I have a zip file and I want to replace one file inside it with another file. So do not need to delete a zip entry just replace the file for the zip entry with another.
Here is what I have tried.
public void replaceConfigurationFile(ZipFile zipFile, ZipOutputStream zos, String pathToNewFile, String configFileToReplaced)
throws IOException {
String zipEntryName;
for(Enumeration<?> e = zipFile.entries(); e.hasMoreElements(); ) {
ZipEntry entryIn = (ZipEntry) e.nextElement();
zipEntryName = entryIn.getName();
if(zipEntryName.endsWith(configFileToReplaced)) {
FileInputStream fis = new FileInputStream(pathToNewFile);
ZipEntry zipEntry = new ZipEntry(zipEntryName);
zos.putNextEntry(zipEntry);
byte[] bytes = new byte[1024];
int length;
while ((length = fis.read(bytes)) >= 0) {
zos.write(bytes, 0, length);
}
zos.closeEntry();
fis.close();
} else {
zos.putNextEntry(entryIn);
InputStream is = zipFile.getInputStream(entryIn);
byte [] buf = new byte[1024];
int len;
while((len = (is.read(buf))) > 0) {
zos.write(buf, 0, len);
}
zos.closeEntry();
}
} // enf of for
}
I have a file zip Entry named :
WEB-INF/classes/config/app-dev.yml
and I have a file at location d drive at location
D:/app-dev.yml
I am able to copy the file in to a different zip file by replacing the file i want to replace . But that is really not needed (to create a different file).
So what should I do to just replace the file with my custom file.
I have searched different posts in Stackoverflow, but unable to find what i need. I read that zip entry cannot be deleted but what about replacing it ? Please help
Your problem is that the new file might be larger than the old file - or smaller! You need to do exactly what you did to allow for the change. The probability that the new file, ZIPped, is exactly the same size as the previous one is virtually nil.
The standard Java ZIP file libraries do not allow you to update an existing ZIP file. You need to use a third party library. A Google search should find you a number of alternatives.
But you need to be aware of a couple of things:
If your application ... or system ... crashes while updating the ZIP file, the file may be corrupted.
When a ZIP file is updated, the replacement version of the file will be appended to the ZIP. The old version of the will still be in the ZIP ... but not in the ZIP file index. Hence updating a ZIP file (without rewriting it) wastes disk space.
Related
While Uncompressing .7z file, Empty folders are ignored, I want to consider Empty folders as well after uncompressing any .7z file.
My Code is as below
SevenZFile sevenZFile = new SevenZFile(new File(filename));
SevenZArchiveEntry entry;
while ((entry = sevenZFile.getNextEntry()) != null){
if (entry.isDirectory()){
continue;
}
File curfile = new File(DestinationPath,entry.getName());
File parent = curfile.getParentFile();
if (!parent.exists()) {
parent.mkdirs();
}
FileOutputStream out = new FileOutputStream(curfile);
byte[] content = new byte[(int) entry.getSize()];
sevenZFile.read(content, 0, content.length);
out.write(content);
out.close();
Your code seems working.
Probably the folder aren't in the "yourfile.7zip" from the beginning.
This is a common issue of 7zip and you have to update your 7zip version.
If the 7Zip contains proper arguments just use:
if (entry.isDirectory()){
new File(DestinationPath,entry.getName()).mkdir();
continue;
}
Since:
A file output stream is an output stream for writing data to a File or
to a FileDescriptor.
That is the proper method to accomplish the task because there's no folder implementation by native library vendor.
I am seeking for most efficient way (in terms of speed) to retrieve some file out of the middle of a ZIP file.
e.g. I have ZIP file, which includes 700 folders (tagged 1 to 700). Each folder equals picture and mp3 file. There is special folder called Info, which contains XML file. Problem is, I need to iterate through this ZIP file to find XML file and then I am displaying images from desired folders. I am using ZipFile approach (thus I am iterating through whole ZIP file, even if I want folder 666, I need to go through 665 items in ZIP file) -> selecting from ZIP file is extremely slow.
I would like to ask you, If you have faced similar issue, how have you solved this? Is there any approach in Java, which turns my ZIP file into virtual folder to browse it much more quicker? Is there any external library, which is the most efficient in terms of time?
Source Code snippet:
try {
FileInputStream fin = new FileInputStream(
"sdcard/external_sd/mtp_data/poi_data/data.zip");
ZipInputStream zin = new ZipInputStream(fin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
// Log.d("ZE", ze.getName());
if (ze.getName().startsWith("body/665/")) {
// Log.d("FILE F", "soubor: "+ze.getName());
if (ze.getName().endsWith(".jpg")
|| ze.getName().endsWith(".JPG")) {
Log.d("OBR", "picture: " + ze.getName());
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
int count;
while ((count = zin.read(buffer)) != -1) {
baos.write(buffer, 0, count);
}
byte[] bytes = baos.toByteArray();
bmp = BitmapFactory.decodeByteArray(bytes, 0,
bytes.length);
photoField.add(bmp);
i++;
}
}
}
}
The ZipFile.getEntry() and ZipFile.getInputStream() methods can be used to access a specific file in a ZIP archive. For example:
ZipFile file = ...
ZipEntry entry = file.getEntry("folder1/picture.jpg");
InputStream in = file.getInputStream(entry);
I would like to create a zip file with Commons VFS2 library. I know how to copy a file when using file prefix but for zip files write and read are not implemented.
fileSystemManager.resolveFile("path comes here")-method fails when I try path zip:/some/file.zip when file.zip is an non-existing zip-file. I can resolve an existing file but non-existing new file fails.
So how to create that new zip file then? I cannot use createFile() because it is not supported and I cannot create the FileObject before this is to be called.
The normal way is to create FileObject with that resolveFile and then call createFile for the object.
The answer to my need is the following code snippet:
// Create access to zip.
FileSystemManager fsManager = VFS.getManager();
FileObject zipFile = fsManager.resolveFile("file:/path/to/the/file.zip");
zipFile.createFile();
ZipOutputStream zos = new ZipOutputStream(zipFile.getContent().getOutputStream());
// add entry/-ies.
ZipEntry zipEntry = new ZipEntry("name_inside_zip");
FileObject entryFile = fsManager.resolveFile("file:/path/to/the/sourcefile.txt");
InputStream is = entryFile.getContent().getInputStream();
// Write to zip.
byte[] buf = new byte[1024];
zos.putNextEntry(zipEntry);
for (int readNum; (readNum = is.read(buf)) != -1;) {
zos.write(buf, 0, readNum);
}
After this you need to close the streams and it works!
In fact, it is possible to create zip files uniquely from Commons-VFS by using the following idio :
destinationFile = fileSystemManager.resolveFile(zipFileName);
// destination is created as a folder, as the inner content of the zip
// is, in fact, a "virtual" folder
destinationFile.createFolder();
// then add files to that "folder" (which is in fact a file)
// and finally close that folder to have a usable zip
destinationFile.close();
// Exception handling is left at user discretion
I am having an issue putting a folder in a zip file I am trying to create. While the path is valid, when I run the code it gives me a File Not Found Exception.
Here is my code
String outFilename = "outfile.zip";
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(outFilename));
byte[] buf = new byte[1024];
File file = new File("workspace");
System.out.println(file.isDirectory());
System.out.println(file.getAbsolutePath());
FileInputStream in = new FileInputStream(file.getAbsolutePath());
out.putNextEntry(new ZipEntry(file.getAbsolutePath()));
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
out.closeEntry();
in.close();
You're trying to read bytes from a directory; it doesn't work like that. The exception says as much, too.
You need to add the directory, then add each file within the directory. If you use the file path you don't need to add the directory explicitly.
I'd be very wary of using the absolute path as the zip entry; better to use a relative path so you can unzip it anywhere and not risk overwriting something you want.
I am having a InputStream Object which is actually a zip file. I want to change it back to zip file and save it. I am using DWR's FileTransfer class object to receive the uploaded data from client.
FileTransfer have 3 methods, getInputStream() is one of them. It returns InputStream from FileTransfer object.
In my case, fileTransfer object holds zip file and as well as InputStream object too.
I have done, lot of searches in google. But i am not able to find one example, that illustrates InputStream to zip conversion.
Update
String zipName = file.getName();
String zipType = file.getMimeType();
InputStream zipStream = file.getInputStream();
ZipInputStream zis = new ZipInputStream(zipStream);
System.out.println("File Name: "+zipName+"\n"+"File Type: "+zipType);
int c;
File f2 = new File(DATA_STORE_LOC+dat+".zip");
path.setPath2(DATA_STORE_LOC+dat+".zip");
FileOutputStream fos = new FileOutputStream(f2);
ZipOutputStream zos = new ZipOutputStream(fos);
c = zis.read();
System.out.println(c);
while ((c = zis.read(BUFFER)) != -1) {
zos.write(BUFFER, 0, c);
}
zos.close();
zis.close();
I tried this code, by thought of a typical file copy program. I know it is false, just tried. It gives me java.util.zip.ZipException: ZIP file must have at least one entry.
Any suggestion would be really appreciative!!!!!
See the examples java2s, input and output. If you have more questions feel free to ask them :)
For clarity, in this input example you should do something like:
// FileInputStream fin = new FileInputStream(args[i]);
ZipInputStream zin = new ZipInputStream(ft.getInputStream());
As Don Roby correctly said, if you just want to copy you need not know the file structure and you could use for example static IOUtils.copy(in, out) to copy the file.
Further, if you do wish to extract the ZIP file contents, you should not plainly copy bytes. The ZIP file has a structure, and you extract Entries from the ZIP file, and not just bytes (see the example). Every Entry is a (compressed) file (or the data thereof) with the original name:
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
System.out.println("Unzipping " + ze.getName());
FileOutputStream fout = new FileOutputStream(ze.getName());
for (int c = zin.read(); c != -1; c = zin.read()) {
...
Please note the javadoc of getNextEntry():
Reads the next ZIP file entry and positions the stream at the beginning of the entry data.
This positioning is crucial to get to the zipped file contents, and not the metadata.
And I do believe that you accidentally remove the first int:
c = zis.read(); // removing the first
while ((c = zis.read(BUFFER)) != -1) { // so you start with the second?
I believe you mix 2 idioms:
c = zis.read();
while(c != -1) {
...
c = zis.read();
}
and:
int c;
while ((c = zis.read(BUFFER)) != -1) { // so you start with the second?
...
}
I think you can see the difference :)
If your input is a an InputStream from a zip file and your desired output is still a zip file with the same contents, you're just doing a file copy operation and shouldn't have to worry about zip at all. You just need to read from the InputStream and write to a FileOutputStream, more or less as you're doing, but without worrying about wrapping either stream in a zip-aware stream.
ZipInputStream is useful if you have to extract the contents of the zip file as separate files, i.e., to programmatically unzip. And on the other side, ZipOutputStream is used if your have the contents and need to combine them into a zip file.