I am trying to copy a file (Base.jar) to the same directory as the running jar file
I keep getting a corrupted jar file, that still holds the correct class structure when opened with winrar. What am I doing wrong? (I have also tried without the ZipInputStream, but that was no help) the byte[] is 20480 because that is size of it on the disk.
my code:
private static void getBaseFile() throws IOException
{
InputStream input = Resource.class.getResourceAsStream("Base.jar");
ZipInputStream zis = new ZipInputStream(input);
byte[] b = new byte[20480];
try {
zis.read(b);
} catch (IOException e) {
}
File dest = new File("Base.jar");
FileOutputStream fos = new FileOutputStream(dest);
fos.write(b);
fos.close();
input.close();
}
InputStream input = Resource.class.getResourceAsStream("Base.jar");
File fileOut = new File("your lib path");
OutputStream out = FileUtils.openOutputStream(fileOut);
IOUtils.copy(in, out);
in.close();
out.close();
and handle exceptions
No need to use ZipInputStream, unless you want to unzip the contents into memory and read.
Just use BufferedInputStream(InputStream) or BufferedReader(InputStreamReader(InputStream)).
did some more googling found this: (Convert InputStream to byte array in Java) worked for me
InputStream is = ...
ByteArrayOutputStream buffer = new ByteArrayOutputStream();
int nRead;
byte[] data = new byte[16384];
while ((nRead = is.read(data, 0, data.length)) != -1) {
buffer.write(data, 0, nRead);
}
buffer.flush();
return buffer.toByteArray();
(it looks very simular to the src for IOUtils.copy())
ZipInputStream is for reading files in the ZIP file format by entry. You need to copy the whole file (resource) that is you need to simply copy all bytes from InputStream no matter what format is. The best way to do it in Java 7 is this:
Files.copy(inputStream, targetPath, optionalCopyOptions);
see API for details
Related
I take a multipartfile (i.e. SAMPLE.csv) in input.
I should zip it (i.e. SAMPLE.zip) and store it via FTP.
public void zipAndStore(MultipartFile file) {
try (ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
InputStream is = file.getInputStream()) {
ZipEntry zipEntry = new ZipEntry("SAMPLE.zip");
zos.putNextEntry(zipEntry);
byte[] bytes = new byte[1024];
int length;
while ((length = is.read(bytes)) >= 0) {
zos.write(bytes, 0, length);
}
zos.closeEntry();
storeFtp("SAMPLE.zip", new ByteArrayInputStream(baos.toByteArray()));
} catch (Exception e) {
}
}
The storeFtp use the org.apache.commons.net.ftp.FTPClient.storeFile(String remote, InputStream local) method.
The problem is that the uploaded file is corrupted and i'm unable to manually decompress.
What's wrong?
A zipfile has a list of DirEntries and a endLocator at the end of the file (after all the ZipFileRecords, i.e. the ZipEntries in the code).
So you probably have to close the zipfile before calling storeFtp() to make sure the DirEntries and the endLocator are written to the file:
zos.closeEntry();
zos.close();
storeFtp("SAMPLE.zip", new ByteArrayInputStream(baos.toByteArray()));
(I don't know Java that well, so I can't check or test the full code)
Also check out this answer.
Need a solution in Java to unzip a huge file(Won't fit in memory) using inputstream. File Object is not available in this case. The file is password protected. Solutions with File object won't help.
I've already tried with 7Zip, But it's not supporting the above case.
When you use streams, you should not read more data than requires. have you tried this?
public void unzip(InputStream is, Cipher cypher) throws IOException {
ZipInputStream zis = new ZipInputStream(new CipherInputStream(is,cypher));
ZipEntry zipEntry = zis.getNextEntry();
byte[] buffer = new byte[1024];
while (zipEntry != null) {
File newFile = new File(zipEntry.getName());
FileOutputStream fos = new FileOutputStream(newFile);
int len;
while ((len = zis.read(buffer)) > 0) {
fos.write(buffer, 0, len);
}
fos.close();
zipEntry = zis.getNextEntry();
}
zis.closeEntry();
zis.close();
}
I had such problems too.
In this repo(https://github.com/r331/memzipenc) you can find a method MemZipDec.unzipFiles(byte[] zipBytes, String password)
I hope it would help.
I have a servlet (JSP) that reads a file from disk and sends it as http response.
The downloaded file has not the same sha1sum that the original and some files seems to be corrupted.
My code is this:
File file = new File((String)application.getAttribute("path")+"/"+item);
if (!file.exists()){
RequestDispatcher dispatcher = application.getRequestDispatcher("/not_found.jsp?item="+item);
dispatcher.forward(request, response);
return;
}
response.setContentType("application/octet-stream");
response.setHeader("Content-Disposition",
"attachment;filename=\""+item+"\"");
Locale.setDefault(new Locale("ES","ve"));
FileInputStream fileIn = new FileInputStream(file);
ServletOutputStream outs = response.getOutputStream();
byte[] outputByte = new byte[4096];
//copy binary contect to output stream
System.out.println(new java.util.Date()+"\tStart download\t"+item+"\t"+request.getRemoteAddr()+"\t"+request.getRemoteHost());
while(fileIn.read(outputByte, 0, 4096) != -1)
{
outs.write(outputByte, 0, 4096);
}
fileIn.close();
outs.flush();
outs.close();
System.out.println(new java.util.Date()+"\tEnd Download\t"+item+"\t"+request.getRemoteAddr()+"\t"+request.getRemoteHost());
Why does downloaded file have different sha1sum as original file and why some but not all downloaded files are corrupt?
You must put the number of read bytes in a variable, and just write as many bytes that you read.
Your implementation always writes the whole buffer of 4096 bytes, no matter of how many bytes that actually were read.
Something like this:
int i = 0;
while( (i=fileIn.read(outbyteByte, 0, 4096)) != -1){
outs.write(outputByte,0,i);
}
I am trying to create a zip file of multiple image files. I have succeeded in creating the zip file of all the images but somehow all the images have been hanged to 950 bytes. I don't know whats going wrong here and now I can't open the images were compressed into that zip file.
Here is my code. Can anyone let me know what's going here?
String path="c:\\windows\\twain32";
File f=new File(path);
f.mkdir();
File x=new File("e:\\test");
x.mkdir();
byte []b;
String zipFile="e:\\test\\test.zip";
FileOutputStream fout=new FileOutputStream(zipFile);
ZipOutputStream zout=new ZipOutputStream(new BufferedOutputStream(fout));
File []s=f.listFiles();
for(int i=0;i<s.length;i++)
{
b=new byte[(int)s[i].length()];
FileInputStream fin=new FileInputStream(s[i]);
zout.putNextEntry(new ZipEntry(s[i].getName()));
int length;
while((length=fin.read())>0)
{
zout.write(b,0,length);
}
zout.closeEntry();
fin.close();
}
zout.close();
This is my zip function I always use for any file structures:
public static File zip(List<File> files, String filename) {
File zipfile = new File(filename);
// Create a buffer for reading the files
byte[] buf = new byte[1024];
try {
// create the ZIP file
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipfile));
// compress the files
for(int i=0; i<files.size(); i++) {
FileInputStream in = new FileInputStream(files.get(i).getCanonicalName());
// add ZIP entry to output stream
out.putNextEntry(new ZipEntry(files.get(i).getName()));
// transfer bytes from the file to the ZIP file
int len;
while((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
// complete the entry
out.closeEntry();
in.close();
}
// complete the ZIP file
out.close();
return zipfile;
} catch (IOException ex) {
System.err.println(ex.getMessage());
}
return null;
}
Change this:
while((length=fin.read())>0)
to this:
while((length=fin.read(b, 0, 1024))>0)
And set buffer size to 1024 bytes:
b=new byte[1024];
I have a silly problem i haven't been able to figure out. Can anyone help me?
My Code is as:
String zipname = "C:/1100.zip";
String output = "C:/1100";
BufferedInputStream bis = null;
BufferedOutputStream bos = null;
try {
ZipFile zipFile = new ZipFile(zipname);
Enumeration<?> enumeration = zipFile.entries();
while (enumeration.hasMoreElements()) {
ZipEntry zipEntry = (ZipEntry) enumeration.nextElement();
System.out.println("Unzipping: " + zipEntry.getName());
bis = new BufferedInputStream(zipFile.getInputStream(zipEntry));
int size;
byte[] buffer = new byte[2048];
It doesn't create a folder but debugging shows all the contents being generated.
In Order to create a folder i used the code
if(!output.exists()){ output.mkdir();} // here i get an error saying filenotfoundexception
bos = new BufferedOutputStream(new FileOutputStream(new File(outPut)));
while ((size = bis.read(buffer)) != -1) {
bos.write(buffer, 0, size);
}
}
} catch (Exception ex) {
ex.printStackTrace();
} finally {
bos.flush();
bos.close();
bis.close();
}
My zip file contains images: a.jpg b.jpg... and in the same hierarchy, I have abc.xml.
I need to extract the content as is in the zip file.
Any helps here.
There are a few problems with your code: Where is outPut declared? output is not a file but a string, so exists() and mkdir() do not exist. Start by declaring output like:
File output = new File("C:/1100");
Furthermore, outPut (with big P) is not declared. It be something like output + File.seprator + zipEntry.getName().
bos = new BufferedOutputStream(new FileOutputStream(output + File.seprator + zipEntry.getName()));
Note that you don't need to pass a File to FileOutputStream, as constructors show in the documentation.
At this point, your code should work if your Zip file does not contain directory. However, when opening the output stream, if zipEntry.getName() has a directory component (for instance somedir/filename.txt), opening the stream will result in a FileNotFoundException, as the parent directory of the file you try to create does not exist. If you want to be able to handle such zip files, you will find your answer in: How to unzip files recursively in Java?