图像/视频/音频成功从Android上传到PHP，但可以'；t打开它们-损坏/损坏

我在Android应用程序中发出此HTTP POST请求：

    private final String delimiter = "--";
    private final String boundary = "SwA"
            + Long.toString(System.currentTimeMillis()) + "SwA";
    private final String charset = "UTF-8";
    private final String lineSpace = "\r\n";
    private final String domain = (domain);

    private HttpURLConnection configureConnectionForMultipart(String url)
            throws MalformedURLException, IOException {
        HttpURLConnection con = (HttpURLConnection) (new URL(url))
                .openConnection();
        con.setRequestMethod("POST");
        con.setDoInput(true);
        con.setDoOutput(true);
        con.setRequestProperty("Connection", "Keep-Alive");
        con.setRequestProperty("Content-Type", "multipart/form-data;boundary="
                + boundary);
        return con;
    }

    private void addFormPart(String paramName, String value, DataOutputStream os)
            throws IOException {
        os.writeBytes(lineSpace + delimiter + boundary + lineSpace);
        os.writeBytes("Content-Disposition: form-data; name=\"" + paramName
                + "\"" + lineSpace);
        os.writeBytes("Content-Type: text/plain; charset=" + charset
                + lineSpace);
        os.writeBytes(lineSpace + value + lineSpace);
        os.flush();
    }

    private void addFilePart(String paramName, File data, DataOutputStream os)
            throws IOException {
        os.writeBytes(lineSpace + delimiter + boundary + lineSpace);
        os.writeBytes("Content-Disposition: form-data; name=\"" + paramName
                + "\"; filename=\"" + data.getAbsolutePath() + "\"" + lineSpace);
        os.writeBytes("Content-Type: application/octet \r\n");
        os.writeBytes("Content-Transfer-Encoding: binary" + lineSpace);
        // os.writeBytes(lineSpace);
        os.flush();

        FileInputStream fis = new FileInputStream(data);
        byte[] buffer = new byte[4096];
        int bytesRead = -1;
        while ((bytesRead = fis.read(buffer)) != -1) {
            os.write(buffer, 0, bytesRead);
        }
        os.writeBytes(lineSpace);
        os.flush();
        fis.close();
    }

    private void finishMultipart(DataOutputStream os) throws IOException {
        // os.writeBytes(lineSpace);
        os.flush();
        os.writeBytes(delimiter + boundary + delimiter + lineSpace);
        os.close();
    }

private class ObjectUploadRunnable implements Runnable {

        private final String _filePath;
        private final String _url = domain + "upload.php";

        public ObjectUploadRunnable(String filePath) {
            _filePath = filePath;
        }

        @Override
        public void run() {

            try {
                HttpURLConnection con = configureConnectionForMultipart(_url);
                con.connect();
                DataOutputStream os = new DataOutputStream(
                        con.getOutputStream());
                File data = new File(_filePath);
                addFilePart("data", data, os);
                finishMultipart(os);
                String response = getResponse(con);
                Log.i("BoxUpload", response);
                con.disconnect();
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }

我通过脚本upload.php在服务器上捕获它：

...
$dir = "/uploads/";
$target_path = $dir.basename($_FILES['data']['name']);

if (move_uploaded_file($_FILES['data']['tmp_name'], $target_path)) {
    echo "file uploaded";
} else {
    echo print_r(error_get_last());
}

一切似乎都成功了，一个大小正确的文件被上传到我的服务器上，在所需的目录中。但是，当我尝试打开文件时，它似乎在某种程度上被损坏或损坏，因为它不会在我尝试的任何应用程序中打开。我上传的图像=jpeg，视频=mp4，音频=mp4。这些文件在上传之前都在客户端上工作。我在POST请求中是否缺少正确编码文件的内容？我以前从来没有上传过文件，所以我希望你能给我一些建议

编辑

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如果这是相关的，我注意到我上传的文件增加了约100kb。可能是我的二进制数据中添加了一些东西，导致文件损坏？

在这里解决了问题。我在addFilePart中注释掉的那一行实际上是必要的。我猜在请求的这一部分中，头信息和二进制数据之间需要有两行。为了清楚起见，它应该是这样的：

private void addFilePart(String paramName, File data, DataOutputStream os)
        throws IOException {
    os.writeBytes(lineSpace + delimiter + boundary + lineSpace);
    os.writeBytes("Content-Disposition: form-data; name=\"" + paramName
            + "\"; filename=\"" + data.getAbsolutePath() + "\"" + lineSpace);
    os.writeBytes("Content-Type: application/octet \r\n");
    os.writeBytes("Content-Transfer-Encoding: binary" + lineSpace);
    os.writeBytes(lineSpace);
    os.flush();

    FileInputStream fis = new FileInputStream(data);
    byte[] buffer = new byte[4096];
    int bytesRead = -1;
    while ((bytesRead = fis.read(buffer)) != -1) {
        os.write(buffer, 0, bytesRead);
    }
    os.writeBytes(lineSpace);
    os.flush();
    fis.close();
}

---

现在一切都很好

你只是假设上传永远不会失败。坏主意。在执行任何其他操作之前，请始终检查$\u文件中的

['errors']

参数。好的，我执行了var\u转储[$\u文件），错误字段显示0个错误。这足够了吗？我认为实际上载中没有任何错误，因为一个大小正确的文件上载到了我想要的位置。我只是无法打开它。