Java.util.zip adding a new file overwrites entire jar?(转)
ZIP and TAR fomats (and the old AR format) allow file append without a full rewrite. However:
- The Java archive classes DO NOT support this mode of operation.
- File append is likely to result in multiple copies of a file in the archive if you append an existing file.
- The ZIP and AR formats have a directory that needs to be rewritten following a file append operation. The standard utilities take precautions when rewriting the directory, but it is possible in theory that you may end up with an archive with a missing or corrupted directory if the append fails.
The ZIP file format was designed to allow appends without a total re-write and is ubiquitous, even on Unix.
http://stackoverflow.com/questions/2993847/append-files-to-an-archive-without-reading-rewriting-the-whole-archive/2993964#2993964
I am using java.util.zip to add some configuration resources into a jar file. when I call addFileToZip() method it overwrites the jar completely, instead of adding the file to the jar. Why I need to write the config to the jar is completely irrelevant. and I do not wish to use any external API's.
EDIT: The jar is not running in the VM and org.cfg.resource is the package I'm trying to save the file to, the file is a standard text document and the jar being edited contains the proper information before this method is used.
My code:
public void addFileToZip(File fileToAdd, File zipFile)
{
ZipOutputStream zos = null;
FileInputStream fis = null;
ZipEntry ze = null;
byte[] buffer = null;
int len; try {
zos = new ZipOutputStream(new FileOutputStream(zipFile));
} catch (FileNotFoundException e) {
} ze = new ZipEntry("org" + File.separator + "cfg" +
File.separator + "resource" + File.separator + fileToAdd.getName());
try {
zos.putNextEntry(ze); fis = new FileInputStream(fileToAdd);
buffer = new byte[(int) fileToAdd.length()]; while((len = fis.read(buffer)) > 0)
{
zos.write(buffer, 0, len);
}
} catch (IOException e) {
}
try {
zos.flush();
zos.close();
fis.close();
} catch (IOException e) {
}
}
The code you showed overrides a file no matter if it would be a zip file or not. ZipOutputStream does not care about existing data. Neither any stream oriented API does.
I would recommend
Create new file using
ZipOutputStream.Open existing with
ZipInputStreamCopy existing entries to new file.
Add new entries.
Replace old file with a new one.
Hopefully in Java 7 we got Zip File System that will save you a lot of work.
We can directly write to files inside zip files
Map<String, String> env = new HashMap<>();
env.put("create", "true");
Path path = Paths.get("test.zip");
URI uri = URI.create("jar:" + path.toUri());
try (FileSystem fs = FileSystems.newFileSystem(uri, env))
{
Path nf = fs.getPath("new.txt");
try (Writer writer = Files.newBufferedWriter(nf, StandardCharsets.UTF_8, StandardOpenOption.CREATE)) {
writer.write("hello");
}
}
http://stackoverflow.com/questions/17500856/java-util-zip-adding-a-new-file-overwrites-entire-jar?lq=1
As others mentioned, it's not possible to append content to an existing zip (or war). However, it's possible to create a new zip on the fly without temporarily writing extracted content to disk. It's hard to guess how much faster this will be, but it's the fastest you can get (at least as far as I know) with standard Java. As mentioned by Carlos Tasada, SevenZipJBindings might squeeze out you some extra seconds, but porting this approach to SevenZipJBindings will still be faster than using temporary files with the same library.
Here's some code that writes the contents of an existing zip (war.zip) and appends an extra file (answer.txt) to a new zip (append.zip). All it takes is Java 5 or later, no extra libraries needed.
public static void addFilesToExistingZip(File zipFile,
File[] files) 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[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 notInFiles = true;
for (File f : files) {
if (f.getName().equals(name)) {
notInFiles = false;
break;
}
}
if (notInFiles) {
// 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
for (int i = 0; i < files.length; i++) {
InputStream in = new FileInputStream(files[i]);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(files[i].getName()));
// 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();
}
public static void addFilesToZip(File source, File[] files)
{
try
{ File tmpZip = File.createTempFile(source.getName(), null);
tmpZip.delete();
if(!source.renameTo(tmpZip))
{
throw new Exception("Could not make temp file (" + source.getName() + ")");
}
byte[] buffer = new byte[1024];
ZipInputStream zin = new ZipInputStream(new FileInputStream(tmpZip));
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(source)); for(int i = 0; i < files.length; i++)
{
InputStream in = new FileInputStream(files[i]);
out.putNextEntry(new ZipEntry(files[i].getName()));
for(int read = in.read(buffer); read > -1; read = in.read(buffer))
{
out.write(buffer, 0, read);
}
out.closeEntry();
in.close();
} for(ZipEntry ze = zin.getNextEntry(); ze != null; ze = zin.getNextEntry())
{
out.putNextEntry(ze);
for(int read = zin.read(buffer); read > -1; read = zin.read(buffer))
{
out.write(buffer, 0, read);
}
out.closeEntry();
} out.close();
tmpZip.delete();
}
catch(Exception e)
{
e.printStackTrace();
}
}
you cannot simply "append" data to a war file or zip file, but it is not because there is an "end of file" indication, strictly speaking, in a war file. It is because the war (zip) format includes a directory, which is normally present at the end of the file, that contains metadata for the various entries in the war file. Naively appending to a war file results in no update to the directory, and so you just have a war file with junk appended to it.
http://stackoverflow.com/questions/2223434/appending-files-to-a-zip-file-with-java
How can I add entries to an existing zip file in Java
You could use zipFile.entries() to get an enumeration of all of the ZipEntry objects in the existing file,
loop through them and add them all to the ZipOutputStream, and then add your new entries in addition.
The function renames the existing zip file to a temporary file and then adds all entries in the existing zip along with the new files, excluding the zip entries that have the same name as one of the new files.
public static void addFilesToExistingZip(File zipFile,
File[] files) 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[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 notInFiles = true;
for (File f : files) {
if (f.getName().equals(name)) {
notInFiles = false;
break;
}
}
if (notInFiles) {
// 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
for (int i = 0; i < files.length; i++) {
InputStream in = new FileInputStream(files[i]);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(files[i].getName()));
// 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();
}
http://stackoverflow.com/questions/3048669/how-can-i-add-entries-to-an-existing-zip-file-in-java?lq=1
Java.util.zip adding a new file overwrites entire jar?(转)的更多相关文章
- java.util.zip.ZipException:ZIP file must have at least one entry
1.错误描述 java.util.zip.ZipException:ZIP file must have at least one entry 2.错误原因 由于在导出文件时,要将导出的文件压缩到压缩 ...
- Tomcat启动报错:org.apache.catalina.LifecycleException: Failed to start component...java.util.zip.ZipException: error in opening zip file
1.项目环境 IntelliJ IDEA2018.1.6 apache-tomcat-8.0.53 基于springboot开发的项目 maven3.5.3 2.出现问题 从svn同步下项目 启动to ...
- [java ] java.util.zip.ZipException: error in opening zip file
严重: Failed to processes JAR found at URL [jar:file:/D:/tools/apache-tomcat-7.0.64_2/webapps/bbs/WEB- ...
- 【POI】解析xls报错:java.util.zip.ZipException: error in opening zip file
今天使用POI解析XLS,报错如下: Servlet.service() for servlet [rest] in context with path [/cetBrand] threw excep ...
- Caused by: java.util.zip.ZipException: zip file is empty
1.问题描述:mybranch分支代码和master分支的代码一模一样,mybranch代码部署到服务器上没有任何问题,而master代码部署到服务器上运行不起来. 2.解决办法: (1)登陆服务器启 ...
- java.util.zip - Recreating directory structure(转)
include my own version for your reference. We use this one to zip up photos to download so it works ...
- [Java 基础] 使用java.util.zip包压缩和解压缩文件
reference : http://www.open-open.com/lib/view/open1381641653833.html Java API中的import java.util.zip ...
- java.util.zip压缩打包文件总结二: ZIP解压技术
一.简述 解压技术和压缩技术正好相反,解压技术要用到的类:由ZipInputStream通过read方法对数据解压,同时需要通过CheckedInputStream设置冗余校验码,如: Checked ...
- java.util.zip压缩打包文件总结一:压缩文件及文件下面的文件夹
一.简述 zip用于压缩和解压文件.使用到的类有:ZipEntry ZipOutputStream 二.具体实现代码 package com.joyplus.test; import java.io ...
随机推荐
- CentOS 如何安装git server + Gitolite 【配置不成功需要再测试2015-8-20】
安装git 关于安装git 可以参考 http://gitolite.com/gitolite/install.html 里面有官方的介绍 1. Git 的工作需要调用 curl,zlib,open ...
- java操作Excel处理数字类型的精度损失问题验证
java操作Excel处理数字类型的精度损失问题验证: 场景: CELL_TYPE_NUMERIC-->CELL_TYPE_STRING--->CELL_TYPE_NUMERIC POI版 ...
- leetcode 编辑距离
class Solution { public: int minDistance(string word1, string word2) { // Start typing your C/C++ so ...
- PHP - 验证用户名
/** * * 函数名:_check_username($user_str,$min_num,$max_num); * 作用:检测用户名是否符合格式 * 参数: * 1:用户名 * 2:不得小于多少位 ...
- PHP - 设置地址栏小图标
效果: 1/把icon图标直接放到根目录. 2/在header标签中写下: <link rel="icon" type="image/x-icon" hr ...
- 2783: [JLOI2012]树( dfs + BST )
直接DFS, 然后用set维护一下就好了.... O(nlogn) ------------------------------------------------------------------ ...
- 基于visual Studio2013解决C语言竞赛题之1023判断排序
题目 解决代码及点评 /* 23. 有10个两位整数,把这些数作以下变化,如果它是素数, 则把它乘以2,若它是偶数则除以2,其余的数减1, 请将变化后的10个数按从小到大 ...
- 编写自定义的JDBC框架与策略模式
本篇根据上一篇利用数据库的几种元数据来仿造Apache公司的开源DbUtils工具类集合来编写自己的JDBC框架.也就是说在本篇中很大程度上的代码都和DbUtils中相似,学完本篇后即更容易了解DbU ...
- OpenSSL---堆栈
堆栈是一种先进后出的数据结构.是一种只允许在其一端进行插入或者删除的线性表.允许插入或删除操作的一端为栈顶,另一端称为栈底.对堆栈的插入和删除操作称为入栈和出栈. 1.1 概述 OpenSSL ...
- Microsoft Visual C++运行库合集下载(静默安装)
Microsoft Visual C++运行库合集下载 CN启示录2013-06-02上传 Microsoft Visual C++运行库合集由国外网友McRip制作,包含了VC2005.VC20 ...