java ZipInputStream getNextEntry 在提取 .zip 文件时为空
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原文地址: http://stackoverflow.com/questions/7561031/
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ZipInputStream getNextEntry is null when extracting .zip files
提问by Omar
I'm trying to extract .zip files and I'm using this code:
我正在尝试提取 .zip 文件，并且正在使用以下代码：
String zipFile = Path + FileName;
FileInputStream fin = new FileInputStream(zipFile);
ZipInputStream zin = new ZipInputStream(fin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
UnzipCounter++;
if (ze.isDirectory()) {
dirChecker(ze.getName());
} else {
FileOutputStream fout = new FileOutputStream(Path
+ ze.getName());
while ((Unziplength = zin.read(Unzipbuffer)) > 0) {
fout.write(Unzipbuffer, 0, Unziplength);
}
zin.closeEntry();
fout.close();
}
}
zin.close();
but the problem is that, while debugging, when the code reaches the while(!=null) part, the zin.getNextEntry() is always null so it doesnt extract anything..
The .zip file is 150kb.. How can I fix this?
但问题是，在调试时，当代码到达 while(!=null) 部分时， zin.getNextEntry() 始终为 null，因此它不会提取任何内容..zip
文件为 150kb.. 我该如何修复这？
The .zip exists
.zip 存在
Code I use to dl the .zip:
我用来 dl .zip 的代码：
URL=intent.getStringExtra("DownloadService_URL");
FileName=intent.getStringExtra("DownloadService_FILENAME");
Path=intent.getStringExtra("DownloadService_PATH");
File PathChecker = new File(Path);
try{
if(!PathChecker.isDirectory())
PathChecker.mkdirs();
URL url = new URL(URL);
URLConnection conexion = url.openConnection();
conexion.connect();
int lenghtOfFile = conexion.getContentLength();
lenghtOfFile/=100;
InputStream input = new BufferedInputStream(url.openStream());
OutputStream output = new FileOutputStream(Path+FileName);
byte data[] = new byte[1024];
long total = 0;
int count = 0;
while ((count = input.read(data)) != -1) {
output.write(data, 0, count);
total += count;
notification.setLatestEventInfo(context, contentTitle, "???? ????? ??? " + FileName + " " + (total/lenghtOfFile), contentIntent);
mNotificationManager.notify(1, notification);
}
output.flush();
output.close();
input.close();
采纳答案by Andrew Thompson
If the Zip is placed in the same directory as this exact source, named "91.zip", it works just fine.
如果 Zip 与这个确切的源放在同一目录中，命名为“91.zip”，它就可以正常工作。
import java.io.*;
import java.util.zip.*;
class Unzip {
public static void main(String[] args) throws Exception {
String Path = ".";
String FileName = "91.zip";
File zipFile = new File(Path, FileName);
FileInputStream fin = new FileInputStream(zipFile);
ZipInputStream zin = new ZipInputStream(fin);
ZipEntry ze = null;
int UnzipCounter = 0;
while ((ze = zin.getNextEntry()) != null) {
UnzipCounter++;
//if (ze.isDirectory()) {
// dirChecker(ze.getName());
//} else {
byte[] Unzipbuffer = new byte[(int) pow(2, 16)];
FileOutputStream fout = new FileOutputStream(
new File(Path, ze.getName()));
int Unziplength = 0;
while ((Unziplength = zin.read(Unzipbuffer)) > 0) {
fout.write(Unzipbuffer, 0, Unziplength);
}
zin.closeEntry();
fout.close();
//}
}
zin.close();
}
}
BTW
顺便提一句
- what is the language in that MP3, Arabic?
- I had to alter the source to get it to compile.
- I used the
Fileconstructor that takes twoStringarguments, to insert the correct separator automatically.
- MP3 中的语言是什么，阿拉伯语？
- 我不得不改变源代码才能编译。
- 我使用
File带有两个String参数的构造函数来自动插入正确的分隔符。
回答by Harry
You might have run into the following problem, which occurs, when reading zip files using a ZipInputStream: Zip files contain entries and additional structure information in a sequence. Furthermore, they contain a registry of all entries at the very end (!) of the file. Only this registry does provide full information about the correct zip file structure. Therefore, reading a zip file in a sequence, by using a stream, sometimes results in a "guess", which can fail. This is a common problem of all zip implementations, not only for java.util.zip. Better approach is to use ZipFile, which determines the structure from the registry at the end of the file. You might want to read http://commons.apache.org/compress/zip.html, which tells a little more details.
在使用 ZipInputStream 读取 zip 文件时，您可能会遇到以下问题： Zip 文件按顺序包含条目和附加结构信息。此外，它们在文件的最后 (!) 包含所有条目的注册表。只有此注册表才提供有关正确 zip 文件结构的完整信息。因此，通过使用流按顺序读取 zip 文件有时会导致“猜测”，这可能会失败。这是所有 zip 实现的常见问题，不仅仅是 java.util.zip。更好的方法是使用 ZipFile，它从文件末尾的注册表中确定结构。您可能想阅读http://commons.apache.org/compress/zip.html，它提供了更多详细信息。
回答by jagdish
Try this code:-
试试这个代码：-
private boolean extractZip(String pathOfZip,String pathToExtract)
{
int BUFFER_SIZE = 1024;
int size;
byte[] buffer = new byte[BUFFER_SIZE];
try {
File f = new File(pathToExtract);
if(!f.isDirectory()) {
f.mkdirs();
}
ZipInputStream zin = new ZipInputStream(new BufferedInputStream(new FileInputStream(pathOfZip), BUFFER_SIZE));
try {
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
String path = pathToExtract +"/"+ ze.getName();
if (ze.isDirectory()) {
File unzipFile = new File(path);
if(!unzipFile.isDirectory()) {
unzipFile.mkdirs();
}
}
else {
FileOutputStream out = new FileOutputStream(path, false);
BufferedOutputStream fout = new BufferedOutputStream(out, BUFFER_SIZE);
try {
while ( (size = zin.read(buffer, 0, BUFFER_SIZE)) != -1 ) {
fout.write(buffer, 0, size);
}
zin.closeEntry();
}catch (Exception e) {
Log.e("Exception", "Unzip exception 1:" + e.toString());
}
finally {
fout.flush();
fout.close();
}
}
}
}catch (Exception e) {
Log.e("Exception", "Unzip exception2 :" + e.toString());
}
finally {
zin.close();
}
return true;
}
catch (Exception e) {
Log.e("Exception", "Unzip exception :" + e.toString());
}
return false;
}
回答by Eric Lindauer
This code works fine for me. Perhaps you need to check that the zipFile String is valid?
这段代码对我来说很好用。也许您需要检查 zipFile 字符串是否有效？
String zipFile = "C:/my.zip";
FileInputStream fin = new FileInputStream(zipFile);
ZipInputStream zin = new ZipInputStream(fin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
System.out.println("got entry " + ze);
}
zin.close();
produced valid results on a 3.3Mb zip file.
在 3.3Mb 的 zip 文件上产生了有效的结果。
回答by Matthew Farwell
This code seems to work correctly for me.
这段代码似乎对我来说工作正常。
Are you sure that your zip file is a valid zip file? If the file does not exist or is not readable then you will get a FileNotFoundException, but if the file is empty or not a valid zip file, then you will get ze == null.
您确定您的 zip 文件是有效的 zip 文件吗？如果文件不存在或不可读，则您将收到 FileNotFoundException，但如果文件为空或不是有效的 zip 文件，则您将收到 ze == null。
while ((ze = zin.getNextEntry()) != null) {
The zip that you specify isn't a valid zip file. The size of the entry is 4294967295
您指定的 zip 不是有效的 zip 文件。条目大小为 4294967295
while ((ze = zin.getNextEntry()) != null) {
System.out.println("ze=" + ze.getName() + " " + ze.getSize());
UnzipCounter++;
This gives:
这给出：
ze=595.mp3 4294967295
...
Exception in thread "main" java.util.zip.ZipException: invalid entry size (expected 4294967295 but got 341297 bytes)
at java.util.zip.ZipInputStream.readEnd(ZipInputStream.java:386)
at java.util.zip.ZipInputStream.read(ZipInputStream.java:156)
at java.io.FilterInputStream.read(FilterInputStream.java:90)
at uk.co.farwell.stackoverflow.ZipTest.main(ZipTest.java:29)
Try your code with a valid zip file.
使用有效的 zip 文件尝试您的代码。
