目前我正在使用下面的代码将zip压缩到另一个zip中.但是当zip文件大小增加时,例如:2GB,程序丢失了内存错误.我已经将xmx增加到1024,仍然prfoblem是相同的.这是任何替代方法处理大文件?
public static void zipFile(File srcFile, File zipFile)
throws FileNotFoundException, IOException {
BufferedInputStream origin = null;
FileOutputStream dest = new FileOutputStream(zipFile);
ZipOutputStream out = new ZipOutputStream(
new BufferedOutputStream(dest));
// out.setMethod(ZipOutputStream.DEFLATED);
byte data[] = new byte[BUFFER];
FileInputStream fi = new FileInputStream(srcFile);
origin = new BufferedInputStream(fi, BUFFER);
ZipEntry entry = new ZipEntry(srcFile.getName());
out.putNextEntry(entry);
int count;
while ((count = origin.read(data, 0, BUFFER)) != -1) {
out.write(data, 0, count);
}
origin.close();
out.close();
}
最佳答案 当你到达它时,你只是复制字节.您不需要将任一文件作为Zip文件处理.只需复制字节即可.