zipoutputstream

问题 In my application there are no of documents(pdf) for a particular tender. I need to create a zip file from those pdf files and allow user to download it. Application is done in JavaEE with struts and mysql. when user clicks the download button this action class gets called. The code does not give any exceptions but it does not prompt user to download anything either. Please help me find what is wrong in the code. Following is the source code of my action class.. public class ActDownloadDocZip

问题 In my application there are no of documents(pdf) for a particular tender. I need to create a zip file from those pdf files and allow user to download it. Application is done in JavaEE with struts and mysql. when user clicks the download button this action class gets called. The code does not give any exceptions but it does not prompt user to download anything either. Please help me find what is wrong in the code. Following is the source code of my action class.. public class ActDownloadDocZip

问题 I keep getting a very annoying OutOfMemory exception on the following code. I'm zipping a lot of small files (PDF, each being 1.5mb approx). At first I was getting the exception afer 25 files zipped, which doesn't seem like a massing archive. Setting up the size of the ZipEntry somehow helped since now I manage to get up to 110 files zipped (I'm debugging under visual studio) Here's my code, maybe there's something wrong with it. Any help would be greatly appreciated. Thanks public static

问题 I am using the following method to filter out META-INF folder and its contents: public static void copyWithoutMetaInf(final String originalZip, final String newZip) throws IOException { final ZipInputStream zip = new ZipInputStream(new FileInputStream(originalZip)); final ZipOutputStream zop = new ZipOutputStream(new FileOutputStream(newZip)); ZipEntry entry; while((entry = zip.getNextEntry()) != null) { if(!entry.getName().contains("META-INF")) { zop.putNextEntry(entry); } } zip.close(); zop

问题 My app is a tender document system where each tender number has one or more pdf files attached. application is done in java ee using struts and mysql. in a database table the paths of each related pdf file for a tender number is stores. I want to get all the pdf files and create a single ZIP file for each tender number so that user can download that zip file and have all the related documents in a single click. I tried Google and found something called ZipOutputStream but i cannot understand

问题 Following is part of the some code. I need to close out resources in finally clause. Do I need to call closeEntry() first or close()? I am getting some error messages. Error closing the zipoutjava.io.IOException: Stream closed at java.util.zip.ZipOutputStream.ensureOpen(ZipOutputStream.java:70) at java.util.zip.ZipOutputStream.closeEntry(ZipOutputStream.java:189) The code ZipOutputStream zos = null; try{ ZipEntry entry = new ZipEntry("file.csv") zipout.putNextEntry(entry); csvBeanWriter = new

问题 I am doing an Android project. I need to zip a complete folder structure of SD card. code is able to zip the folder if any files exists inside the folder, else skips that folder. My Folder structure is like: mnt/sdcard/Movies/Telugu/Files. mnt/sdcard/Movies/English --> English is empty folder I did not see English folder in output zip file. My code: public void zip() { try { // create a ZipOutputStream to zip the data to ZipOutputStream zos = new ZipOutputStream(new FileOutputStream( _zipFile

问题 I want to compress an InputStream using ZipOutputStream and then get the InputStream from compressed ZipOutputStream without saving file on disc. Is that possible? 回答1: I figured it out: public InputStream getCompressed( InputStream is ) throws IOException { byte data[] = new byte[2048]; ByteArrayOutputStream bos = new ByteArrayOutputStream(); ZipOutputStream zos = new ZipOutputStream( bos ); BufferedInputStream entryStream = new BufferedInputStream( is, 2048); ZipEntry entry = new ZipEntry(

问题 Follow up of Question: Java: how to compress a byte[] using ZipOutputStream without intermediate file I can zip data without an intermediate file (or memory file). I now need to zip chunks of data and add them to a single zip file. I am using a single ZipOutputStream as suggested in the previous question. String infile = "test.txt"; FileInputStream in = new FileInputStream(infile); String outfile = "test.txt.zip"; FileOutputStream out = new FileOutputStream(outfile); ByteArrayOutputStream

问题 I have two examples : Example 1: try (ByteArrayOutputStream baous = new ByteArrayOutputStream(); FileOutputStream fouscrx = new FileOutputStream(new File(output, "example"))) { try (ZipOutputStream zous = new ZipOutputStream(baous)) { for (File file: files) { try (FileInputStream fis = new FileInputStream(file)) { ZipEntry zipEntry = new ZipEntry(file.getPath().substring(output.getPath().length() + 1)); zous.putNextEntry(zipEntry); byte[] bytes = new byte[2048]; int length; while ((length =