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2014/10/16 ¤W¤È 08:28:11
android Programming Glossary: zipinputstreamOpen a file from archive without temporary extraction http://stackoverflow.com/questions/12863731/open-a-file-from-archive-without-temporary-extraction extraction. Is there a working solution for passing ZipInputStream to an Activity to open it as a file Is there another workaround.. you could create a ContentProvider that serves up the ZipInputStream . In this sample project I demonstrate how to create a ContentProvider.. via AssetManager . In your case you would use your ZipInputStream . Note that my sample project assumes it is being run on a device..
Unzip file from zip archive of multiple files using ZipFile class http://stackoverflow.com/questions/2974798/unzip-file-from-zip-archive-of-multiple-files-using-zipfile-class improve this question You can use the AssetManager and ZipInputStream http developer.android.com reference android content res AssetManager.html.. reference android content res AssetManager.html ZipInputStream in null try final String zipPath data sample.zip Context.getAssets.. String zipPath data sample.zip Context.getAssets in new ZipInputStream getAssets .open zipPath for ZipEntry entry in.getNextEntry entry..
How to unzip files programmatically in Android? http://stackoverflow.com/questions/3382996/how-to-unzip-files-programmatically-in-android unpackZip String path String zipname InputStream is ZipInputStream zis try String filename is new FileInputStream path zipname.. filename is new FileInputStream path zipname zis new ZipInputStream new BufferedInputStream is ZipEntry ze byte buffer new byte..
How to speed up unzipping time in Java / Android? http://stackoverflow.com/questions/4504291/how-to-speed-up-unzipping-time-in-java-android unzip try FileInputStream fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin File rootfolder new File directory.. fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin File rootfolder new File directory rootfolder.mkdirs ZipEntry..
FileOutputStream throws FileNotFoundException when UnZipping http://stackoverflow.com/questions/7353871/fileoutputstream-throws-filenotfoundexception-when-unzipping FileName FileInputStream fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry.. fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry null if ze.isDirectory..
How to zip and unzip the files? http://stackoverflow.com/questions/7485114/how-to-zip-and-unzip-the-files try File f new File location if f.isDirectory f.mkdirs ZipInputStream zin new ZipInputStream new FileInputStream zipFile try ZipEntry.. if f.isDirectory f.mkdirs ZipInputStream zin new ZipInputStream new FileInputStream zipFile try ZipEntry ze null while ze zin.getNextEntry..
Unzip a zipped file on sd card in Android application http://stackoverflow.com/questions/7697466/unzip-a-zipped-file-on-sd-card-in-android-application import java.util.zip.ZipEntry import java.util.zip.ZipInputStream @author jon public class Decompress private String _zipFile.. unzip try FileInputStream fin new FileInputStream _zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry.. fin new FileInputStream _zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry null Log.v Decompress..
Download and Extract Zip File in Android http://stackoverflow.com/questions/9324103/download-and-extract-zip-file-in-android YOU HAVE SET STREAM CONTENT LENGTH OF THE SAME.. ZipInputStream zipInputStream new ZipInputStream urlConnection.getInputStream.. OF THE SAME.. ZipInputStream zipInputStream new ZipInputStream urlConnection.getInputStream Iterate over all the files and..
Open a file from archive without temporary extraction http://stackoverflow.com/questions/12863731/open-a-file-from-archive-without-temporary-extraction preview files if there's not enough space to make a temporary extraction. Is there a working solution for passing ZipInputStream to an Activity to open it as a file Is there another workaround for this problem Thanks in advance. java android file zip.. android file zip share improve this question In principle you could create a ContentProvider that serves up the ZipInputStream . In this sample project I demonstrate how to create a ContentProvider supporting openFile that uses a pipe created by ParcelFileDescriptor.createPipe.. In my case I am sending an asset on which I get an InputStream via AssetManager . In your case you would use your ZipInputStream . Note that my sample project assumes it is being run on a device that has a PDF viewer since it is serving a PDF out of..
Unzip file from zip archive of multiple files using ZipFile class http://stackoverflow.com/questions/2974798/unzip-file-from-zip-archive-of-multiple-files-using-zipfile-class to the ZipFile constructor android unzip zipfile share improve this question You can use the AssetManager and ZipInputStream http developer.android.com reference android content res AssetManager.html ZipInputStream in null try final String zipPath.. can use the AssetManager and ZipInputStream http developer.android.com reference android content res AssetManager.html ZipInputStream in null try final String zipPath data sample.zip Context.getAssets in new ZipInputStream getAssets .open zipPath for ZipEntry.. content res AssetManager.html ZipInputStream in null try final String zipPath data sample.zip Context.getAssets in new ZipInputStream getAssets .open zipPath for ZipEntry entry in.getNextEntry entry null entry in.getNextEntry handle the zip entry catch..
How to unzip files programmatically in Android? http://stackoverflow.com/questions/3382996/how-to-unzip-files-programmatically-in-android bit. The increase in performance is perceptible. private boolean unpackZip String path String zipname InputStream is ZipInputStream zis try String filename is new FileInputStream path zipname zis new ZipInputStream new BufferedInputStream is ZipEntry.. String zipname InputStream is ZipInputStream zis try String filename is new FileInputStream path zipname zis new ZipInputStream new BufferedInputStream is ZipEntry ze byte buffer new byte 1024 int count while ze zis.getNextEntry null zapis do souboru..
How to speed up unzipping time in Java / Android? http://stackoverflow.com/questions/4504291/how-to-speed-up-unzipping-time-in-java-android this problem My unzip method looks like this public void unzip try FileInputStream fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin File rootfolder new File directory rootfolder.mkdirs ZipEntry ze null while ze zin.getNextEntry.. method looks like this public void unzip try FileInputStream fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin File rootfolder new File directory rootfolder.mkdirs ZipEntry ze null while ze zin.getNextEntry null if ze.isDirectory..
FileOutputStream throws FileNotFoundException when UnZipping http://stackoverflow.com/questions/7353871/fileoutputstream-throws-filenotfoundexception-when-unzipping doInBackground String... params try String zipFile Path FileName FileInputStream fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry null if ze.isDirectory dirChecker ze.getName else.. params try String zipFile Path FileName FileInputStream fin new FileInputStream zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry null if ze.isDirectory dirChecker ze.getName else FileOutputStream fout..
How to zip and unzip the files? http://stackoverflow.com/questions/7485114/how-to-zip-and-unzip-the-files void unzip String zipFile String location throws IOException try File f new File location if f.isDirectory f.mkdirs ZipInputStream zin new ZipInputStream new FileInputStream zipFile try ZipEntry ze null while ze zin.getNextEntry null String path location.. String location throws IOException try File f new File location if f.isDirectory f.mkdirs ZipInputStream zin new ZipInputStream new FileInputStream zipFile try ZipEntry ze null while ze zin.getNextEntry null String path location ze.getName if ze.isDirectory..
Unzip a zipped file on sd card in Android application http://stackoverflow.com/questions/7697466/unzip-a-zipped-file-on-sd-card-in-android-application import java.io.FileInputStream import java.io.FileOutputStream import java.util.zip.ZipEntry import java.util.zip.ZipInputStream @author jon public class Decompress private String _zipFile private String _location public Decompress String zipFile String.. _zipFile zipFile _location location _dirChecker public void unzip try FileInputStream fin new FileInputStream _zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry null Log.v Decompress Unzipping ze.getName if ze.isDirectory.. location _dirChecker public void unzip try FileInputStream fin new FileInputStream _zipFile ZipInputStream zin new ZipInputStream fin ZipEntry ze null while ze zin.getNextEntry null Log.v Decompress Unzipping ze.getName if ze.isDirectory _dirChecker..
Download and Extract Zip File in Android http://stackoverflow.com/questions/9324103/download-and-extract-zip-file-in-android IN THAT CASE JUST PASS THAT INPUTSTEAM OVER HERE...MAKE SURE YOU HAVE SET STREAM CONTENT LENGTH OF THE SAME.. ZipInputStream zipInputStream new ZipInputStream urlConnection.getInputStream Iterate over all the files and folders for zipEntry.. OVER HERE...MAKE SURE YOU HAVE SET STREAM CONTENT LENGTH OF THE SAME.. ZipInputStream zipInputStream new ZipInputStream urlConnection.getInputStream Iterate over all the files and folders for zipEntry zipInputStream.getNextEntry zipEntry..
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