Question about the upload component

General
1

Hello!
I have probably a silly question according to the upload component of vaadin. I used the code from the sample for my upload component.

I want to do the following:

  • The file should be only uploaded to the server temporarily, because I directly want to read the file and parse the content into a String. For that I already wrote a class which takes the filename as constructor and then proceeds. Now I need your help how I can give the right filename to the constructor.

In usual webapplications the file will be stored in a temporarily folder on the webserver and needs to be copied explicitly to the final destination by the webapplicationcode. I believe in Vaadin this will be similiar.

What do I have to change in the code that I can reach my goal?


public class OpenCSVFile implements Receiver {

	private String fileName;
    private String mtype;
    private int counter;
    
    FileOutputStream fileOutput = null;
    private String actualFilename;

    /**
     * return an OutputStream that simply counts lineends
     */
    public OutputStream receiveUpload(String filename, String MIMEType) {
    	
    	counter = 0;
        fileName = filename;
        mtype = MIMEType;
        
        //File file = new File("/tmp/uploads/" + filename);
        //ReadCsvFile csvFile = new ReadCsvFile(file);
        //csvFile.readFile();

        
        return new OutputStream() {
            private static final int searchedByte = '\n';

            @Override
            public void write(int b) throws IOException {
            	
            	
            	            	
            	System.out.println(":"+b);
                if (b == searchedByte) {
                    counter++;
                }
            }
        };
    }

    public String getFileName() {
        return fileName;
    }

    public String getMimeType() {
        return mtype;
    }

    public int getLineBreakCount() {
    	//File file = new File(fileName);
        //ReadCsvFile csvFile = new ReadCsvFile(file);
        //csvFile.readFile();
        System.out.println("Hello, fileName: "+getFileName());
    	
        return counter;
    }

Thank yout for any help!!!

2

If you just want to read the file, and not store it, could you just skip the part of storing the file and just directly convert the outputstream into a string?


http://stackoverflow.com/questions/216894/get-an-outputstream-into-a-string

Google for more conversion options.

3

Thank you for your answer. After googeling I found a code which seams to work for me:

public class UploadBuffer implements StreamSource, Upload.Receiver {
    ByteArrayOutputStream outputBuffer = null;

    String mimeType;

    String fileName;

    public UploadBuffer() {

    }

    public InputStream getStream() {
        if (outputBuffer == null) {
            return null;
        }
        return new ByteArrayInputStream(outputBuffer.toByteArray());
    }

    /**
     * @see com.vaadin.ui.Upload.Receiver#receiveUpload(String,
     *      String)
     */
    public OutputStream receiveUpload(String filename, String MIMEType) {
        fileName = filename;
        mimeType = MIMEType;
        outputBuffer = new ByteArrayOutputStream();
        return outputBuffer;
    }

    /**
     * Returns the fileName.
     * 
     * @return String
     */
    public String getFileName() {
        return fileName;
    }

    /**
     * Returns the mimeType.
     * 
     * @return String
     */
    public String getMimeType() {
        return mimeType;
    }

}

This code works perfectly for me but I have another question. In the gui I have to following lines of code:

private void createHeader() {
		
		upload.addListener(new Upload.FinishedListener() {
            public void uploadFinished(FinishedEvent event) {
            	StreamResource resource = new StreamResource(uploadBuffer, uploadBuffer.getFileName(),main.getApplication());
            	Link userlist = new Link("Download ",resource);
            	main.addComponent(userlist);
            	System.out.println("List: "+userlist.getApplication());
            }
        });
		main.addComponent(upload);
		//uploadLayout.setComponentAlignment(upload, Alignment.MIDDLE_CENTER);
	}

With that code I now have a link which opens the document for me, but as I said before my parser needs the filename. Until now I didn´t find a method from the link which simply returns the ULR to the path. I thought I just take the path and give it to my parser.

Is this a working solution, and if yes how can I simply returns the path of the file?

Thank you for your help and time!

4

Ok, now the following code is working, it takes the file and convert it into a string. Then you can do with the content whatever you want:

@SuppressWarnings("serial")
public class UploadBuffer implements Receiver {
	
	private String fileName;
    private String mtype;
    private int counter;
    private StringBuilder string = new StringBuilder();
    
	@Override
	public OutputStream receiveUpload(String filename, String MIMEType) {
		
		 fileName = filename;
         mtype = MIMEType;
         return new OutputStream()
         {
        	 
			@Override
             public void write(int b) throws IOException {
                 string.append((char) b );
             }
             
             //Netbeans IDE automatically overrides this toString()
             public String toString(){
                 return string.toString();
             }
         };
	}
	
	 public String getFileName() {
         return fileName;
     }

     public String getMimeType() {
         return mtype;
     }

     public int getLineBreakCount() {
         return counter;
     }
    public String getContent(){
    	return string.toString();
    }
}

Thanks for your help!