I have a project called "MyApp". MyApp will use a java library I am creating called "MyLibrary". I am writing a function in "MyLibrary" to unzip a zip file in "MyApp"(or whatever app is using "MyLibrary") "resources" dir.
Reading https://community.oracle.com/blogs/kohsuke/2007/04/25/how-convert-javaneturl-javaiofile I cannot create a File via a path because it is not a "physical file". I was using zip4j, but its constructor takes a File or String rather than an InputStream. So I cannot do this:
ZipFile zipfile = new
ZipFile("src/main/resources/compressed.zip");
downloadedZipfile.extractAll("src/main/resources");
java.io.File javadoc and http://www.mkyong.com/java/how-to-convert-inputstream-to-file-in-java/ indicate there isn't a way to convert an InputStream to a File.
Is there another way to access the zip file in the project that is using my library? Thank you in advance.
UPDATE:
The zipIn lacks entries, so the while loop won't extract the file out.
InputStream in = getInputStream("", JSON_FILENAME);
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
try {
while ((entry = zipIn.getNextEntry()) != null) {
String filepath = entry.getName();
if(!entry.isDirectory()) {
extractFile(zipIn, filepath);
}
else {
File dir = new File(filepath);
dir.mkdir();
}
zipIn.closeEntry();
}
zipIn.close();
} catch (IOException e) {
e.printStackTrace();
}
private void extractFile(ZipInputStream zipIn, String filepath) {
BufferedOutputStream bos = null;
try {
bos = new BufferedOutputStream(new FileOutputStream(filepath));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
byte[] bytesIn = new byte[BUFFER_SIZE];
int read = 0;
try {
while ((read = zipIn.read(bytesIn)) != -1) {
bos.write(bytesIn, 0, read);
}
bos.close();
} catch (IOException e) {
e.printStackTrace();
}
}
Will the file be extracted into the home directory of "MyApp" when using this code that belongs to "MyLibrary"?
If you have a InputStream to your virtual ZIP file you can use java.util.zip.ZipInputStream to read the ZIP entries:
InputStream in = ...
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
// handle the entry
}
I am using ZipOutputStream,FileOutputStream and FileInputStream.
First I created a folder with one file. It successfully created. Then I tried to create zip files. Dynamically, it creates file first time correctly but at second time , third time it gives error while opening it.
Error: zip [path/././file.zip] Cannot open The process cannot access the file because it is being used by another process.
I created following code in java,
My Code:
demopath+="/myzip"+po.getPoid();
createDir(demopath);
createFileForFamilies("My content", demopath+"/file");
this.zipDirectory(new File(demopath), demopath+".zip");
My file creator function:
public String createFileForFamilies(String content, String path) {
FileOutputStream fop = null;
File file;
try {
file = new File(path);
fop = new FileOutputStream(file);
// if file doesnt exists, then create it
if (!file.exists()) {
file.createNewFile();
}
// get the content in bytes
byte[] contentInBytes = content.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
return ("Done");
} catch (IOException e) {
System.err.println(e);
return ("Done");
} finally {
try {
if (fop != null) {
fop.close();
}
} catch (IOException e) {
System.err.println(e);
return ("Abort");
}
}
}
My Zip creation function:
public void zipDirectory(File dir, String zipDirName) {
try {
populateFilesList(dir);
//now zip files one by one
//create ZipOutputStream to write to the zip file
FileOutputStream fos = new FileOutputStream(zipDirName);
ZipOutputStream zos = new ZipOutputStream(fos);
for (String filePath : filesListInDir) {
System.out.println("Zipping " + filePath);
//for ZipEntry we need to keep only relative file path, so we used substring on absolute path
ZipEntry ze = new ZipEntry(filePath.substring(dir.getAbsolutePath().length() + 1, filePath.length()));
zos.putNextEntry(ze);
//read the file and write to ZipOutputStream
FileInputStream fis = new FileInputStream(filePath);
byte[] buffer = new byte[1024];
int len;
while ((len = fis.read(buffer)) > 0) {
zos.write(buffer, 0, len);
}
zos.closeEntry();
fis.close();
}
zos.close();
fos.close();
} catch (IOException e) {
e.printStackTrace();
}
}
Thank you Boris...
This is a solution:
Map<String, String> env = new HashMap<>();
env.put("create", "true");
// locate file system by using the syntax
// defined in java.net.JarURLConnection
URI uri = URI.create("jar:file:/"+zipPath+".zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
java.nio.file.Path externalTxtFile;
java.nio.file.Path pathInZipfile ;
externalTxtFile = Paths.get(gamesPath);
pathInZipfile = zipfs.getPath("/file.txt");
Files.copy(externalTxtFile, pathInZipfile,
StandardCopyOption.REPLACE_EXISTING);
}
How do I upload a photo using a URL in the playframework?
I was thinking like this:
URL url = new URL("http://www.google.ru/intl/en_com/images/logo_plain.png");
BufferedImage img = ImageIO.read(url);
File newFile = new File("google.png");
ImageIO.write(img, "png", newFile);
But maybe there's another way. In the end I have to get the File and file name.
Example controller:
public static Result uploadPhoto(String urlPhoto){
Url url = new Url(urlPhoto); //doSomething
//get a picture and write to a temporary file
File tempPhoto = myUploadPhoto;
uploadFile(tempPhoto); // Here we make a copy of the file and save it to the file system.
return ok('something');
}
To get that photo you can use The play WS API, the code behind is an example extracted from the play docs in the section Processing large responses, I recommend you to read the full docs here
final Promise<File> filePromise = WS.url(url).get().map(
new Function<WSResponse, File>() {
public File apply(WSResponse response) throws Throwable {
InputStream inputStream = null;
OutputStream outputStream = null;
try {
inputStream = response.getBodyAsStream();
// write the inputStream to a File
final File file = new File("/tmp/response.txt");
outputStream = new FileOutputStream(file);
int read = 0;
byte[] buffer = new byte[1024];
while ((read = inputStream.read(buffer)) != -1) {
outputStream.write(buffer, 0, read);
}
return file;
} catch (IOException e) {
throw e;
} finally {
if (inputStream != null) {inputStream.close();}
if (outputStream != null) {outputStream.close();}
}
}
}
);
Where url is :
String url = "http://www.google.ru/intl/en_com/images/logo_plain.png"
This is as suggested in play documentation for large files:
*
When you are downloading a large file or document, WS allows you to
get the response body as an InputStream so you can process the data
without loading the entire content into memory at once.
*
Pretty much the same as the above answer then some...
Route: POST /testFile 'location of your controller goes here'
Request body content: {"url":"http://www.google.ru/intl/en_com/images/logo_plain.png"}
Controller(using code from JavaWS Processing large responses):
public static Promise<Result> saveFile() {
//you send the url in the request body in order to avoid complications with encoding
final JsonNode body = request().body().asJson();
// use new URL() to validate... not including it for brevity
final String url = body.get("url").asText();
//this one's copy/paste from Play Framework's docs
final Promise<File> filePromise = WS.url(url).get().map(response -> {
InputStream inputStream = null;
OutputStream outputStream = null;
try {
inputStream = response.getBodyAsStream();
final File file = new File("/temp/image");
outputStream = new FileOutputStream(file);
int read = 0;
byte[] buffer = new byte[1024];
while ((read = inputStream.read(buffer)) != -1) {
outputStream.write(buffer, 0, read);
}
return file;
} catch (IOException e) {
throw e;
} finally {
if (inputStream != null) {
inputStream.close();
}
if (outputStream != null) {
outputStream.close();
}
}
}); // copy/paste ended
return filePromise.map(file -> (Result) ok(file.getName() + " saved!")).recover(
t -> (Result) internalServerError("error -> " + t.getMessage()));
}
And that's it...
In order to serve the file after the upload phase you can use this answer(I swear I'm not promoting myself...): static asset serving from absolute path in play framework 2.3.x
I'm trying to create a folder in internal storage called "unzip" and then unzip a file into the internal storage "unzip folder". Not sure what i'm doing wrong? If you could please explain what's wrong it'd be great! Thanks
Edit: The issues is that I don't think the folder is being created and the file is also not being unzipped.
public void send(View view) {
try {
File mydir = this.getDir("unzip", Context.MODE_PRIVATE);//create folder in internal storage
unzip(getFilesDir().getAbsolutePath(), job_no, getFilesDir() + "/unzip/");
} catch (IOException e) {
}
}
public void unzip(String filepath, String filename, String unzip_path) throws IOException {
InputStream is = new FileInputStream(filepath + filename);
ZipInputStream zis = new ZipInputStream(new BufferedInputStream(is));
try {
ZipEntry ze;
while ((ze = zis.getNextEntry()) != null) {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
int count;
String filename_temp = ze.getName();
File fmd = new File(filepath + filename_temp);
if (!fmd.getParentFile().exists()) {
fmd.getParentFile().mkdirs();
}
FileOutputStream fout = new FileOutputStream(unzip_path + filename_temp);
while ((count = zis.read(buffer)) != -1) {
baos.write(buffer, 0, count);
byte[] bytes = baos.toByteArray();
fout.write(bytes);
baos.reset();
}
fout.close();
//}
}
} finally {
zis.close();
}
}
This will get absolute storage device.
final static String MEDIA_PATH = Environment.getExternalStorageDirectory().getPath() + "/";
I have a dynamic text file that picks content from a database according to the user's query. I have to write this content into a text file and zip it in a folder in a servlet. How should I do this?
Look at this example:
StringBuilder sb = new StringBuilder();
sb.append("Test String");
File f = new File("d:\\test.zip");
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(f));
ZipEntry e = new ZipEntry("mytext.txt");
out.putNextEntry(e);
byte[] data = sb.toString().getBytes();
out.write(data, 0, data.length);
out.closeEntry();
out.close();
This will create a zip in the root of D: named test.zip which will contain one single file called mytext.txt. Of course you can add more zip entries and also specify a subdirectory like this:
ZipEntry e = new ZipEntry("folderName/mytext.txt");
You can find more information about compression with Java here.
Java 7 has ZipFileSystem built in, that can be used to create, write and read file from zip file.
Java Doc: ZipFileSystem Provider
Map<String, String> env = new HashMap<>();
// Create the zip file if it doesn't exist
env.put("create", "true");
URI uri = URI.create("jar:file:/codeSamples/zipfs/zipfstest.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
Path externalTxtFile = Paths.get("/codeSamples/zipfs/SomeTextFile.txt");
Path pathInZipfile = zipfs.getPath("/SomeTextFile.txt");
// Copy a file into the zip file
Files.copy(externalTxtFile, pathInZipfile, StandardCopyOption.REPLACE_EXISTING);
}
To write a ZIP file, you use a ZipOutputStream. For each entry that you want to place into the ZIP file, you create a ZipEntry object. You pass the file name to the ZipEntry constructor; it sets the other parameters such as file date and decompression method. You can override these settings if you like. Then, you call the putNextEntry method of the ZipOutputStream to begin writing a new file. Send the file data to the ZIP stream. When you are done, call closeEntry. Repeat for all the files you want to store. Here is a code skeleton:
FileOutputStream fout = new FileOutputStream("test.zip");
ZipOutputStream zout = new ZipOutputStream(fout);
for all files
{
ZipEntry ze = new ZipEntry(filename);
zout.putNextEntry(ze);
send data to zout;
zout.closeEntry();
}
zout.close();
Here is an example code to compress a Whole Directory(including sub files and sub directories), it's using the walk file tree feature of Java NIO.
import java.io.FileOutputStream;
import java.io.IOException;
import java.nio.file.*;
import java.nio.file.attribute.BasicFileAttributes;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;
public class ZipCompress {
public static void compress(String dirPath) {
final Path sourceDir = Paths.get(dirPath);
String zipFileName = dirPath.concat(".zip");
try {
final ZipOutputStream outputStream = new ZipOutputStream(new FileOutputStream(zipFileName));
Files.walkFileTree(sourceDir, new SimpleFileVisitor<Path>() {
#Override
public FileVisitResult visitFile(Path file, BasicFileAttributes attributes) {
try {
Path targetFile = sourceDir.relativize(file);
outputStream.putNextEntry(new ZipEntry(targetFile.toString()));
byte[] bytes = Files.readAllBytes(file);
outputStream.write(bytes, 0, bytes.length);
outputStream.closeEntry();
} catch (IOException e) {
e.printStackTrace();
}
return FileVisitResult.CONTINUE;
}
});
outputStream.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
To use this, just call
ZipCompress.compress("target/directoryToCompress");
and you'll get a zip file directoryToCompress.zip
Single file:
String filePath = "/absolute/path/file1.txt";
String zipPath = "/absolute/path/output.zip";
try (ZipOutputStream zipOut = new ZipOutputStream(new FileOutputStream(zipPath))) {
File fileToZip = new File(filePath);
zipOut.putNextEntry(new ZipEntry(fileToZip.getName()));
Files.copy(fileToZip.toPath(), zipOut);
}
Multiple files:
List<String> filePaths = Arrays.asList("/absolute/path/file1.txt", "/absolute/path/file2.txt");
String zipPath = "/absolute/path/output.zip";
try (ZipOutputStream zipOut = new ZipOutputStream(new FileOutputStream(zipPath))) {
for (String filePath : filePaths) {
File fileToZip = new File(filePath);
zipOut.putNextEntry(new ZipEntry(fileToZip.getName()));
Files.copy(fileToZip.toPath(), zipOut);
}
}
Spring boot controller, zip the files in a directory, and can be downloaded.
#RequestMapping(value = "/files.zip")
#ResponseBody
byte[] filesZip() throws IOException {
File dir = new File("./");
File[] filesArray = dir.listFiles();
if (filesArray == null || filesArray.length == 0)
System.out.println(dir.getAbsolutePath() + " have no file!");
ByteArrayOutputStream bo = new ByteArrayOutputStream();
ZipOutputStream zipOut= new ZipOutputStream(bo);
for(File xlsFile:filesArray){
if(!xlsFile.isFile())continue;
ZipEntry zipEntry = new ZipEntry(xlsFile.getName());
zipOut.putNextEntry(zipEntry);
zipOut.write(IOUtils.toByteArray(new FileInputStream(xlsFile)));
zipOut.closeEntry();
}
zipOut.close();
return bo.toByteArray();
}
This is how you create a zip file from a source file:
String srcFilename = "C:/myfile.txt";
String zipFile = "C:/myfile.zip";
try {
byte[] buffer = new byte[1024];
FileOutputStream fos = new FileOutputStream(zipFile);
ZipOutputStream zos = new ZipOutputStream(fos);
File srcFile = new File(srcFilename);
FileInputStream fis = new FileInputStream(srcFile);
zos.putNextEntry(new ZipEntry(srcFile.getName()));
int length;
while ((length = fis.read(buffer)) > 0) {
zos.write(buffer, 0, length);
}
zos.closeEntry();
fis.close();
zos.close();
}
catch (IOException ioe) {
System.out.println("Error creating zip file" + ioe);
}
public static void main(String args[])
{
omtZip("res/", "omt.zip");
}
public static void omtZip(String path,String outputFile)
{
final int BUFFER = 2048;
boolean isEntry = false;
ArrayList<String> directoryList = new ArrayList<String>();
File f = new File(path);
if(f.exists())
{
try {
FileOutputStream fos = new FileOutputStream(outputFile);
ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(fos));
byte data[] = new byte[BUFFER];
if(f.isDirectory())
{
//This is Directory
do{
String directoryName = "";
if(directoryList.size() > 0)
{
directoryName = directoryList.get(0);
System.out.println("Directory Name At 0 :"+directoryName);
}
String fullPath = path+directoryName;
File fileList = null;
if(directoryList.size() == 0)
{
//Main path (Root Directory)
fileList = f;
}else
{
//Child Directory
fileList = new File(fullPath);
}
String[] filesName = fileList.list();
int totalFiles = filesName.length;
for(int i = 0 ; i < totalFiles ; i++)
{
String name = filesName[i];
File filesOrDir = new File(fullPath+name);
if(filesOrDir.isDirectory())
{
System.out.println("New Directory Entry :"+directoryName+name+"/");
ZipEntry entry = new ZipEntry(directoryName+name+"/");
zos.putNextEntry(entry);
isEntry = true;
directoryList.add(directoryName+name+"/");
}else
{
System.out.println("New File Entry :"+directoryName+name);
ZipEntry entry = new ZipEntry(directoryName+name);
zos.putNextEntry(entry);
isEntry = true;
FileInputStream fileInputStream = new FileInputStream(filesOrDir);
BufferedInputStream bufferedInputStream = new BufferedInputStream(fileInputStream, BUFFER);
int size = -1;
while( (size = bufferedInputStream.read(data, 0, BUFFER)) != -1 )
{
zos.write(data, 0, size);
}
bufferedInputStream.close();
}
}
if(directoryList.size() > 0 && directoryName.trim().length() > 0)
{
System.out.println("Directory removed :"+directoryName);
directoryList.remove(0);
}
}while(directoryList.size() > 0);
}else
{
//This is File
//Zip this file
System.out.println("Zip this file :"+f.getPath());
FileInputStream fis = new FileInputStream(f);
BufferedInputStream bis = new BufferedInputStream(fis,BUFFER);
ZipEntry entry = new ZipEntry(f.getName());
zos.putNextEntry(entry);
isEntry = true;
int size = -1 ;
while(( size = bis.read(data,0,BUFFER)) != -1)
{
zos.write(data, 0, size);
}
}
//CHECK IS THERE ANY ENTRY IN ZIP ? ----START
if(isEntry)
{
zos.close();
}else
{
zos = null;
System.out.println("No Entry Found in Zip");
}
//CHECK IS THERE ANY ENTRY IN ZIP ? ----START
}catch(Exception e)
{
e.printStackTrace();
}
}else
{
System.out.println("File or Directory not found");
}
}
}
Given exportPath and queryResults as String variables, the following block creates a results.zip file under exportPath and writes the content of queryResults to a results.txt file inside the zip.
URI uri = URI.create("jar:file:" + exportPath + "/results.zip");
Map<String, String> env = Collections.singletonMap("create", "true");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
Path filePath = zipfs.getPath("/results.txt");
byte[] fileContent = queryResults.getBytes();
Files.write(filePath, fileContent, StandardOpenOption.CREATE);
}
You have mainly to create two functions. First is writeToZipFile() and second is createZipfileForOutPut .... and then call the createZipfileForOutPut('file name of .zip')` …
public static void writeToZipFile(String path, ZipOutputStream zipStream)
throws FileNotFoundException, IOException {
System.out.println("Writing file : '" + path + "' to zip file");
File aFile = new File(path);
FileInputStream fis = new FileInputStream(aFile);
ZipEntry zipEntry = new ZipEntry(path);
zipStream.putNextEntry(zipEntry);
byte[] bytes = new byte[1024];
int length;
while ((length = fis.read(bytes)) >= 0) {
zipStream.write(bytes, 0, length);
}
zipStream.closeEntry();
fis.close();
}
public static void createZipfileForOutPut(String filename) {
String home = System.getProperty("user.home");
// File directory = new File(home + "/Documents/" + "AutomationReport");
File directory = new File("AutomationReport");
if (!directory.exists()) {
directory.mkdir();
}
try {
FileOutputStream fos = new FileOutputStream("Path to your destination" + filename + ".zip");
ZipOutputStream zos = new ZipOutputStream(fos);
writeToZipFile("Path to file which you want to compress / zip", zos);
zos.close();
fos.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
There is another option by using zip4j at https://github.com/srikanth-lingala/zip4j
Creating a zip file with single file in it / Adding single file to an existing zip
new ZipFile("filename.zip").addFile("filename.ext");
Or
new ZipFile("filename.zip").addFile(new File("filename.ext"));
Creating a zip file with multiple files / Adding multiple files to an existing zip
new ZipFile("filename.zip").addFiles(Arrays.asList(new File("first_file"), new File("second_file")));
Creating a zip file by adding a folder to it / Adding a folder to an existing zip
new ZipFile("filename.zip").addFolder(new File("/user/myuser/folder_to_add"));
Creating a zip file from stream / Adding a stream to an existing zip
new ZipFile("filename.zip").addStream(inputStream, new ZipParameters());
I know this question is answered but if you have a list of strings and you want to create a separate file for each string in the archive, you can use the snippet below.
public void zipFileTest() throws IOException {
Map<String, String> map = Map.ofEntries(
new AbstractMap.SimpleEntry<String, String>("File1.txt", "File1 Content"),
new AbstractMap.SimpleEntry<String, String>("File2.txt", "File2 Content"),
new AbstractMap.SimpleEntry<String, String>("File3.txt", "File3 Content")
);
createZipFileFromStringContents(map, "archive.zip");
}
public void createZipFileFromStringContents(Map<String, String> map, String zipfilePath) throws IOException {
FileOutputStream fout = new FileOutputStream(zipfilePath);
ZipOutputStream zout = new ZipOutputStream(fout);
for (Map.Entry<String, String> entry : map.entrySet()) {
String fileName = entry.getKey();
ZipEntry zipFile = new ZipEntry(fileName);
zout.putNextEntry(zipFile);
String fileContent = entry.getValue();
zout.write(fileContent.getBytes(), 0, fileContent.getBytes().length);
zout.closeEntry();
}
zout.close();
}
It will create a zip file with the structure as in the below image:
Here is my working solution:
public static byte[] createZipFile(Map<String, FileData> files) throws IOException {
try(ByteArrayOutputStream tZipFile = new ByteArrayOutputStream()) {
try (ZipOutputStream tZipFileOut = new ZipOutputStream(tZipFile)) {
for (Map.Entry<String, FileData> file : files.entrySet()) {
ZipEntry zipEntry = new ZipEntry(file.getValue().getFileName());
tZipFileOut.putNextEntry(zipEntry);
tZipFileOut.write(file.getValue().getBytes());
}
}
return tZipFile.toByteArray();
}
}
public class FileData {
private String fileName;
private byte[] bytes;
public String getFileName() {
return this.fileName;
}
public byte[] getBytes() {
return this.bytes;
}
}
This will create byte[] of ZIP file which contains one or more compressed files. I've used this method inside controller method and write bytes[] of ZIP file into response to download ZIP file(s) from server.
Since it took me a while to figure it out, I thought it would be helpful to post my solution using Java 7+ ZipFileSystem
openZip(runFile);
addToZip(filepath); //loop construct;
zipfs.close();
private void openZip(File runFile) throws IOException {
Map<String, String> env = new HashMap<>();
env.put("create", "true");
env.put("encoding", "UTF-8");
Files.deleteIfExists(runFile.toPath());
zipfs = FileSystems.newFileSystem(URI.create("jar:" + runFile.toURI().toString()), env);
}
private void addToZip(String filename) throws IOException {
Path externalTxtFile = Paths.get(filename).toAbsolutePath();
Path pathInZipfile = zipfs.getPath(filename.substring(filename.lastIndexOf("results"))); //all files to be stored have a common base folder, results/ in my case
if (Files.isDirectory(externalTxtFile)) {
Files.createDirectories(pathInZipfile);
try (DirectoryStream<Path> ds = Files.newDirectoryStream(externalTxtFile)) {
for (Path child : ds) {
addToZip(child.normalize().toString()); //recursive call
}
}
} else {
// copy file to zip file
Files.copy(externalTxtFile, pathInZipfile, StandardCopyOption.REPLACE_EXISTING);
}
}
public static void zipFromTxt(String zipFilePath, String txtFilePath) {
Assert.notNull(zipFilePath, "Zip file path is required");
Assert.notNull(txtFilePath, "Txt file path is required");
zipFromTxt(new File(zipFilePath), new File(txtFilePath));
}
public static void zipFromTxt(File zipFile, File txtFile) {
ZipOutputStream out = null;
FileInputStream in = null;
try {
Assert.notNull(zipFile, "Zip file is required");
Assert.notNull(txtFile, "Txt file is required");
out = new ZipOutputStream(new FileOutputStream(zipFile));
in = new FileInputStream(txtFile);
out.putNextEntry(new ZipEntry(txtFile.getName()));
int len;
byte[] buffer = new byte[1024];
while ((len = in.read(buffer)) > 0) {
out.write(buffer, 0, len);
out.flush();
}
} catch (Exception e) {
log.info("Zip from txt occur error,Detail message:{}", e.toString());
} finally {
try {
if (in != null) in.close();
if (out != null) {
out.closeEntry();
out.close();
}
} catch (Exception e) {
log.info("Zip from txt close error,Detail message:{}", e.toString());
}
}
}
Using Jeka https://jeka.dev JkPathTree, it's quite straightforward.
Path wholeDirToZip = Paths.get("dir/to/zip");
Path zipFile = Paths.get("file.zip");
JkPathTree.of(wholeDirToZip).zipTo(zipFile);
If you want decompress without software better use this code. Other code with pdf files sends error on manually decompress
byte[] buffer = new byte[1024];
try {
FileOutputStream fos = new FileOutputStream("123.zip");
ZipOutputStream zos = new ZipOutputStream(fos);
ZipEntry ze= new ZipEntry("file.pdf");
zos.putNextEntry(ze);
FileInputStream in = new FileInputStream("file.pdf");
int len;
while ((len = in.read(buffer)) > 0) {
zos.write(buffer, 0, len);
}
in.close();
zos.closeEntry();
zos.close();
} catch(IOException ex) {
ex.printStackTrace();
}