Android:如何将.mp3文件上传到http服务器？

Question

我想将.mp3文件(仅)从设备上传到我的服务器.

我想浏览媒体数据的路径并选择任何mp3文件并上传.

我怎样才能做到这一点？

Answer 1

我的最终工作JAVA和PHP代码,用于将Android的SD卡上传文件到我自己的Web服务器.

Java/Android代码:

private void doFileUpload() {

    HttpURLConnection conn = null;
    DataOutputStream dos = null;
    DataInputStream inStream = null;
    String existingFileName = Environment.getExternalStorageDirectory().getAbsolutePath() + "/mypic.png";
    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";
    int bytesRead, bytesAvailable, bufferSize;
    byte[] buffer;
    int maxBufferSize = 1 * 1024 * 1024;
    String responseFromServer = "";
    String urlString = "http://mywebsite.com/directory/upload.php";

    try {

        //------------------ CLIENT REQUEST
        FileInputStream fileInputStream = new FileInputStream(new File(existingFileName));
        // open a URL connection to the Servlet
        URL url = new URL(urlString);
        // Open a HTTP connection to the URL
        conn = (HttpURLConnection) url.openConnection();
        // Allow Inputs
        conn.setDoInput(true);
        // Allow Outputs
        conn.setDoOutput(true);
        // Don't use a cached copy.
        conn.setUseCaches(false);
        // Use a post method.
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Connection", "Keep-Alive");
        conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
        dos = new DataOutputStream(conn.getOutputStream());
        dos.writeBytes(twoHyphens + boundary + lineEnd);
        dos.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + existingFileName + "\"" + lineEnd);
        dos.writeBytes(lineEnd);
        // create a buffer of maximum size
        bytesAvailable = fileInputStream.available();
        bufferSize = Math.min(bytesAvailable, maxBufferSize);
        buffer = new byte[bufferSize];
        // read file and write it into form...
        bytesRead = fileInputStream.read(buffer, 0, bufferSize);

        while (bytesRead > 0) {

            dos.write(buffer, 0, bufferSize);
            bytesAvailable = fileInputStream.available();
            bufferSize = Math.min(bytesAvailable, maxBufferSize);
            bytesRead = fileInputStream.read(buffer, 0, bufferSize);

        }

        // send multipart form data necesssary after file data...
        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
        // close streams
        Log.e("Debug", "File is written");
        fileInputStream.close();
        dos.flush();
        dos.close();

    } catch (MalformedURLException ex) {
        Log.e("Debug", "error: " + ex.getMessage(), ex);
    } catch (IOException ioe) {
        Log.e("Debug", "error: " + ioe.getMessage(), ioe);
    }

    //------------------ read the SERVER RESPONSE
    try {

        inStream = new DataInputStream(conn.getInputStream());
        String str;

        while ((str = inStream.readLine()) != null) {

            Log.e("Debug", "Server Response " + str);

        }

        inStream.close();

    } catch (IOException ioex) {
        Log.e("Debug", "error: " + ioex.getMessage(), ioex);
    }
}

要在服务器上运行的相关PHP代码(upload.php):

<?php
// Where the file is going to be placed 
$target_path = "uploads/";

/* Add the original filename to our target path.  
Result is "uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name']). 
    " has been uploaded";
    chmod ("uploads/".basename( $_FILES['uploadedfile']['name']), 0644);
} else{
    echo "There was an error uploading the file, please try again!";
    echo "filename: " .  basename( $_FILES['uploadedfile']['name']);
    echo "target_path: " .$target_path;
}
?>

注意事项.
1)我在SD卡的根目录中有"mypic.png".如果您通过Mass Storage USB视图查看Android设备,则会将该文件放在您遇到的第一个目录中.

2)必须在手机上关闭USB大容量存储!或者只是从正在编写代码的计算机上完全拔掉它,以确保是这种情况.

3)我必须在与我的php文件相同的目录中创建一个"uploads"文件夹.

4)您显然必须将我编写的网址http://mywebsite.com/directory/upload.php更改为您自己的网站.

Answer 2

感谢Keaton的好建议.

我已经安排了一些Java代码,以便它可以使用并添加对其他URL参数的支持:

public class HttpMultipartUpload {
    static String lineEnd = "\r\n";
    static String twoHyphens = "--";
    static String boundary = "AaB03x87yxdkjnxvi7";

    public static String upload(URL url, File file, String fileParameterName, HashMap<String, String> parameters)
            throws IOException {
        HttpURLConnection conn = null;
        DataOutputStream dos = null;
        DataInputStream dis = null;
        FileInputStream fileInputStream = null;

        byte[] buffer;
        int maxBufferSize = 20 * 1024;
        try {
            //------------------ CLIENT REQUEST
            fileInputStream = new FileInputStream(file);

            // open a URL connection to the Servlet
            // Open a HTTP connection to the URL
            conn = (HttpURLConnection) url.openConnection();
            // Allow Inputs
            conn.setDoInput(true);
            // Allow Outputs
            conn.setDoOutput(true);
            // Don't use a cached copy.
            conn.setUseCaches(false);
            // Use a post method.
            conn.setRequestMethod("POST");
            conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);

            dos = new DataOutputStream(conn.getOutputStream());

            dos.writeBytes(twoHyphens + boundary + lineEnd);
            dos.writeBytes("Content-Disposition: form-data; name=\"" + fileParameterName
                    + "\"; filename=\"" + file.toString() + "\"" + lineEnd);
            dos.writeBytes("Content-Type: text/xml" + lineEnd);
            dos.writeBytes(lineEnd);

            // create a buffer of maximum size
            buffer = new byte[Math.min((int) file.length(), maxBufferSize)];
            int length;
            // read file and write it into form...
            while ((length = fileInputStream.read(buffer)) != -1) {
                dos.write(buffer, 0, length);
            }

            for (String name : parameters.keySet()) {
                dos.writeBytes(lineEnd);
                dos.writeBytes(twoHyphens + boundary + lineEnd);
                dos.writeBytes("Content-Disposition: form-data; name=\"" + name + "\"" + lineEnd);
                dos.writeBytes(lineEnd);
                dos.writeBytes(parameters.get(name));
            }

            // send multipart form data necessary after file data...
            dos.writeBytes(lineEnd);
            dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
            dos.flush();
        } finally {
            if (fileInputStream != null) fileInputStream.close();
            if (dos != null) dos.close();
        }

        //------------------ read the SERVER RESPONSE
        try {
            dis = new DataInputStream(conn.getInputStream());
            StringBuilder response = new StringBuilder();

            String line;
            while ((line = dis.readLine()) != null) {
                response.append(line).append('\n');
            }

            return response.toString();
        } finally {
            if (dis != null) dis.close();
        }
    }
}