I'm trying to zip files from server into a folder using ZipOutputStream.
After archive download it can't be opened after double click. Error "The compressed (zipped) folder is invalid" occures. But if I open it from context menu - > 7zip -> open file it works normal. What can be reason of the problem?
sourceFileName="./file.txt"'
sourceFile = new File(sourceFileName);
try {
// set the content type and the filename
responce.setContentType("application/zip");
response.addHeader("Content-Disposition", "attachment; filename=" + sourceFileName + ".zip");
responce.setContentLength((int) sourceFile.length());
// get a ZipOutputStream, so we can zip our files together
ZipOutputStream outZip = new ZipOutputStream((responce.getOutputStream());
// Add ZIP entry to output stream.
outZip.putNextEntry(new ZipEntry(sourceFile.getName()));
int length = 0;
byte[] bbuf = new byte[(int) sourceFile.length()];
DataInputStream in = new DataInputStream(new FileInputStream(sourceFile));
while ((in != null) && ((length = in.read(bbuf)) != -1)) {
outZip.write(bbuf, 0, length);
}
outZip.closeEntry();
in.close();
outZip.flush();
outZip.close();
7Zip can open a wide variety of zip formats, and is relatively tolerant of oddities. Windows double-click requires a relatively specific format and is far less tolerant.
You need to look up the zip format and then look at your file (and "good" ones) with a hex editor (such as Hex Editor Neo), to see what may be wrong.
(One possibility is that you're using the wrong compression algorithm. And there are several other variations to consider as well, particularly whether or not you generate a "directory".)
It could be that a close is missing. It could be that the path encoding in the zip cannot be handled by Windows. It might be that Windows has difficulty with the directory structure, or that a path name contains a (back)slash. So it is detective work, trying different files. If you immediately stream the zip to the HTTP response, then finish has to be called i.o. close.
After the code being posted:
The problem is the setContentLength giving the original file size. But when given, it should give the compressed size.
DataInputStream is not needed, and one should here do a readFully.
responce.setContentType("application/zip");
response.addHeader("Content-Disposition", "attachment; filename=file.zip");
//Path sourcePath = sourceFile.toPath();
Path sourcePath = Paths.get(sourceFileName);
ZipOutputStream outZip = new ZipOutputStream((responce.getOutputStream(),
StandardCharsets.UTF-8);
outZip.putNextEntry(new ZipEntry(sourcePath.getFileName().toString()));
Files.copy(sourcePath, outZip);
outZip.closeEntry();
Either finish or closethe zip at the end.
outZip.finish();
//outZip.close();
in.close();
I am not sure (about the best code style) whether to close the response output stream already oneself.
But when not closing finish() must be called, flush() will not suffice, as at the end data is written to the zip.
For file names with for instance Cyrillic letters, it would be best to add a Unicode charset like UTF-8. In fact let UTF-8 be the Esperanto standard world-wide.
A last note: if only one file one could use GZipOutputstream for file.txt.gz or query the browser's capabilities (request parameters) and deliver it compressed as file.txt.
Related
How to download all files in the file directory when clicking the export or download at the same time？
At present, all the files in the file directory have been obtained, then all the files are placed in the list, and then the stream is written after traversing all the files. However, when importing the second file, it will report cannot reset buffer after response has been committed
The source of the problem is in this code： // response.reset();
Code：
String filePath = "/code/data/";
// Get all file addresses of the directory
List<String> filePathList = getFilePath(filePath);
//Create thread pool
for (String str : filePathList){
download(request, response, str);
}
private void download(HttpServletRequest request,
HttpServletResponse response,String filePath) {
File file = new File(filePath);
//Gets the file name.
String fileName = file.getName();
InputStream fis = null;
try {
fis = new FileInputStream(file);
request.setCharacterEncoding("UTF-8");
String agent = request.getHeader("User-Agent").toUpperCase();
if ((agent.indexOf("MSIE") > 0) || ((agent.indexOf("RV") != - 1) &&
(agent.indexOf("FIREFOX") == -1))) {
fileName = URLEncoder.encode(fileName, "UTF-8");
} else {
fileName = new String(fileName.getBytes("UTF-8"), "ISO8859-1");
}
// response.reset();
response.setCharacterEncoding("UTF-8");
response.setContentType("application/force-download");
// Set forced download not to open
response.addHeader("Content-Disposition",
"attachment; filename=" + fileName);
response.setHeader("Content-Length", String.valueOf(file.length()));
byte[] b = new byte[1024];
int len;
while ((len = fis.read(b)) != - 1) {
response.getOutputStream().write(b, 0, len);
}
response.flushBuffer();
fis.close();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
}
What are the good solutions Thanks
I have not read your code in detail because the bad formatting makes my head hurt.
However, from a superficial reading, it looks like this server-side code is trying to deliver multiple files in response to a single HTTP request.
AFAIK, that is not possible. The HTTP request / response model does not support this. It certainly does not allow a servlet to:
change response headers after the response output stream has been opened
do anything after the response output stream has been closed.
(Your code appears to be trying to do both of those things!)
So, you have to do it differently. Here are some possibilities:
On the server side, assemble all of the files to be downloaded into (say) a temporary ZIP file and then send that. Leave it to the user to unpack the ZIP file ... or not ... as they want.
This is often the best approach. Imagine how annoyed you would be if a few thousand separate files unexpectedly landed in your web browser's Downloads folder.
As 1. and also do something on the client side to transparently unpack the files from the ZIP and put them in the right place in the client's file system.
The "something" could be custom javascript embedded in the web page, or a custom client implemented in Java ... or any other language. (But in the former case, there may be a security issue in allowing sandboxed javascript to write files in arbitrary places without the user confirming each file ... tedious.)
You might be able to send a "multipart" document as the response. However from what I have read, most browsers don't support multipart for downloads; e.g. some browsers will discard all but the last part. (Note: multipart is not designed for this purpose ...)
Change things so that an HTTP request only downloads one file at a time from the directory, and add some client-side stuff to 1) fetch a list of files from the server and iterate the list, fetching each file.
See also: Download multiple files with a single action
I have a very confusing problem and hope that I can get some ideas here.
My problem is very simple, but I didn't find a solution yet.
I want to create a simple ZIP File with ZipEntry's in it. The ZipEntry's are created by a given byte array (saved in a Postgres-DB with Hibernate).
When I put this byte array into my ZipOutputStream.write(..) the ZIP File created is always corrupt. What am I doing wrong?
The ZIP File is transferred to a FTP-Server afterwards.
ByteArrayOutputStream bos = new ByteArrayOutputStream();
final ZipOutputStream zipOut = new ZipOutputStream(bos);
String filename = "test.zip";
for(final Attachment attachment : transportDoc.getAttachments()) {
log.debug("Adding "+attachment.getFileName()+" to ZIP file /tmp/"+filename);
ZipEntry ze = new ZipEntry(attachment.getFileName());
zipOut.putNextEntry(ze);
zipOut.write(attachment.getFileContent());
zipOut.flush();
zipOut.closeEntry();
}
zipOut.close();
org.apache.commons.io.FileUtils.writeByteArrayToFile(new File("/tmp/"+filename), bos.toByteArray());
I am confused, because when I replaced
zipOut.write(attachment.getFileContent()); //This is the byte array from db
with
zipOut.write("Bla bla".getBytes());
it worked!
But the byte array from the DB can't be corrupt, because it can be written to a file with
org.apache.commons.io.FileUtils.writeByteArrayToFile(new File("/tmp/test.png"), attachment.getFileContent());
with no problem. It is a correct file.
I hope you have some ideas left.
Thanks in advance.
EDIT:
I tried to repair the ZIP file offline and then this messages appears:
zip warning: no end of stream entry found: cglhnngplpmhipfg.png
(This png file is the byte-Array-File)
Simple unzip-command output the following:
unzip created.zip
Archive: created.zip
error [created.zip]: missing 2 bytes in zipfile
(attempting to process anyway)
error [created.zip]: attempt to seek before beginning of zipfile
(please check that you have transferred or created the zipfile in the
appropriate BINARY mode and that you have compiled UnZip properly)
(attempting to re-compensate)
replace cglhnngplpmhipfg.png? [y]es, [n]o, [A]ll, [N]one, [r]ename: y
inflating: cglhnngplpmhipfg.png
error: invalid compressed data to inflate
file #2: bad zipfile offset (local header sig): 24709
(attempting to re-compensate)
inflating: created.xml
EDIT 2:
When I write this file to the Filesystem and add this file to the ZIP by an InputStream it doesn't work either! But the File on the Filesystem is ok. I can open the Image with no problem. Its very confusing
File tmpAttachment = new File("/tmp/"+filename+attachment.getFileName());
FileUtils.writeByteArrayToFile(tmpAttachment, attachment.getFileContent());
FileInputStream inTmp = new FileInputStream(tmpAttachment);
int len;
byte[] buffer = new byte[1024];
while ((len = inTmp.read(buffer)) > 0) {
zipOut.write(buffer, 0, len);
}
inTmp.close();
EDIT 3:
This problem only appears when I try to add "complex" files like png or pdf. If I put a txt-file in it, it works.
The problem was NOT in the Zip-Library itself.
It was the transmission to an external FTP Server with wrong mode. (Not binary).
Thanks all for your help.
Try closeEntry() before flush(). Also you can try to explicitly specify the size of the entry using ze.setSize(attachment.getFileContent().length).
I want to download multiple zip files as one zip file for a request.
I have zip file paths like C, https://test12.zip etc. So how can I download these files as a one zip file. I have been searching this for a while. All i got is examples for downloading multiple files(local) and zip them. This is what i tried for downloading one file. For multiple files it won't work.
URL url = new URL("https://test12.zip");
URLConnection connection = url.openConnection();
InputStream stream = connection.getInputStream();
BufferedOutputStream outs = new BufferedOutputStream(response.getOutputStream());
int len;
byte[] buf = new byte[1024];
while ((len = stream.read(buf)) > 0) {
outs.write(buf, 0, len);
}
outs.close();
Any help would be much appreciated.
A ZIP file consists of two parts: First the compressed file entries (filename, attributes and data) and at the end of the file there is a central directory containing a list of all entries, again with filename and attributes.
Hence, you can not directly combine or concatenate zip files. In Java you can only decompress the downloaded zip files on-the-fly (without storing them in the file-system) and at the same time using the decompressed content to create a new combined ZIP file:
First create a ZipOutputStream for the zip file you want to create.
Then use the InputStream of each download and use it with a ZipInputStream.
Iterates through all the entries in every ZipInputStream and for each entry create a new identical entry in the ZipOutputStream and copy the content from the ZipInputStream to the ZipOutputStream.
How to use ZipInputStream see for example: https://stackoverflow.com/a/36648504/150978
Note that this process requires to decompress and afterwards re-compress the file content. Depending on the archive size this can result in a high utilization of one CPU core.
I have to implement an application that permits printing the content of all files within a tar.gz file.
For Example:
if I have three files like this in a folder called testx:
A.txt contains the words "God Save The queen"
B.txt contains the words "Ubi maior, minor cessat"
C.txt.gz is a file compressed with gzip that contain the file c.txt with the words "Hello America!!"
So I compress testx, obtain the compressed tar file: testx.tar.gz.
So with my Java application I would like to print in the console:
"God Save The queen"
"Ubi maior, minor cessat"
"Hello America!!"
I have implemented the ZIP version and it works well, but keeping tar library from apache ant http://commons.apache.org/compress/, I noticed that it is not easy like ZIP java utils.
Could someone help me?
I have started looking on the net to understand how to accomplish my aim, so I have the following code:
GZIPInputStream gzipInputStream=null;
gzipInputStream = new GZIPInputStream( new FileInputStream(fileName));
TarInputStream is = new TarInputStream(gzipInputStream);
TarEntry entryx = null;
while((entryx = is.getNextEntry()) != null) {
if (entryx.isDirectory()) continue;
else {
System.out.println(entryx.getName());
if ( entryx.getName().endsWith("txt.gz")){
is.copyEntryContents(out);
// out is a OutputStream!!
}
}
}
So in the line is.copyEntryContents(out), it is possible to save on a file the stream passing an OutputStream, but I don't want it! In the zip version after keeping the first entry, ZipEntry, we can extract the stream from the compressed root folder, testx.tar.gz, and then create a new ZipInputStream and play with it to obtain the content.
Is it possible to do this with the tar.gz file?
Thanks.
surfing the net, i have encountered an interesting idea at : http://hype-free.blogspot.com/2009/10/using-tarinputstream-from-java.html.
After converting ours TarEntry to Stream, we can adopt the same idea used with Zip Files like:
InputStream tmpIn = new StreamingTarEntry(is, entryx.getSize());
// use BufferedReader to get one line at a time
BufferedReader gzipReader = new BufferedReader(
new InputStreamReader(
new GZIPInputStream(
inputZip )));
while (gzipReader.ready()) { System.out.println(gzipReader.readLine()); }
gzipReader.close();
SO with this code you could print the content of the file testx.tar.gz ^_^
To not have to write to a File you should use a ByteArrayOutputStream and use the public String toString(String charsetName)
with the correct encoding.
I have a database file in res/raw/ folder. I am calling Resources.openRawResource() with the file name as R.raw.FileName and I get an input stream, but I have an another database file in device, so to copy the contents of that db to the device db I use:
BufferedInputStream bi = new BufferedInputStream(is);
and FileOutputStream, but I get an exception that database file is corrupted. How can I proceed?
I try to read the file using File and FileInputStream and the path as /res/raw/fileName, but that also doesn't work.
Yes, you should be able to use openRawResource to copy a binary across from your raw resource folder to the device.
Based on the example code in the API demos (content/ReadAsset), you should be able to use a variation of the following code snippet to read the db file data.
InputStream ins = getResources().openRawResource(R.raw.my_db_file);
ByteArrayOutputStream outputStream=new ByteArrayOutputStream();
int size = 0;
// Read the entire resource into a local byte buffer.
byte[] buffer = new byte[1024];
while((size=ins.read(buffer,0,1024))>=0){
outputStream.write(buffer,0,size);
}
ins.close();
buffer=outputStream.toByteArray();
A copy of your file should now exist in buffer, so you can use a FileOutputStream to save the buffer to a new file.
FileOutputStream fos = new FileOutputStream("mycopy.db");
fos.write(buffer);
fos.close();
InputStream.available has severe limitations and should never be used to determine the length of the content available for streaming.
http://developer.android.com/reference/java/io/FileInputStream.html#available():
"[...]Returns an estimated number of bytes that can be read or skipped without blocking for more input. [...]Note that this method provides such a weak guarantee that it is not very useful in practice."
You have 3 solutions:
Go through the content twice, first just to compute content length, second to actually read the data
Since Android resources are prepared by you, the developer, hardcode its expected length
Put the file in the /asset directory and read it through AssetManager which gives you access to AssetFileDescriptor and its content length methods. This may however give you the UNKNOWN value for length, which isn't that useful.