My application is receiving email through SMTP server. There are one or more attachments in the email and email attachment return as byte[] (using sun javamail api).
I am trying to zip the attachment files on the fly without writing them to disk first.
What is/are possible way to achieve this outcome?
You can use Java's java.util.zip.ZipOutputStream to create a zip file in memory. For example:
public static byte[] zipBytes(String filename, byte[] input) throws IOException {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
ZipEntry entry = new ZipEntry(filename);
entry.setSize(input.length);
zos.putNextEntry(entry);
zos.write(input);
zos.closeEntry();
zos.close();
return baos.toByteArray();
}
I have the same problem but i needed a many files in a zip.
protected byte[] listBytesToZip(Map<String, byte[]> mapReporte) throws IOException {
String extension = ".pdf";
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
for (Entry<String, byte[]> reporte : mapReporte.entrySet()) {
ZipEntry entry = new ZipEntry(reporte.getKey() + extension);
entry.setSize(reporte.getValue().length);
zos.putNextEntry(entry);
zos.write(reporte.getValue());
}
zos.closeEntry();
zos.close();
return baos.toByteArray();
}
zos.closeEntry() be inside the for loop?You can create a zip file from byte array and return to ui streamedContent
public StreamedContent getXMLFile() {
try {
byte[] blobFromDB= null;
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
String fileName= "fileName";
ZipEntry entry = new ZipEntry(fileName+".xml");
entry.setSize(byteArray.length);
zos.putNextEntry(entry);
zos.write(byteArray);
zos.closeEntry();
zos.close();
InputStream is = new ByteArrayInputStream(baos.toByteArray());
StreamedContent zipedFile= new DefaultStreamedContent(is, "application/zip", fileName+".zip", Charsets.UTF_8.name());
return fileDownload;
} catch (IOException e) {
LOG.error("IOException e:{} ",e.getMessage());
} catch (Exception ex) {
LOG.error("Exception ex:{} ",ex.getMessage());
}
}
byte[] createReport() {
try {
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
ZipArchiveOutputStream zipOutputStream = new
ZipArchiveOutputStream(byteArrayOutputStream);
zipOutputStream.setMethod(ZipArchiveOutputStream.STORED);
zipOutputStream.setEncoding(ENCODING);
String text= "text";
byte[] textBytes = text.getBytes(StandardCharsets.UTF_8);
ArchiveEntry zipEntryReportObject = newStoredEntry("file.txt", textBytes);
zipOutputStream.putArchiveEntry(zipEntryReportObject);
zipOutputStream.write(textBytes);
zipOutputStream.closeArchiveEntry();
zipOutputStream.close();
return byteArrayOutputStream.toByteArray();
} catch (IOException e) {
return null;
}
and
ArchiveEntry newStoredEntry(String name, byte[] data) {
ZipArchiveEntry zipEntry = new ZipArchiveEntry(name);
zipEntry.setSize(data.length);
zipEntry.setCompressedSize(zipEntry.getSize());
CRC32 crc32 = new CRC32();
crc32.update(data);
zipEntry.setCrc(crc32.getValue());
return zipEntry;
}
Maybe the java.util.zip package might help you
Since you're asking about how to convert from byte array I think (not tested) you can use the ByteArrayInputStream method
int read(byte[] b, int off, int len)
Reads up to len bytes of data into an array of bytes from this input stream.
that you will feed to
ZipInputStream This class implements an input stream filter for reading files in the ZIP file format.
You have to use a ZipOutputStream for that.
http://java.sun.com/javase/6/docs/api/java/util/zip/ZipOutputStream.html
ByteArrayInputStream bais = new ByteArrayInputStream(retByte);
ZipInputStream zis = new ZipInputStream(bais);
zis.getNextEntry();
Scanner sc = new Scanner(zis);
while (sc.hasNextLine()) {
System.out.println("-->:" +sc.nextLine());
}
zis.closeEntry();
zis.close();
public static void createZip(byte[] data) throws ZipException {
ZipInputStream zipStream = new ZipInputStream(new ByteArrayInputStream(data));
ZipParameters parameters = new ZipParameters();
parameters.setFileNameInZip("bank.zip");
new ZipFile("F:\\ssd\\bank.zip").addStream(new ByteArrayInputStream(data), parameters);
}
