[英]Best Way to copy a Zip File via Java
经过一番研究:
和一些谷歌研究我想出了这个java函数:
static void copyFile(File zipFile, File newFile) throws IOException {
ZipFile zipSrc = new ZipFile(zipFile);
ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(newFile));
Enumeration srcEntries = zipSrc.entries();
while (srcEntries.hasMoreElements()) {
ZipEntry entry = (ZipEntry) srcEntries.nextElement();
ZipEntry newEntry = new ZipEntry(entry.getName());
zos.putNextEntry(newEntry);
BufferedInputStream bis = new BufferedInputStream(zipSrc
.getInputStream(entry));
while (bis.available() > 0) {
zos.write(bis.read());
}
zos.closeEntry();
bis.close();
}
zos.finish();
zos.close();
zipSrc.close();
}
这段代码正在运行......但它并不好看和干净......任何人都有一个好主意或一个例子?
编辑:
我想能够添加一些类型的验证,如果zip存档得到正确的结构...所以复制它像普通文件而不考虑其内容对我不起作用...或者您希望之后检查...我不确定这个
你只想复制完整的zip文件? 打开并阅读zip文件不需要...只需复制它就像复制其他文件一样。
public final static int BUF_SIZE = 1024; //can be much bigger, see comment below
public static void copyFile(File in, File out) throws Exception {
FileInputStream fis = new FileInputStream(in);
FileOutputStream fos = new FileOutputStream(out);
try {
byte[] buf = new byte[BUF_SIZE];
int i = 0;
while ((i = fis.read(buf)) != -1) {
fos.write(buf, 0, i);
}
}
catch (Exception e) {
throw e;
}
finally {
if (fis != null) fis.close();
if (fos != null) fos.close();
}
}
我的解决方案
import java.io.*;
import javax.swing.*;
public class MovingFile
{
public static void copyStreamToFile() throws IOException
{
FileOutputStream foutOutput = null;
String oldDir = "F:/UPLOADT.zip";
System.out.println(oldDir);
String newDir = "F:/NewFolder/UPLOADT.zip"; // name as the destination file name to be done
File f = new File(oldDir);
f.renameTo(new File(newDir));
}
public static void main(String[] args) throws IOException
{
copyStreamToFile();
}
}
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