我想将一些压缩数据放入远程存储库 要将数据放在此存储库中,我只能使用将资源名称及其内容作为String的方法。 (比如data.txt +“你好世界”) 存储库正在调用文件系统但不是,所以我不能直接使用File。
我希望能够做到以下几点:
当我尝试获取压缩文件的字符串表示时,问题出现在第3步。
这是一个示例类,使用zip *流并模拟显示我的问题的存储库 创建的zip文件正常工作,但在“序列化”后,它会被破坏 (示例类使用jakarta commons.io)
非常感谢你的帮助。
package zip;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.zip.ZipEntry;
import java.util.zip.ZipInputStream;
import java.util.zip.ZipOutputStream;
import org.apache.commons.io.FileUtils;
/**
* Date: May 19, 2010 - 6:13:07 PM
*
* @author Guillaume AME.
*/
public class ZipMe {
public static void addOrUpdate(File zipFile, File ... files) throws IOException {
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
if (files != null) {
for (File file : files) {
InputStream in = new FileInputStream(file);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(file.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
}
tempFile.delete();
out.close();
}
public static void main(String[] args) throws IOException {
final String zipArchivePath = "c:/temp/archive.zip";
final String tempFilePath = "c:/temp/data.txt";
final String resultZipFile = "c:/temp/resultingArchive.zip";
File zipArchive = new File(zipArchivePath);
FileUtils.touch(zipArchive);
File tempFile = new File(tempFilePath);
FileUtils.writeStringToFile(tempFile, "hello world");
addOrUpdate(zipArchive, tempFile);
//archive.zip exists and contains a compressed data.txt that can be read using winrar
//now simulate writing of the zip into a in memory cache
String archiveText = FileUtils.readFileToString(zipArchive);
FileUtils.writeStringToFile(new File(resultZipFile), archiveText);
//resultingArchive.zip exists, contains a compressed data.txt, but it can not
//be read using winrar: CRC failed in data.txt. The file is corrupt
}
}
答案 0 :(得分:2)
Zip文件是二进制文件。 Java中的字符串处理是文本的,可能会破坏它所看到的CRLF,零字节和EOF标记。在阅读和重写zip文件时,我建议您尝试使用readFileToByteArray
和writeByteArrayToFile
作为实验。如果这样可行,那么我怀疑字符串处理是罪魁祸首。
答案 1 :(得分:1)
服务器发送字符串表示形式 data.zip到存储库
因此,您希望获得zip(即二进制)流的字符串(即文本)表示。
Base64是最常用的方式。
一个流行的Java实现来自Apache commons(codec组件)