I have a password protected zip file [in the form of a base64 encoded data and the name of the zip file] which contains a single xml. I wish to parse that xml without writing anything to disk. What is the way to do this in Zip4j? Following is what I tried.
String docTitle = request.getDocTitle();
byte[] decodedFileData = Base64.getDecoder().decode(request.getBase64Data());
InputStream inputStream = new ByteArrayInputStream(decodedFileData);
try (ZipInputStream zipInputStream = new ZipInputStream(inputStream, password)) {
while ((localFileHeader = zipInputStream.getNextEntry()) != null) {
String fileTitle = localFileHeader.getFileName();
File extractedFile = new File(fileTitle);
try (InputStream individualFileInputStream = org.apache.commons.io.FileUtils.openInputStream(extractedFile)) {
// Call parser
parser.parse(localFileHeader.getFileName(),
individualFileInputStream));
} catch (IOException e) {
// Handle IOException
}
}
} catch (IOException e) {
// Handle IOException
}
Which is throwing me java.io.FileNotFoundException: File 'xyz.xml' does not exist at line FileUtils.openInputStream(extractedFile). Can you please suggest me the right way to do this?
ZipInputStream keeps all content of a zip file. Each call of zipInputStream.getNextEntry() delivers the content of each file and moves "pointer" to the next entry (file). You also can read the file (ZipInputStream.read) before moving to the next entry.
Your case:
byte[] decodedFileData = Base64.getDecoder().decode(request.getBase64Data());
InputStream inputStream = new ByteArrayInputStream(decodedFileData);
try (ZipInputStream zipInputStream = new ZipInputStream(inputStream, password)) {
ZipEntry zipEntry = null;
while ((zipEntry = zipInputStream.getNextEntry()) != null) {
byte[] fileContent = IOUtils.toByteArray(zipInputStream);
parser.parse(zipEntry.getName(),
new ByteArrayInputStream(fileContent)));
}
} catch (Exception e) {
// Handle Exception
}
Related
I'm trying to zip a message (String) into a zip file, and then set it as the body of the exchange object in Apache Camel, so that one of the downstream services (also using Apache Camel) is able to extract the zip file using the exchange.getIn().getBody() method.
The first part is good, I'm able to set a zip file to the body, but when I try to retrieve this on the other side of the queue (Active MQ), the exchange.getIn().getBody(ZipFile.class) returns null. In fact, the body itself is null.
Why would that be?
I have tried sending a normal String in the body, and that worked fine. The File (ZipFile) doesn't set, wonder why.
Here are snippets of the code -
Route -
from(some_route)
.bean(SomeClass.class, "zipAndSend")
.to("activemq:queue:" + somequeue)
To zip a file -
public void zipAndSend(Exchange exchange) throws Exception {
String incomingMessage;
try {
incomingMessage = exchange.getIn().getBody().toString();
File file = ZipUtil.createFile(incomingMessage);
String zipFilePath = file.getPath().replace(".xml", ".zip");
ZipFile zipFile = ZipUtil.zipFile(file.getPath(), zipFilePath);
exchange.getOut().setHeader("Compressed", "Y");
exchange.getOut().setHeader("ZipFilePath", zipFilePath);
exchange.getOut().setBody(zipFile);
//the body is set correctly here, so far so good
} catch (Exception e) {
e.printStackTrace(); //other operations
}
}
public static File createFile(String incomingMessaage) {
String fileName = "C:\\Project\\ZipUnzipTest\\incoming.xml";
File file = new File(fileName);
try {
FileUtils.writeStringToFile(file, incomingMessaage);
} catch (Exception e) {
//log.error("Error in Writing Message into file " + fileName, e);
String errorFile = fileName.replace("work", "error");
}
return file;
}
Over in the other service (end of the queue), I am overriding the process() method like below, to be able to extract the message (String) back from the file inside the zipped file.
public void process(Exchange exchange) throws WorkflowDBException,
Exception {
try {
ZipFile zipFile = exchange.getIn().getBody(ZipFile.class); //NPE as body is null
String zipFilePath = exchange.getIn().getHeader("ZipFilePath").toString();
File inFile = ZipUtil.unzipFile(zipFile, "C:\\Project\\ZipUnzipTest\\Output\\WF", true);
String incomingMessage;
incomingMessage = FileUtils.readFileToString(inFile, "UTF-8");
} catch (Exception e) {e.printStackTrace();}
}
Dependencies -
<dependency>
<groupId>net.lingala.zip4j</groupId>
<artifactId>zip4j</artifactId>
<version>1.3.2</version>
</dependency>
<dependency>
<groupId>commons-io</groupId>
<artifactId>commons-io</artifactId>
<version>1.3.2</version>
</dependency>
I expect the content of the body to be same in the in and out space. Alas, it isn't.
It turns out, Camel (amongst other frameworks) handles bytes[] rather well. I converted the ZipFile into a byte array, and used it to set the body of the exchange object.
incomingMessage = exchange.getIn().getBody().toString();
File file = ZipUtil.createFile(incomingMessage);
String zipFilePath = file.getPath().replace(".xml", ".zip");
ZipFile zipFile = ZipUtil.zipFile(file.getPath(), zipFilePath);
messageData = FileUtils.readFileToByteArray(new File(zipFilePath));
exchange.getOut().setHeader("Compressed", "Y");
exchange.getOut().setHeader("ZipFilePath", zipFilePath);
exchange.getOut().setBody(messageData);
And while reading it, I used the ZipInputStream to get the Zip Entry from the ByteArrayInputStream.
byte [] bytes = exchange.getIn().getBody(byte[].class);
ZipInputStream zipStream = new ZipInputStream(new ByteArrayInputStream(bytes));
try {
StringBuilder s = new StringBuilder();
StringBuilder temp = new StringBuilder();
byte[] buffer = new byte[1024];
int read = 0;
ZipEntry entry;
while ((entry = zipStream.getNextEntry())!= null) {
while ((read = zipStream.read(buffer, 0, 1024)) >= 0) {
s.append(new String(buffer, 0, read));
}
temp = temp.append(s);
s.setLength(0);
}
return temp.toString();
} catch (Exception e) {
e.printStackTrace();
}
Yes, that's it.
Still open to other ways to solve this :)
I must get file content from ZIP archive (only one file, I know its name) using SFTP. The only thing I'm having is ZIP's InputStream. Most examples show how get content using this statement:
ZipFile zipFile = new ZipFile("location");
But as I said, I don't have ZIP file on my local machine and I don't want to download it. Is an InputStream enough to read?
UPD: This is how I do:
import java.util.zip.ZipInputStream;
import com.jcraft.jsch.Channel;
import com.jcraft.jsch.ChannelSftp;
import com.jcraft.jsch.JSch;
import com.jcraft.jsch.Session;
public class SFTP {
public static void main(String[] args) {
String SFTPHOST = "host";
int SFTPPORT = 3232;
String SFTPUSER = "user";
String SFTPPASS = "mypass";
String SFTPWORKINGDIR = "/dir/work";
Session session = null;
Channel channel = null;
ChannelSftp channelSftp = null;
try {
JSch jsch = new JSch();
session = jsch.getSession(SFTPUSER, SFTPHOST, SFTPPORT);
session.setPassword(SFTPPASS);
java.util.Properties config = new java.util.Properties();
config.put("StrictHostKeyChecking", "no");
session.setConfig(config);
session.connect();
channel = session.openChannel("sftp");
channel.connect();
channelSftp = (ChannelSftp) channel;
channelSftp.cd(SFTPWORKINGDIR);
ZipInputStream stream = new ZipInputStream(channelSftp.get("file.zip"));
ZipEntry entry = zipStream.getNextEntry();
System.out.println(entry.getName); //Yes, I got its name, now I need to get content
} catch (Exception ex) {
ex.printStackTrace();
} finally {
session.disconnect();
channelSftp.disconnect();
channel.disconnect();
}
}
}
Below is a simple example on how to extract a ZIP File, you will need to check if the file is a directory. But this is the simplest.
The step you are missing is reading the input stream and writing the contents to a buffer which is written to an output stream.
// Expands the zip file passed as argument 1, into the
// directory provided in argument 2
public static void main(String args[]) throws Exception
{
if(args.length != 2)
{
System.err.println("zipreader zipfile outputdir");
return;
}
// create a buffer to improve copy performance later.
byte[] buffer = new byte[2048];
// open the zip file stream
InputStream theFile = new FileInputStream(args[0]);
ZipInputStream stream = new ZipInputStream(theFile);
String outdir = args[1];
try
{
// now iterate through each item in the stream. The get next
// entry call will return a ZipEntry for each file in the
// stream
ZipEntry entry;
while((entry = stream.getNextEntry())!=null)
{
String s = String.format("Entry: %s len %d added %TD",
entry.getName(), entry.getSize(),
new Date(entry.getTime()));
System.out.println(s);
// Once we get the entry from the stream, the stream is
// positioned read to read the raw data, and we keep
// reading until read returns 0 or less.
String outpath = outdir + "/" + entry.getName();
FileOutputStream output = null;
try
{
output = new FileOutputStream(outpath);
int len = 0;
while ((len = stream.read(buffer)) > 0)
{
output.write(buffer, 0, len);
}
}
finally
{
// we must always close the output file
if(output!=null) output.close();
}
}
}
finally
{
// we must always close the zip file.
stream.close();
}
}
Code excerpt came from the following site:
http://www.thecoderscorner.com/team-blog/java-and-jvm/12-reading-a-zip-file-from-java-using-zipinputstream#.U4RAxYamixR
Well, I've done this:
zipStream = new ZipInputStream(channelSftp.get("Port_Increment_201405261400_2251.zip"));
zipStream.getNextEntry();
sc = new Scanner(zipStream);
while (sc.hasNextLine()) {
System.out.println(sc.nextLine());
}
It helps me to read ZIP's content without writing to another file.
The ZipInputStream is an InputStream by itself and delivers the contents of each entry after each call to getNextEntry(). Special care must be taken, not to close the stream from which the contents is read, since it is the same as the ZIP stream:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(zipIn);
zipIn.closeEntry();
}
}
private void readContents(InputStream contentsIn) throws IOException {
byte contents[] = new byte[4096];
int direct;
while ((direct = contentsIn.read(contents, 0, contents.length)) >= 0) {
System.out.println("Read " + direct + "bytes content.");
}
}
When delegating reading contents to other logic, it can be necessary to wrap the ZipInputStream with a FilterInputStream to close only the entry instead of the whole stream as in:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(new FilterInputStream(zipIn) {
#Override
public void close() throws IOException {
zipIn.closeEntry();
}
});
}
}
OP was close. Just need to read the bytes. The call to getNextEntry positions the stream at the beginning of the entry data (docs). If that's the entry we want (or the only entry), then the InputStream is in the right spot. All we need to do is read that entry's decompressed bytes.
byte[] bytes = new byte[(int) entry.getSize()];
int i = 0;
while (i < bytes.length) {
// .read doesn't always fill the buffer we give it.
// Keep calling it until we get all the bytes for this entry.
i += zipStream.read(bytes, i, bytes.length - i);
}
So if these bytes really are text, then we can decode those bytes to a String. I'm just assuming utf8 encoding.
new String(bytes, "utf8")
Side note: I personally use apache commons-io IOUtils to cut down on this kind of lower level stuff. The docs for ZipInputStream.read seem to imply that read will stop at the end of the current zip entry. If that is true, then reading the current textual entry is one line with IOUtils.
String text = IOUtils.toString(zipStream)
Unzip archive (zip) with preserving file structure into given directory.
Note; this code use deps on "org.apache.commons.io.IOUtils"), but you can replace it by yours custom 'read-stream' code
public static void unzipDirectory(File archiveFile, File destinationDir) throws IOException
{
Path destPath = destinationDir.toPath();
try (ZipInputStream zis = new ZipInputStream(new FileInputStream(archiveFile)))
{
ZipEntry zipEntry;
while ((zipEntry = zis.getNextEntry()) != null)
{
Path resolvedPath = destPath.resolve(zipEntry.getName()).normalize();
if (!resolvedPath.startsWith(destPath))
{
throw new IOException("The requested zip-entry '" + zipEntry.getName() + "' does not belong to the requested destination");
}
if (zipEntry.isDirectory())
{
Files.createDirectories(resolvedPath);
} else
{
if(!Files.isDirectory(resolvedPath.getParent()))
{
Files.createDirectories(resolvedPath.getParent());
}
try (FileOutputStream outStream = new FileOutputStream(resolvedPath.toFile()))
{
IOUtils.copy(zis, outStream);
}
}
}
}
}
Here a more generic solution to process a zip inputstream with a BiConsumer. It's nearly the same solution that was used by haui
private void readZip(InputStream is, BiConsumer<ZipEntry,InputStream> consumer) throws IOException {
try (ZipInputStream zipFile = new ZipInputStream(is);) {
ZipEntry entry;
while((entry = zipFile.getNextEntry()) != null){
consumer.accept(entry, new FilterInputStream(zipFile) {
#Override
public void close() throws IOException {
zipFile.closeEntry();
}
});
}
}
}
You can use it by just calling
readZip(<some inputstream>, (entry, is) -> {
/* don't forget to close this stream after processing. */
is.read() // ... <- to read each entry
});
If content of your ZIP consist of 1 file (for example, zipped content of HTTP response), you can read text content using Kotlin as follows:
#Throws(IOException::class)
fun InputStream.readZippedContent() = ZipInputStream(this).use { stream ->
stream.nextEntry?.let { stream.bufferedReader().readText() } ?: String()
}
This extension function unzips first ZIP entry of Zip file and read content as plain text.
Usage:
val inputStream: InputStream = ... // your zipped InputStream
val textContent = inputStream.readZippedContent()
I've been trying to tackle this problem for a day or two and can't seem to figure out precisely how to add text files to a zip file, I was able to figure out how to add these text files to a 7zip file which was insanely easy, but a zip file seems to me much more complicated for some reason. I want to return a zip file for user reasons btw.
Here's what I have now:
(I know the code isn't too clean at the moment, I plan to tackle that after getting the bare functionality down).
private ZipOutputStream addThreadDumpsToZipFile(File file, List<Datapoint<ThreadDump>> allThreadDumps, List<Datapoint<String>> allThreadDumpTextFiles) {
ZipOutputStream threadDumpsZipFile = null;
try {
//creat new zip file which accepts input stream
//TODO missing step: create text files containing each thread dump then add to zip
threadDumpsZipFile = new ZipFile(new FileOutputStream(file));
FileInputStream fileInputStream = null;
try {
//add data to each thread dump entry
for(int i=0; i<allThreadDumpTextFiles.size();i++) {
//create file for each thread dump
File threadDumpFile = new File("thread_dump_"+i+".txt");
FileUtils.writeStringToFile(threadDumpFile,allThreadDumpTextFiles.get(i).toString());
//add entry/file to zip file (creates block to add input to)
ZipEntry threadDumpEntry = new ZipEntry("thread_dump_"+i); //might need to add extension here?
threadDumpsZipFile.putNextEntry(threadDumpEntry);
//add the content to this entry
fileInputStream = new FileInputStream(threadDumpFile);
byte[] byteBuffer = new byte[(int) threadDumpFile.length()]; //see if this sufficiently returns length of data
int bytesRead = -1;
while ((bytesRead = fileInputStream.read(byteBuffer)) != -1) {
threadDumpsZipFile.write(byteBuffer, 0, bytesRead);
}
}
threadDumpsZipFile.flush();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
fileInputStream.close();
} catch(Exception e) {
}
}
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return threadDumpsZipFile;
}
As you can sort of guess, I have a set of Thread Dumps that I want to add to my zip file and return to the user.
Let me know if you guys need any more info!
PS: There might be some bugs in this question, I just realized with some breakpoints that the threadDumpFile.length() won't really work.
Look forward to your replies!
Thanks,
Arsa
Here's a crack at it. I think you'll want to keep the file extensions when you make your ZipEntry objects. See if you can implement the below createTextFiles() function; the rest of this works -- I stubbed that method to return a single "test.txt" file with some dummy data to verify.
void zip()
{
try {
FileOutputStream fos = new FileOutputStream("yourZipFile.zip");
ZipOutputStream zos = new ZipOutputStream(fos);
File[] textFiles = createTextFiles(); // should be an easy step
for (int i = 0; i < files.length; i++) {
addToZipFile(file[i].getName(), zos);
}
zos.close();
fos.close();
} catch (Exception e) {
e.printStackTrace();
}
}
void addToZipFile(String fileName, ZipOutputStream zos) throws Exception {
File file = new File(fileName);
FileInputStream fis = new FileInputStream(file);
ZipEntry zipEntry = new ZipEntry(fileName);
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();
}
I must get file content from ZIP archive (only one file, I know its name) using SFTP. The only thing I'm having is ZIP's InputStream. Most examples show how get content using this statement:
ZipFile zipFile = new ZipFile("location");
But as I said, I don't have ZIP file on my local machine and I don't want to download it. Is an InputStream enough to read?
UPD: This is how I do:
import java.util.zip.ZipInputStream;
import com.jcraft.jsch.Channel;
import com.jcraft.jsch.ChannelSftp;
import com.jcraft.jsch.JSch;
import com.jcraft.jsch.Session;
public class SFTP {
public static void main(String[] args) {
String SFTPHOST = "host";
int SFTPPORT = 3232;
String SFTPUSER = "user";
String SFTPPASS = "mypass";
String SFTPWORKINGDIR = "/dir/work";
Session session = null;
Channel channel = null;
ChannelSftp channelSftp = null;
try {
JSch jsch = new JSch();
session = jsch.getSession(SFTPUSER, SFTPHOST, SFTPPORT);
session.setPassword(SFTPPASS);
java.util.Properties config = new java.util.Properties();
config.put("StrictHostKeyChecking", "no");
session.setConfig(config);
session.connect();
channel = session.openChannel("sftp");
channel.connect();
channelSftp = (ChannelSftp) channel;
channelSftp.cd(SFTPWORKINGDIR);
ZipInputStream stream = new ZipInputStream(channelSftp.get("file.zip"));
ZipEntry entry = zipStream.getNextEntry();
System.out.println(entry.getName); //Yes, I got its name, now I need to get content
} catch (Exception ex) {
ex.printStackTrace();
} finally {
session.disconnect();
channelSftp.disconnect();
channel.disconnect();
}
}
}
Below is a simple example on how to extract a ZIP File, you will need to check if the file is a directory. But this is the simplest.
The step you are missing is reading the input stream and writing the contents to a buffer which is written to an output stream.
// Expands the zip file passed as argument 1, into the
// directory provided in argument 2
public static void main(String args[]) throws Exception
{
if(args.length != 2)
{
System.err.println("zipreader zipfile outputdir");
return;
}
// create a buffer to improve copy performance later.
byte[] buffer = new byte[2048];
// open the zip file stream
InputStream theFile = new FileInputStream(args[0]);
ZipInputStream stream = new ZipInputStream(theFile);
String outdir = args[1];
try
{
// now iterate through each item in the stream. The get next
// entry call will return a ZipEntry for each file in the
// stream
ZipEntry entry;
while((entry = stream.getNextEntry())!=null)
{
String s = String.format("Entry: %s len %d added %TD",
entry.getName(), entry.getSize(),
new Date(entry.getTime()));
System.out.println(s);
// Once we get the entry from the stream, the stream is
// positioned read to read the raw data, and we keep
// reading until read returns 0 or less.
String outpath = outdir + "/" + entry.getName();
FileOutputStream output = null;
try
{
output = new FileOutputStream(outpath);
int len = 0;
while ((len = stream.read(buffer)) > 0)
{
output.write(buffer, 0, len);
}
}
finally
{
// we must always close the output file
if(output!=null) output.close();
}
}
}
finally
{
// we must always close the zip file.
stream.close();
}
}
Code excerpt came from the following site:
http://www.thecoderscorner.com/team-blog/java-and-jvm/12-reading-a-zip-file-from-java-using-zipinputstream#.U4RAxYamixR
Well, I've done this:
zipStream = new ZipInputStream(channelSftp.get("Port_Increment_201405261400_2251.zip"));
zipStream.getNextEntry();
sc = new Scanner(zipStream);
while (sc.hasNextLine()) {
System.out.println(sc.nextLine());
}
It helps me to read ZIP's content without writing to another file.
The ZipInputStream is an InputStream by itself and delivers the contents of each entry after each call to getNextEntry(). Special care must be taken, not to close the stream from which the contents is read, since it is the same as the ZIP stream:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(zipIn);
zipIn.closeEntry();
}
}
private void readContents(InputStream contentsIn) throws IOException {
byte contents[] = new byte[4096];
int direct;
while ((direct = contentsIn.read(contents, 0, contents.length)) >= 0) {
System.out.println("Read " + direct + "bytes content.");
}
}
When delegating reading contents to other logic, it can be necessary to wrap the ZipInputStream with a FilterInputStream to close only the entry instead of the whole stream as in:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(new FilterInputStream(zipIn) {
#Override
public void close() throws IOException {
zipIn.closeEntry();
}
});
}
}
OP was close. Just need to read the bytes. The call to getNextEntry positions the stream at the beginning of the entry data (docs). If that's the entry we want (or the only entry), then the InputStream is in the right spot. All we need to do is read that entry's decompressed bytes.
byte[] bytes = new byte[(int) entry.getSize()];
int i = 0;
while (i < bytes.length) {
// .read doesn't always fill the buffer we give it.
// Keep calling it until we get all the bytes for this entry.
i += zipStream.read(bytes, i, bytes.length - i);
}
So if these bytes really are text, then we can decode those bytes to a String. I'm just assuming utf8 encoding.
new String(bytes, "utf8")
Side note: I personally use apache commons-io IOUtils to cut down on this kind of lower level stuff. The docs for ZipInputStream.read seem to imply that read will stop at the end of the current zip entry. If that is true, then reading the current textual entry is one line with IOUtils.
String text = IOUtils.toString(zipStream)
Unzip archive (zip) with preserving file structure into given directory.
Note; this code use deps on "org.apache.commons.io.IOUtils"), but you can replace it by yours custom 'read-stream' code
public static void unzipDirectory(File archiveFile, File destinationDir) throws IOException
{
Path destPath = destinationDir.toPath();
try (ZipInputStream zis = new ZipInputStream(new FileInputStream(archiveFile)))
{
ZipEntry zipEntry;
while ((zipEntry = zis.getNextEntry()) != null)
{
Path resolvedPath = destPath.resolve(zipEntry.getName()).normalize();
if (!resolvedPath.startsWith(destPath))
{
throw new IOException("The requested zip-entry '" + zipEntry.getName() + "' does not belong to the requested destination");
}
if (zipEntry.isDirectory())
{
Files.createDirectories(resolvedPath);
} else
{
if(!Files.isDirectory(resolvedPath.getParent()))
{
Files.createDirectories(resolvedPath.getParent());
}
try (FileOutputStream outStream = new FileOutputStream(resolvedPath.toFile()))
{
IOUtils.copy(zis, outStream);
}
}
}
}
}
Here a more generic solution to process a zip inputstream with a BiConsumer. It's nearly the same solution that was used by haui
private void readZip(InputStream is, BiConsumer<ZipEntry,InputStream> consumer) throws IOException {
try (ZipInputStream zipFile = new ZipInputStream(is);) {
ZipEntry entry;
while((entry = zipFile.getNextEntry()) != null){
consumer.accept(entry, new FilterInputStream(zipFile) {
#Override
public void close() throws IOException {
zipFile.closeEntry();
}
});
}
}
}
You can use it by just calling
readZip(<some inputstream>, (entry, is) -> {
/* don't forget to close this stream after processing. */
is.read() // ... <- to read each entry
});
If content of your ZIP consist of 1 file (for example, zipped content of HTTP response), you can read text content using Kotlin as follows:
#Throws(IOException::class)
fun InputStream.readZippedContent() = ZipInputStream(this).use { stream ->
stream.nextEntry?.let { stream.bufferedReader().readText() } ?: String()
}
This extension function unzips first ZIP entry of Zip file and read content as plain text.
Usage:
val inputStream: InputStream = ... // your zipped InputStream
val textContent = inputStream.readZippedContent()
Simple question,
I'm writing a series of text files into a zip, just wrapping a fileoutputstream in a zipoutputstream and then in a printwriter.
public static int saveData(File outfile, DataStructure input) {
//variables
ArrayList<String> out = null;
FileOutputStream fileout = null;
ZipOutputStream zipout = null;
PrintWriter printer = null;
//parameter tests
try {
fileout = new FileOutputStream(outfile);
zipout = new ZipOutputStream(fileout);
printer = new PrintWriter(zipout);
} catch (Exception e) {
e.printStackTrace();
return util.FILE_INVALID;
}
for(DataItem data : input){
//process the data into a list of strings
try {
zipout.putNextEntry(new ZipEntry( dataFileName ));
for(String s : out) {
printer.println(s);
}
zipout.closeEntry();
} catch (Exception e) {
try {
fileout.close();
} catch (Exception x) {
x.printStackTrace();
return util.CRITICAL_ERROR;
}
e.printStackTrace();
return util.CRITICAL_ERROR;
}
}
try {
fileout.close();
} catch (Exception e) {
e.printStackTrace();
return util.CRITICAL_ERROR;
}
return util.SUCCESS;
}
Previously in the app i've been developing I've just been saving to the current directory for testing and I know in the case of a file already existing that the file will be overwritten (and have been exploiting this). What I dont know is the behaviour for zips. Will it overwrite entries of the same name? Or will it simply overwrite the whole zip file (which would be convenient for my purposes.
K.Barad
As Joel said, If you try to add a duplicate ZipEntry you will get an exception. If you want to replace the current entry you need to delete it and re-insert it.
You might want to do something like here below to achieve it:
private ZipFile addFileToExistingZip(File zipFile, File versionFile) throws IOException{
// get a temp file
File tempFile = File.createTempFile(zipFile.getName(), null);
// delete it, otherwise you cannot rename your existing zip to it.
tempFile.delete();
boolean renameOk=zipFile.renameTo(tempFile);
if (!renameOk)
{
throw new RuntimeException("could not rename the file "+zipFile.getAbsolutePath()+" to "+tempFile.getAbsolutePath());
}
byte[] buf = new byte[4096 * 1024];
ZipInputStream zin = new ZipInputStream(new FileInputStream(tempFile));
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipFile));
ZipEntry entry = zin.getNextEntry();
while (entry != null) {
String name = entry.getName();
boolean toBeDeleted = false;
if (versionFile.getName().indexOf(name) != -1) {
toBeDeleted = true;
}
if(!toBeDeleted){
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(name));
// Transfer bytes from the ZIP file to the output file
int len;
while ((len = zin.read(buf)) > 0) {
out.write(buf, 0, len);
}
}
entry = zin.getNextEntry();
}
// Close the streams
zin.close();
// Compress the files
InputStream in = new FileInputStream(versionFile);
String fName = versionFile.getName();
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(fName));
// 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();
tempFile.delete();
return new ZipFile(zipFile);
}
The above code worked for me where the need was to add a new zip entry to an existing zip file. If the entry is already present inside the zip, then overwrite it.
Comments/improvements in the code are welcome!
Thanks!
If you try to add a duplicate ZipEntry you will get an exception. If you want to replace the current entry you need to delete it and re-insert it. I suspect the exception you get is much the same as this one.