我收到以下IOException:
java.io.IOException: Access is denied
at java.io.WinNTFileSystem.createFileExclusively(Native Method)
at java.io.File.createNewFile(File.java:850)
at zipUnzipper.main(zipUnzipper.java:41)
尝试运行以下代码时:
public class zipUnzipper {
public zipUnzipper() {
}
public static void main(String[] args){
//Unzip to temp folder. Add all files to mFiles. Print names of all files in mFfiles.
File file = new File("C:\\aZipFile.zip");
String filename = file.getName();
String filePathName = new String();
int o = filename.lastIndexOf('.');
filename = filename.substring(0,o);
try {
ZipFile zipFile = new ZipFile (file.getAbsoluteFile());
Enumeration entries = zipFile.entries();
while(entries.hasMoreElements()) {
ZipEntry zipEntry = (ZipEntry) entries.nextElement();
System.out.println("Unzipping: " + zipEntry.getName());
BufferedInputStream bis = new BufferedInputStream(zipFile.getInputStream(zipEntry));
byte[] buffer = new byte[2048];
filePathName = "C:\\TEMP\\"+filename+"\\";
File fileToWrite = new File(filePathName+ zipEntry.getName());
fileToWrite.mkdirs();
fileToWrite.createNewFile();
FileOutputStream fos = new FileOutputStream(fileToWrite);
BufferedOutputStream bos = new BufferedOutputStream( fos , buffer.length );
int size;
while ((size = bis.read(buffer, 0, buffer.length)) != -1) {
bos.write(buffer, 0, size);
}
bos.flush();
bos.close();
bis.close();
}
zipFile.close();
File folder = new File (filePathName);
File [] mFiles = folder.listFiles();
for (int x=0; x<mFiles.length; x++) {
System.out.println(mFiles[x].getAbsolutePath());
}
} catch (ZipException ze) {
ze.printStackTrace();
} catch (IOException ioe) {
ioe.printStackTrace();
}
}
}
在我看来,由于某种原因,JVM无法创建新文件。如果文件已经存在,代码运行得非常好。是否存在某种访问文件,它指示JVM是否可以创建新文件,或者我只是做错了什么?
非常感谢任何帮助: - )
我正在运行Java 1.4并且已经在Windows XP中的JDeveloper中进行了测试。
答案 0 :(得分:11)
问题是这些调用相互衔接:
fileToWrite.mkdirs(); //creates a directory e.g. C:\temp\foo\x
fileToWrite.createNewFile(); //attempts to create a file C:\temp\foo\x
创建操作失败,因为您刚刚创建了一个与您要创建的文件同名的目录。
您想要做的是:
fileToWrite.getParentFile().mkdirs()
此外,调用createNewFile()
是不必要的。
根据您的代码。以下“解压缩”一个zip文件:
import java.io.*;
import java.util.zip.ZipFile;
import java.util.zip.ZipEntry;
import java.util.Enumeration;
public class Unzipper {
public static void main(String[] args)
throws IOException {
final File file = new File(args[0]);
final ZipFile zipFile = new ZipFile(file);
final byte[] buffer = new byte[2048];
final File tmpDir = new File(System.getProperty("java.io.tmpdir"), zipFile.getName());
if(!tmpDir.mkdir() && tmpDir.exists()) {
System.err.println("Cannot create: " + tmpDir);
System.exit(0);
}
for(final Enumeration entries = zipFile.entries(); entries.hasMoreElements();) {
final ZipEntry zipEntry = (ZipEntry) entries.nextElement();
System.out.println("Unzipping: " + zipEntry.getName());
final InputStream is = zipFile.getInputStream(zipEntry);
final File fileToWrite = new File(tmpDir, zipEntry.getName());
final File folder = fileToWrite.getParentFile();
if(!folder.mkdirs() && !folder.exists()) {
System.err.println("Cannot create: " + folder);
System.exit(0);
}
if(!zipEntry.isDirectory()) {
//No need to use buffered streams since we're doing our own buffering
final FileOutputStream fos = new FileOutputStream(fileToWrite);
int size;
while ((size = is.read(buffer)) != -1) {
fos.write(buffer, 0, size);
}
fos.close();
is.close();
}
}
zipFile.close();
}
}
免责声明:我没有超出基本要求进行测试。
答案 1 :(得分:2)
你为什么打电话给createNewFile()
?只需创建FileOutputStream。
答案 2 :(得分:0)
也可能是在启动应用程序的环境中,您无法访问尝试创建文件的位置的权限。以管理员身份启动应用程序或在项目文件夹中创建文件。