCurrentDirectory(),
file_path, file_name);
var memory = new MemoryStream();
using (var stream = new FileStream(path, FileMode.Open))
{
await stream.CopyToAsync(memory);
}
memory.Position = 0;
return File(memory, GetContentType(path), Path.GetFileName(path));
}
I call this: https://localhost:44384/Home/Downloadapp?filename=CCI_Zoom_iOS_Mobile_plist to download the plist successfully and have https://localhost:44384/Home/Downloadapp?filename=CCI_Zoom_iOS_Mobile_ipa in the plist string to call and serve the ipa file.
However, When I call
itms-services://?action=download-manifest&url=https://localhost:44384/Home/Downloadapp?filename=CCI Zoom_iOS_Mobile_plist
using a URL.Action call/also append itms-services.... in the beginning of the URL, it says "Open this page in itunes" and then nothing happens.
On an ios device, normally it should give an option to install the application but in my case nothing is happening.
What dot net core (2.1) code shall I write to generate a link that serves a plist file to download an ios application?
p.s. I am new to .net core so I need your help on this. It would be much appreciated.
Thanks
I want to deliver large Excel files using a webservice or httphandler.
As the Excel files can be very big in size, I want to split them up into smaller files, to decrease the memory footprint.
So I will have a master excelfile that contains the column headers and data.
And further files which will only contain data.
During download, I want to stream the master excel file first and then append all other related excel files as one download stream.
I don't want to zip them! It should be one file at the end
Is this possible?
Master excel file with headers:
All other files will look like this (without headers):
This will indeed return crap:
void Main()
{
CombineMultipleFilesIntoSingleFile();
}
// Define other methods and classes here
private static void CombineMultipleFilesIntoSingleFile(string outputFilePath= #"C:\exceltest\main.xlsx", string inputDirectoryPath = #"C:\exceltest", string inputFileNamePattern="*.xlsx")
{
string[] inputFilePaths = Directory.GetFiles(inputDirectoryPath, inputFileNamePattern);
Console.WriteLine("Number of files: {0}.", inputFilePaths.Length);
using (var outputStream = File.Create(outputFilePath))
{
foreach (var inputFilePath in inputFilePaths)
{
using (var inputStream = File.OpenRead(inputFilePath))
{
// Buffer size can be passed as the second argument.
inputStream.CopyTo(outputStream);
}
Console.WriteLine("The file {0} has been processed.", inputFilePath);
}
}
}
When you are requesting for file, do not download it at first request.
Request for file names to be downloaded in AJAX request.
For each file name received, prepare its path to the server.
Create hidden iFrames for each file path and specify src as file path for each file to be downloaded.
When iFrame's src attribute is set, it will navigate to the file path and each iFrame will download single file, so multiple iFrame downloads multiple files.
You cannot download multiple files in single request. As if you will append the stream of multiple files, it will create a garbage file, a single garbage file.
I am using the Export plugin in Grails for generating PDF/Excel report. I am able to generate single report on PDF/Excel button click. But, now I want to generate multiple reports on single button click. I tried for loop, method calls but no luck.
Reference links are ok. I don't expect entire code, needs reference only.
If you take a look at the source code for the ExportService in the plugin you will notice there are various export methods. Two of which support OutputStreams. Using either of these methods (depending on your requirements for the other parameters) will allow you to render a report to an output stream. Then using those output streams you can create a zip file which you can deliver to the HTTP client.
Here is a very rough example, which was written off the top of my head so it's really just an idea rather than working code:
// Assumes you have a list of maps.
// Each map will have two keys.
// outputStream and fileName
List files = []
// call the exportService and populate your list of files
// ByteArrayOutputStream outputStream = new ByteArrayOutputStream()
// exportService.export('pdf', outputStream, ...)
// files.add([outputStream: outputStream, fileName: 'whatever.pdf'])
// ByteArrayOutputStream outputStream2 = new ByteArrayOutputStream()
// exportService.export('pdf', outputStream2, ...)
// files.add([outputStream: outputStream2, fileName: 'another.pdf'])
// create a tempoary file for the zip file
File tempZipFile = File.createTempFile("temp", "zip")
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(tempZipFile))
// set the compression ratio
out.setLevel(Deflater.BEST_SPEED);
// Iterate through the list of files adding them to the ZIP file
files.each { file ->
// Associate an input stream for the current file
ByteArrayInputStream input = new ByteArrayInputStream(file.outputStream.toByteArray())
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(file.fileName))
// Transfer bytes from the current file to the ZIP file
org.apache.commons.io.IOUtils.copy(input, out);
// Close the current entry
out.closeEntry()
// Close the current input stream
input.close()
}
// close the ZIP file
out.close()
// next you need to deliver the zip file to the HTTP client
response.setContentType("application/zip")
response.setHeader("Content-disposition", "attachment;filename=WhateverFilename.zip")
org.apache.commons.io.IOUtils.copy((new FileInputStream(tempZipFile), response.outputStream)
response.outputStream.flush()
response.outputStream.close()
That should give you an idea of how to approach this. Again, the above is just for demonstration purposes and isn't production ready code, nor have I even attempted to compile it.
I am new to Struts and working on File Upload using Struts.
Client:
It is Java Program which hits my Strut app by using apache HttpClient API and provides me
File.
Client as per need sometime gives me .wav file and sometime .zip file and sometime both.
Server:
Struts app which got the request from client app and upload the file.
Here, problem comes as I upload the file, it get uploaded using ".tmp" extension, which I want to get uploaded with the same extension what client has passed.
Or there is any other way by which we can check what is the extension of the file client has sent....?
I am stuck in this problem and not able to go ahead.
Please Find the code attached and tell me what modification I have to do:
Server Code:
MultiPartRequestWrapper multiWrapper=null;
File baseFile=null;
System.out.println("inside do post");
multiWrapper = ((MultiPartRequestWrapper)request);
Enumeration e = multiWrapper.getFileParameterNames();
while (e.hasMoreElements()) {
// get the value of this input tag
String inputValue = (String) e.nextElement();
// Get a File object for the uploaded File
File[] file = multiWrapper.getFiles(inputValue);
// If it's null the upload failed
if (file != null) {
FileInputStream fis=new FileInputStream(file[0]);
System.out.println(file[0].getAbsolutePath());
System.out.println(fis);
int ch;
while((ch=fis.read())!=-1){
System.out.print((char)ch);
}
}
}
System.out.println("III :"+multiWrapper.getParameter("method"));
Client code:
HttpClient client = new HttpClient();
MultipartPostMethod mPost = new MultipartPostMethod(url);
File zipFile = new File("D:\\a.zip");
File wavFile = new File("D:\\b.wav");
mPost.addParameter("recipientFile", zipFile);
mPost.addParameter("promptFile", wavFile);
mPost.addParameter("method", "addCampaign");
statusCode1 = client.executeMethod(mPost);
actually Client is written long back and cant be modified and I want to identify something at server side only to find the extension.
Please help, Thanks.
Struts2 File Uploader interceptor when uploading file pass the content type information to the Action class and one can easily find the file type by comparing contentType with MIME type.
If you want to can create a map with key as content type and file type as its value like
map.Add("image/bmp",".bmp", )
map.Add("image/gif",".gif", )
map.Add("image/jpeg",".jpeg", )
and can easily fetch the type based on the extension provides.Hope this will help you.
I have looked online with mixed results, but is there a way to programmatically extract a zip file on the BB? Very basic my app will display different encrypted file types, and those files are delivered in a zip file. My idea was to have the user browse to the file on their SDCard, select it, and I extract what i need as a stream from the file. is this possible?
Use GZIPInputStream
Example:
try
{
InputStream inputStream = httpConnection.openInputStream();
GZIPInputStream gzis = new GZIPInputStream(inputStream);
StringBuffer sb = new StringBuffer();
char c;
while ((c = (char)gzis.read()) != -1)
{
sb.append(c);
}
String data = sb.toString();
gzis.close();
}
catch(IOException ioe)
{
}
Just two things:
In BB API there are only GZip and ZLib support, and no multiple files compression support, so it's not possible to compress several files and extract only one of them.
Up to my experience, such functionality will fly on simulator, but may be really performance-killing on real device
See How to retrieve data from a attached zip file in Blackberry application?
PS Actually you can implement custom multi-entries stream and parse it after decompress, but that seems to be useless, if you want this archive format to be supported in other applications.