将包含多个文件的Inputstream放入一个ZipEntry中

Question

我想将一个File数组压缩到一个zipfile并将其发送到浏览器。每个Inputstream的File是一个shapefile，实际上由多个文件（.shp，.dbf，.shx，...）组成。

当我只使用以下代码发送一个File时，它可以正常工作，并返回一个包含所有所需文件的zipfile。

用于发送单个文件的代码

FileInputStream is = new FileInputStream(files.get(0));

response.setContentType("application/octet-stream");
response.setHeader("Content-Disposition", "attachment; filename=" + getCurrentUser(request).getNiscode() + ".zip");

while (is.available() > 0) {
    response.getOutputStream().write(is.read());
}

is.close();
if (response.getOutputStream() != null) {
    response.getOutputStream().flush();
    response.getOutputStream().close();
}

当我尝试将所有文​​件一起发送时，会返回带有所需文件夹的zipfile，但在每个文件夹中只存在一个只有.file扩展名的元素。它与ZipOutputStream的条目有关？

用于发送所有文件的代码

byte[] zip = this.zipFiles(files, Ids);

response.setContentType("application/zip");
response.setHeader("Content-Disposition", "attachment; filename="test.zip");

response.getOutputStream().write(zip);
response.flushBuffer();

private byte[] zipFiles(ArrayList<File> files, String[] Ids) throws IOException {
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ZipOutputStream zos = new ZipOutputStream(baos);

        int count = 0;
        for (File file : files) {
            FileInputStream fis = new FileInputStream(file);

            zos.putNextEntry(new ZipEntry(Ids[count] + "/"));
            zos.putNextEntry(new ZipEntry(Ids[count] + "/" + file.getName()));

            while (fis.available() > 0) {
                zos.write(fis.read());
            }
            zos.closeEntry();
            fis.close();
            count ++;
        }
        zos.flush();
        baos.flush();
        zos.close();
        baos.close();

        return baos.toByteArray();
    }

Answer 1

根据您的代码，您的files数组中的每个文件似乎都是一个zip文件

当你以后再做zipFiles时，你正在制作一个zip文件，其文件夹中包含更多的zip文件。你显然不希望这样，但是你想要一个包含所有包含zip文件内容的文件夹的zip文件。

基于位于“Reading data from multiple zip files and combining them to one”的Thanador现有答案，我设计了以下解决方案，还包括目录和正确的流处理：

/**
 * Combine multiple zipfiles together
 * @param files List of file objects pointing to zipfiles
 * @param ids List of file names to use in the final output
 * @return The byte[] object representing the final output
 * @throws IOException When there was a problem reading a zipfile
 * @throws NullPointerException When there either input is or contains a null value
 */
private byte[] zipFiles(ArrayList<File> files, String[] ids) throws IOException {
    ByteArrayOutputStream baos = new ByteArrayOutputStream();

    // Buffer to copy multiple bytes at once, this is generally faster that just reading and writing 1 byte at a time like your previous code does
    byte[] buf = new byte[16 * 1024];
    int length = files.size();
    assert length == ids.length;
    try (ZipOutputStream zos = new ZipOutputStream(baos)) {
        for (int i = 0; i < length; i++) {
            try (ZipInputStream inStream = new ZipInputStream(new FileInputStream(files.get(i))) {
                ZipEntry entry;
                while ((entry = inStream.getNextEntry()) != null) {
                    zos.putNextEntry(new ZipEntry(ids[i] + "/" + entry.getName()));
                    int readLength;
                    while ((readLength = inStream.read(buf)) > 0) {
                        zos.write(buf, 0, readLength);
                    }
                }
            }
        }
    }

    return baos.toByteArray();
}

• 从技术上讲，它直接写入你从response.getOutputStream()获得的输出流的速度更快，内存效率更高，但我在上面的例子中没有这样做，所以你可以更轻松地在代码中实现该方法
• 如果你关闭一个流，它会自动刷新它，我正在使用try-with-resources来关闭它们

Answer 2

尝试以下解决方案

private byte[] zipFiles(ArrayList<File> files, String[] Ids) throws IOException {
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ZipOutputStream zos = new ZipOutputStream(baos);
    for (File file : files) {
        ZipEntry ze= new ZipEntry(file.getName());
        zos.putNextEntry(ze);

        FileInputStream fis = new FileInputStream(file);

        int len;
        while ((len = fis .read(buffer)) > 0) {
            zos.write(buffer, 0, len);
        }

        fis .close();
    }
    byte[] byteArray = baos.toByteArray();
    zos.flush();
    baos.flush();
    zos.close();
    baos.close();

    return byteArray;
}

IDK为什么你把count变量和为什么你放两次zos.putNextEntry(new ZipEntry())。

Answer 3

 public static void compressListFiles(List<Pair<String, String>> filePairList, ZipOutputStream zipOut) throws Exception {
    for (Pair<String, String> pair : filePairList) {
        File fileToZip = new File(pair.getValue());
        String fileId = pair.getKey();
        FileInputStream fis = new FileInputStream(fileToZip);
        ZipEntry zipEntry = new ZipEntry(fileId + "/" + fileToZip.getName());
        zipOut.putNextEntry(zipEntry);
        byte[] bytes = new byte[1024];
        int length;
        while ((length = fis.read(bytes)) >= 0) {
            zipOut.write(bytes, 0, length);
        }
        fis.close();
    }
}