Content type problem for 'does not appear to be a proper arcive'
Hi all,
The following code will create a ZipOutputStream using ByteArrayOutputStream (not FileOutputStream) and attach the outputstream to a MIME multipart email using ByteArrayDataSource (so the file never exists physically).
It works and sends the email but the zip 'does not appear to be a valid archive' even though it looks about the right size. If I do the same and use FileOutputStream to create a physical file it works OK. I think it is an issue with the content type, or maybe this just isn't possible!
I have tried:
byteArray.toByteArray(),
byteArray.toString().getBytes() and application/zip & application/unknown.
Can anyone help or suggest an alternative way to create a 'in-memory' zip file that can then be emailed?
Many thanks
// Specify files to be zipped
String[] filesToZip = new String[3];
filesToZip[0] = "C:\\Program Files\\NetBeans3.6\\firstfile.txt";
filesToZip[1] = "C:\\Program Files\\NetBeans3.6\\secondfile.txt";
filesToZip[2] = "C:\\Program Files\\NetBeans3.6\\thirdfile.txt";
byte[] buffer = new byte[18024];
// Specify zip file name
String zipFileName= eq_rt.getReportName() + ".zip";
try {
// Create ZipOutputStream to store the FileOutputStream
//ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipFileName));
ByteArrayOutputStream byteArray = new ByteArrayOutputStream();
ZipOutputStream out = new ZipOutputStream(byteArray);
// Set the compression ratio
out.setLevel(Deflater.DEFAULT_COMPRESSION);
// iterate through the array of files, adding each to the zip file
for (int a = 0; a < filesToZip.length; a++) {
System.out.println(a);
// // Associate a file input stream for the current file
FileInputStream in = new FileInputStream(filesToZip[a]);
// This ROCKS as it is passing a array into the text file .getBytes() seems
// to be the KEY in getting ByteArrayInputStream to WORK
String strSocketInput = "TAIWAN";
ByteArrayInputStream baIn = new ByteArrayInputStream(strSocketInput.getBytes());
//ByteArrayInputStream baIn = new ByteArrayInputStream( getAttachementNoFormat(eq_rt.getStoredProc() ) );
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(filesToZip[a]));
// Transfer bytes from the current file to the ZIP file
int len;
while ((len = baIn.read(buffer)) > 0)
out.write(buffer, 0, len);
// Close the current entry
out.closeEntry();
// Close the current file input stream
baIn.close();
// DataSource sourcezip = new FileDataSource(zipFileName);
//DataSource sourcezip = new ByteArrayDataSource(byteArray.toByteArray(), zipFileName, "application/octet-stream");
DataSource sourcezip = new ByteArrayDataSource(byteArray.toByteArray(), zipFileName, "application/octet-stream" );
// Create a new MIME bodypart
BodyPart attachment = new MimeBodyPart();
attachment.setDataHandler(new DataHandler(sourcezip));
attachment.setFileName(zipFileName);
/* attach the attachemnts to the mail */
multipart.addBodyPart(attachment);
// Close the ZipOutPutStream
out.close();
Many thanks Dr Clap. Moving the Closing the ZipOutputStream before I attached it to the email solved my problem.
/* Close the ZipOutPutStream (very important to close the zip before you attach it to the email) Thanks DrClap */
out.close();
/* Create a datasource for email attachment */
// DataSource sourcezip = new FileDataSource(zipFileName);
DataSource sourcezip = new ByteArrayDataSource(byteArray.toByteArray(), zipFileName, "application/zip" );
/* Create a new MIME bodypart */
BodyPart attachment = new MimeBodyPart();
attachment.setDataHandler(new DataHandler(sourcezip));
attachment.setFileName(zipFileName);
/* attach the attachemnts to the mail */
multipart.addBodyPart(attachment);
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