I have added a button in flash I want this button to save a mp3 file from the server when a user clicks on it inside a browser.
One way to do this is:
import flash.net.FileReference;
import flash.utils.ByteArray;
var fileReference:FileReference = new FileReference();
var data:ByteArray; // load up your data into this ByteArray
fileReference.save(data, "outputfile.dat");
But how do we load a mp3 file from server in variable 'data'?
Just found FileReference.download() on google
Related
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.
Currently I have a basic WebView set up with JavaFX and a html file that I need to load in my local home folder. However, when I set a URL object to the home folder and try to load that URL using the WebEngine's load function, the page does not load. The page content.html is perfectly fine. Here is my code:
String contentURL = new URL("file://" + System.getProperty("user.home") + "/content.html").toExternalForm();
webEngine.load(contentURL);
System.out.println(contentURL);
Put an extra slash to file protocol; "file:///". For more info follow this wiki.
I would like to find out, how can I email a .txt file that is located in the application storage directory.
Thanks
Jared
Use applicationStorageDirectory and navigateToURL with mailto like so,
(Note: I am not going in depth with reading text file. I can help though if you need help. Putting it only to give some idea here.)
//Find the target file
var fileToFind:File = File.applicationStorageDirectory.resolvePath("filename.txt");
//Extract content from file
var fileStream:FileStream = new FileStream();
fileStream.openAsync(fileToFind, FileMode.READ);
//fileStream.addEventListener(Event.COMPLETE, showContent);
//fileStream.addEventListener(IOErrorEvent.IO_ERROR, alertError);
//To mail
navigateToURL(newURLRequest("mailto:someEmail#gmail.com"+"?subject=Subject"+"&body="+ textfileContents + "));
--
If you want to send attachment or HTML formatted email, see this link,
SMTPMailer
.
Best luck.
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.
Trying to fix an old .asp site to work on an ipad. One of the features is the users ability to download their search results into an excel worksheet. The code uses:
Response.ContentType = "application/vnd.ms-excel"
Response.AddHeader "Content-Disposition", "attachment;filename=results.xls"
Response.CharSet = "iso-8859-1"
When viewing the site on the ipad, when the link is click for the page with the code above it does nothing, just spins. Is it the fact that I am trying to export the data as excel, I have read in some posts how it is the encoding! Should I convert the code to export the results page as a csv file and then allow the user to open it in anything they want/have available? What's the best way to do it to hit the most devices...
Thanks
In the past i'd a same scenario so what i did:
FILE: DOWNLOAD.ASP
<%
' get the file to download
myFile = request.querystring("File")
myFullPath = "c:\name_folder\" & myFile ' example of full path and filename
' set headers
Response.ContentType = "application/octet-stream"
Response.AddHeader "Content-Disposition", "attachment; filename=" & myFile
' send the file using the stream as
Set adoStream = CreateObject("ADODB.Stream")
adoStream.Open()
adoStream.Type = 1
adoStream.LoadFromFile(myFullPath)
Response.BinaryWrite adoStream.Read()
adoStream.Close
Set adoStream = Nothing
%>
FILE: HTML
Download Excel file
This example is full working with Ipad using the native browser Safari.
The file Result.xls is downloaded and loaded in the Viewer whitout the capability to be edit.
My iPad users use the App QuickOffice to let the file be saved in a virtual folder, rename the file, delete, ... but they cant edit the file, that App is just for manage the files and isnt required for download the file.
If your user need also edit the XLS file on the iPad i suggest to use (for example) the Google App Document, it let the user to edit and manage the file directly in the browser.
Hope it help