In my case I have to download images from the resources folder in my web app. Right now I am using the following code to download images through URL.
url = new URL(properties.getOesServerURL() + "//resources//WebFiles//images//" + imgPath);
filename = url.getFile();
is = url.openStream();
os = new FileOutputStream(sClientPhysicalPath + "//resources//WebFiles//images//" + imgPath);
b = new byte[2048];
while ((length = is.read(b)) != -1) {
os.write(b, 0, length);
}
But I want a single operation to read all images at once and create a zip file for this.
I don't know so much about the use of sequence input streams and zip input streams so if it is possible through these, please let me know.
The only way I can see you being able to do this is something like the following:
try {
ZipOutputStream zip = new ZipOutputStream(new FileOutputStream("C:/archive.zip"));
//GetImgURLs() is however you get your image URLs
for(URL imgURL : GetImgURLs()) {
is = imgURL.openStream();
zip.putNextEntry(new ZipEntry(imgURL.getFile()));
int length;
byte[] b = new byte[2048];
while((length = is.read(b)) > 0) {
zip.write(b, 0, length);
}
zip.closeEntry();
is.close();
}
zip.close();
}
Ref: ZipOutputStream Example
The url should return zip file. Else you have to take one by one and create a zip using your program
Related
Scenario: Uncompress a tar file using Apache commons.
Problem: The tar i am using is a build tar which gets deployed into a web server. This tar contains duplicate entries like below.
appender_class.xml
APPENDER_CLASS.xml
when extracting using the below code only appender_class.xml is extracted but i want both the files how can i do that ? Renaming in fly is fine but how can i accomplish that?
public static void untar(File[] files) throws Exception {
String path = files[0].toString();
File tarPath = new File(path);
TarEntry entry;
TarInputStream inputStream = null;
FileOutputStream outputStream = null;
try {
inputStream = new TarInputStream(new FileInputStream(tarPath));
while (null != (entry = inputStream.getNextEntry())) {
int bytesRead;
System.out.println("tarpath:" + tarPath.getName());
System.out.println("Entry:" + entry.getName());
String pathWithoutName = path.substring(0, path.indexOf(tarPath.getName()));
System.out.println("pathname:" + pathWithoutName);
if (entry.isDirectory()) {
File directory = new File(pathWithoutName + entry.getName());
directory.mkdir();
continue;
}
byte[] buffer = new byte[1024];
outputStream = new FileOutputStream(pathWithoutName + entry.getName());
while ((bytesRead = inputStream.read(buffer, 0, 1024)) > -1) {
outputStream.write(buffer, 0, bytesRead);
}
System.out.println("Extracted " + entry.getName());
}
}
Try opening your FileOutputstream like this instead:
File outputFile = new File(pathWithoutName + entry.getName());
for(int i = 2; outputFile.exists(); i++) {
outputFile = new File(pathWithoutName + entry.getName() + i);
}
outputStream = new FileOutputStream(outputFile);
It should generate a file called APPENDER_CLASS.xml2 if it encounters a previously created file called APPENDER_CLASS.xml. If a APPENDER_CLASS.xml2 exists it will create a APPENDER_CLASS.xml3, ad infinitum.
File.exists() takes case sensitivity into account (windows filenames are case insensitive, whereas unix, linux and mac are case sensitive). Thus with the above code on case insensitive filesystems the file would be renamed and on case sensitive filesystems the file would not be renamed.
I have to edit the html files of epubs programmatically so what I did was to unzip the .epub and create a parser to make the necessary edits for the html files. However, when I convert them back into an .epub using my code, EpubChecker shows that:
Error: Required META-INF/container.xml resource is missing
When I uncompressed my edited .epub, the container.xml is present and is not missing.
I understand that the mimetype and META-INF has to be zipped first. Here is my code to convert the files back to epub:
FileOutputStream fos = new FileOutputStream(zipFile);
ZipOutputStream zos = new ZipOutputStream(fos);
System.out.println("Output to Zip : " + zipFile);
writeMimeType(zos);
ZipEntry container = new ZipEntry("META-INF\\container.xml");
zos.putNextEntry(container);
FileInputStream inMime2 = new FileInputStream(SOURCE_FOLDER + File.separator + "META-INF\\container.xml");
int len2;
while((len2 = inMime2.read(buffer)) > 0){
zos.write(buffer, 0, len2);
}
inMime2.close();
for(String file : this.fileList){
if(!file.toString().equals("mimetype") && !file.toString().equals("META-INF\\container.xml")){
System.out.println("File Added : " + file);
ZipEntry ze= new ZipEntry(file);
zos.putNextEntry(ze);
FileInputStream in =
new FileInputStream(SOURCE_FOLDER + File.separator + file);
int len;
while ((len = in.read(buffer)) > 0) {
zos.write(buffer, 0, len);
}
in.close();
}
}
zos.closeEntry();
zos.close();
When I manually zip the directory using WinRar, no errors are seen and it works properly. I don't know what I am doing wrong. Can somebody please help me?Thank you.
Looks like you're on Windows, so your FileInputStream(SOURCE_FOLDER + File.separator + "META-INF\\container.xml"); statement is correct for the OS, but I'd guess you need to change the other 2 strings to use the forward slash for the zipentry path.
ZipEntry container = new ZipEntry("META-INF\\container.xml");
try instead as
ZipEntry container = new ZipEntry("META-INF/container.xml");
and change
if(!file.toString().equals("mimetype") && !file.toString().equals("META-INF\\container.xml")){
to
if(!file.toString().equals("mimetype") && !file.toString().equals("META-INF/container.xml")){
accordingly.
You may need to adjust your other ZipEntry's as well. From the ZIP spec (section "4.4.17 file name"):
All slashes MUST be forward slashes '/' as opposed to backwards
slashes '\'
NOTE: This is a followup to my question here.
I have a program that takes the contents of a directory and bundles everything into a JAR file. The code I use to do this is here:
try
{
FileOutputStream stream = new FileOutputStream(target);
JarOutputStream jOS = new JarOutputStream(stream);
LinkedList<File> fileList = new LinkedList<File>();
buildList(directory, fileList);
JarEntry jarAdd;
String basePath = directory.getAbsolutePath();
byte[] buffer = new byte[4096];
for(File file : fileList)
{
String path = file.getPath().substring(basePath.length() + 1);
path.replaceAll("\\\\", "/");
jarAdd = new JarEntry(path);
jarAdd.setTime(file.lastModified());
jOS.putNextEntry(jarAdd);
FileInputStream in = new FileInputStream(file);
while(true)
{
int nRead = in.read(buffer, 0, buffer.length);
if(nRead <= 0)
break;
jOS.write(buffer, 0, nRead);
}
in.close();
}
jOS.close();
stream.close();
So, all is well and good and the jar gets created, and when I explore its contents with 7-zip it has all the files I need. However, when I try to access the contents of the Jar via a URLClassLoader (the jar is not on the classpath and I don't intend it to be), I get null pointer exceptions.
The odd thing is, when I use a Jar that I've exported from Eclipse, I can access the contents of it in the way I want. This leads me to believe that I'm somehow not creating the Jar correctly, and am leaving something out. Is there anything missing from the method up above?
I figured it out based on this question - the problem was me not properly handling backslashes.
Fixed code is here:
FileOutputStream stream = new FileOutputStream(target);
JarOutputStream jOS = new JarOutputStream(stream);
LinkedList<File> fileList = new LinkedList<File>();
buildList(directory, fileList);
JarEntry entry;
String basePath = directory.getAbsolutePath();
byte[] buffer = new byte[4096];
for(File file : fileList)
{
String path = file.getPath().substring(basePath.length() + 1);
path = path.replace("\\", "/");
entry = new JarEntry(path);
entry.setTime(file.lastModified());
jOS.putNextEntry(entry);
FileInputStream in = new FileInputStream(file);
while(true)
{
int nRead = in.read(buffer, 0, buffer.length);
if(nRead <= 0)
break;
jOS.write(buffer, 0, nRead);
}
in.close();
jOS.closeEntry();
}
jOS.close();
stream.close();
I have a lot of zips with a large amount of Files which i have to unzip to the SD-Card.
Currently I am using the util.zip Package with Zipentry for each File inside the zips. This works fine but is very slow.
So I wonder if there is a lib which can handle those Files faster than the normal zip of Java/Android.
Edit: The zips are archives which are about 5 to 10MB large and contain about 50 to 100 jpg-Files whiche are each part of a picture. I need to extract all zips to a specific folder.
Slow means that the same files are extracted on an IPhone in fraction of the time.
Edit 2 the code:
ZipInputStream zin = new ZipInputStream(fd);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
System.out.println("Unzipping " + ze.getName());
if (ze.isDirectory()) {
File f = new File(targetFilePath + ze.getName());
if (!f.isDirectory()) {
f.mkdirs();
}
} else {
int size;
byte[] buffer = new byte[2048];
FileOutputStream outStream = new FileOutputStream(targetFilePath + ze.getName() + ".tile");
BufferedOutputStream bufferOut = new BufferedOutputStream(outStream, buffer.length);
while ((size = zin.read(buffer, 0, buffer.length)) != -1) {
bufferOut.write(buffer, 0, size);
}
bufferOut.flush();
bufferOut.close();
}
}
zin.close();
The best case you could do is go NDK way. AFAIK, Google even provides supported headers/API in NDK on zlib.
I have written a Java web application that allows a user to download files from a server. These files are quite large and so are zipped together before download.
It works like this:
1. The user gets a list of files that match his/her criteria
2. If the user likes a file and wants to download he/she selects it by checking a checkbox
3. The user then clicks "download"
4. The files are then zipped and stored on a servera
5. The user this then presented with a page which contains a link to the downloadable zip filea
6. However on downloading the zip file the file that is downloaded is 0 bytes in sizea
I have checked the remote server and the zip file is being created properly, all that is left is to serve the file the user somehow, can you see where I might be going wrong, or suggest a better way to serve the zip file.
The code that creates the link is:
<%
String zipFileURL = (String) request.getAttribute("zipFileURL"); %>
<p>Zip File Link</p>
The code that creates the zipFileURL variable is:
public static String zipFiles(ArrayList<String> fileList, String contextRootPath) {
//time-stamping
Date date = new Date();
Timestamp timeStamp = new Timestamp(date.getTime());
Iterator fileListIterator = fileList.iterator();
String zipFileURL = "";
try {
String ZIP_LOC = contextRootPath + "WEB-INF" + SEP + "TempZipFiles" + SEP;
BufferedInputStream origin = null;
zipFileURL = ZIP_LOC
+ "FITS." + timeStamp.toString().replaceAll(":", ".").replaceAll(" ", ".") + ".zip";
FileOutputStream dest = new FileOutputStream(ZIP_LOC
+ "FITS." + timeStamp.toString().replaceAll(":", ".").replaceAll(" ", ".") + ".zip");
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(
dest));
// out.setMethod(ZipOutputStream.DEFLATED);
byte data[] = new byte[BUFFER];
while(fileListIterator.hasNext()) {
String fileName = (String) fileListIterator.next();
System.out.println("Adding: " + fileName);
FileInputStream fi = new FileInputStream(fileName);
origin = new BufferedInputStream(fi, BUFFER);
ZipEntry entry = new ZipEntry(fileName);
out.putNextEntry(entry);
int count;
while ((count = origin.read(data, 0, BUFFER)) != -1) {
out.write(data, 0, count);
}
origin.close();
}
out.close();
} catch (Exception e) {
e.printStackTrace();
}
return zipFileURL;
}
A URL cannot access any files (directly) under WEB-INF. I'd suggest using a servlet to return the file from whatever location it was saved to
Would also suggest saving the file outside the context of your webapp