8

我知道这在这里和那里都有答案，但我无法让它们中的任何一个起作用。有没有人知道一个很好的参考，或者这个教程，也许也在这里发布？

我需要做的是：

1）提供一个按钮，打开相机应用程序。我已经这样做了startResultActivity()

2）用户拍摄照片，然后返回应用程序，保存照片，最好在 ImageView 中预览。我尝试了一些东西，但我无法在模拟设备中进行测试。

3) 按下“发送”按钮，应用程序将图片发送到 HTTP POST。有了“多部分”，不管是什么。php 开发人员不希望我将图片作为从位图数组转换的字符串发送。

对此的任何帮助将不胜感激。谢谢 ！

4

1 回答 1

14

此链接对于单击、保存和获取图像的路径应该绰绰有余： 捕获图像

这是我为通过 HTTP POST 上传图像而编写的类：

public class MultipartServer {

private static final String TAG = "MultipartServer";
private static String crlf = "\r\n";
private static String twoHyphens = "--";
private static String boundary =  "*****";
private static String avatarPath = null;

public static String postData(URL url, List<NameValuePair> nameValuePairs) throws IOException {

    HttpURLConnection connection = (HttpURLConnection) url.openConnection();
    connection.setReadTimeout(10000);
    connection.setConnectTimeout(15000);
    connection.setRequestMethod("POST");
    connection.setUseCaches(false);
    connection.setDoInput(true);
    connection.setDoOutput(true);

    connection.setRequestProperty("Connection", "Keep-Alive");
    connection.setRequestProperty("Cache-Control", "no-cache");
    connection.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);

    String avatarName = null;
    StringBuilder query = new StringBuilder();
    boolean first = true;
    for (NameValuePair pair : nameValuePairs) {
        if (first)
            first = false;
        else
            query.append("&");
        query.append(URLEncoder.encode(pair.getName(), "UTF-8"));
        query.append("=");
        query.append(URLEncoder.encode(pair.getValue(), "UTF-8"));
        if ((avatarName = pair.getName()).equals("avatar")) {
            avatarPath = pair.getValue();
        }

    }

    FileInputStream inputStream;
    OutputStream outputStream = connection.getOutputStream();
    DataOutputStream dataOutputStream = new DataOutputStream(outputStream);

    dataOutputStream.writeBytes(query.toString());

    // Write Avatar (if any)
    if(avatarName != null && avatarPath != null) {
        dataOutputStream.writeBytes(twoHyphens + boundary + crlf);
        dataOutputStream.writeBytes("Content-Disposition: form-data; name=\"" + avatarName + "\";filename=\"" + new File(avatarPath).getName() + "\";" + crlf);
        dataOutputStream.writeBytes(crlf);

        /*Bitmap avatar = BitmapFactory.decodeFile(avatarPath);
        avatar.compress(CompressFormat.JPEG, 75, outputStream);
        outputStream.flush();*/

        inputStream = new FileInputStream(avatarPath);
        byte[] data = new byte[1024];
        int read;
        while((read = inputStream.read(data)) != -1)
            dataOutputStream.write(data, 0, read);
        inputStream.close();

        dataOutputStream.writeBytes(crlf);
        dataOutputStream.writeBytes(twoHyphens + boundary + twoHyphens + crlf);
    }

    dataOutputStream.flush();
    dataOutputStream.close();

    String responseMessage = connection.getResponseMessage();
    Log.d(TAG, responseMessage);

    InputStream in = connection.getInputStream();
    BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(in, "UTF-8"));

    StringBuilder response = new StringBuilder();
    char []b = new char[512];
    int read;
    while((read = bufferedReader.read(b))!=-1) {
        response.append(b, 0, read);
    }

    connection.disconnect();
    Log.d(TAG, response.toString());
    return response.toString();
}
}

用法非常简单：调用此静态方法并传递图像的路径，例如：

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("avatar", imagePath));

最后：

MultipartServer.postData(url, nameValuePairs);

并且不要忘记在单独的线程中调用此函数，否则您将获得 NetworkOnMainThreadException.. :)

更新

我建议不要重新发明轮子并改用OkHttp。请查看食谱页面。免责声明：我不是该项目的贡献者，但我喜欢它。感谢 Square 团队。

于 2013-11-04T06:30:20.187 回答