Question

我有一个应用程序可以拍照并向我提供给定图片的uri。接下来，我想做的是将此图片发送到我的网络服务。

但是，一旦我尝试了，我得到了一个我之前遇到的错误，这与asynctask有关。所以试图解决它，我有一个HttpManager类，它包含有关如何连接到webservice，url本身以及它处理图像uri的位置的信息。

public static String uploadImageToWebservice(String uri, String imageUri) {
    HttpURLConnection connection = null;
    int responseCode = 0;
    String image = "";
    try {

        URL url = new URL(uri + imageUri);
        connection = (HttpURLConnection) url.openConnection();
        connection.setDoOutput(true);
        connection.setInstanceFollowRedirects(false);
        connection.setRequestMethod("GET");
        connection.setRequestProperty("Content-Type", "text/plain");

        /*
        * String userpassword = adminUser + ":" + adminPassword; String
        * encodedAuthorization = DatatypeConverter
        * .printBase64Binary(userpassword.getBytes("UTF-8"));
        * connection.setRequestProperty("Authorization", "Basic " +
        * encodedAuthorization);
        */

        InputStream is = connection.getInputStream();
        image = is.toString();
        responseCode = connection.getResponseCode();
        connection.disconnect();
    } catch (IOException e) {
        Log.e("URL", "failure response from server >" + e.getMessage()
                + "<");

    } finally {
        if (connection != null) {
            connection.disconnect();
        }

    }
    return image;
}

这就是我在活动中处理这种方法的方法。

private void submitImage(String uri) {
    HttpAsyncTask at = new HttpAsyncTask();
    at.execute(uri);

}

    private class HttpAsyncTask extends AsyncTask<String, String, String> {

        @Override
        protected String doInBackground(String... params) {
            String image = HttpManager.uploadImageToWebservice(params[0],
                    params[1]);
            return image;
        }

    }

然后我使用Web服务的Uri在oncreate中调用submitImage方法。

但我有点被困在图像本身的uri放在哪里也可以发送。我只是觉得我错过了什么，我无法弄清楚它在哪里。希望它能够理解所有这一切。

提前致谢！

Answer 1

使用代码上传图片

            HttpURLConnection conn = null;
            DataOutputStream dos = null;
            String lineEnd = "\r\n";
            String twoHyphens = "--";
            String boundary = "*****";
            int bytesRead, bytesAvailable, bufferSize;
            byte[] buffer;
            int maxBufferSize = 1 * 1024 * 1024;

                   FileInputStream fileInputStream = new FileInputStream("sourcefile");
                    URL url = new URL("URL");

                    // Open a HTTP connection to the URL
                    conn = (HttpURLConnection) url.openConnection();
                    conn.setDoInput(true); // Allow Inputs
                    conn.setDoOutput(true); // Allow Outputs
                    conn.setUseCaches(false); // Don't use a Cached Copy
                    conn.setRequestMethod("POST");
                    conn.setRequestProperty("Connection", "Keep-Alive");
                    conn.setRequestProperty("ENCTYPE", "multipart/form-data");
                    conn.setRequestProperty("Content-Type",
                            "multipart/form-data;boundary=" + boundary);
                    conn.setRequestProperty("bill", sourceFileUri);

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

                    dos.writeBytes(twoHyphens + boundary + lineEnd);
                    dos.writeBytes("Content-Disposition: form-data; name=\"bill\";filename=\""
                            + sourceFileUri + "\"" + 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);

                    // Responses from the server (code and message)
                    serverResponseCode = conn.getResponseCode();
                    // String serverResponseMessage = conn
                    // .getResponseMessage();

                    if (serverResponseCode == 200) {
                        statud = "uploaded";
                        // messageText.setText(msg);
                        // Toast.makeText(ctx, "File Upload Complete.",
                        // Toast.LENGTH_SHORT).show();
                        // System.out.println("Uploaded successfyuly http: 200");
                        // recursiveDelete(mDirectory1);

                    }

                    // close the streams //
                    fileInputStream.close();
                    dos.flush();
                    dos.close();

php code

<?php

   $uploads_dir = './images/';
   $tmp_name = $_FILES['bill']['tmp_name'];
   $pic_name = $_FILES['bill']['name'];

   if (move_uploaded_file($tmp_name, $uploads_dir.$pic_name )) {
   echo "uploaded"; 
   }

?>

使用asynctask将图像发送到webservice

1 个答案: