Question

这是我的班级将捕获的相机图像上传到服务器它运作良好，但我想知道如何在上传后获取图像路径或图像Uri。我需要上传图像路径并稍后在mysql数据库中保存，请帮我。

public class UploadActivity extends Activity {
    // LogCat tag
    private static final String TAG = NewReclamationActivity.class.getSimpleName();


    private String filePath = null;


    long totalSize = 0;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);


        // Receiving the data from previous activity
        Intent i = getIntent();

        // image path that is captured in previous activity
        filePath = i.getStringExtra("filePath");
        // bitmap factory
        BitmapFactory.Options options = new BitmapFactory.Options();

        // down sizing image as it throws OutOfMemory Exception for larger
        // images
        options.inSampleSize = 8;

        BitmapFactory.decodeFile(filePath, options);


        new UploadFileToServer().execute();


    }



    /**
     * Uploading the file to server
     * */
    private class UploadFileToServer extends AsyncTask<Void, Integer, String> {



        protected String doInBackground(Void... params) {
            return uploadFile();
        }

        @SuppressWarnings("deprecation")
        private String uploadFile() {
            String responseString ;

            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(Configimage.FILE_UPLOAD_URL);

            try {
                AndroidMultiPartEntity entity = new AndroidMultiPartEntity(
                        new ProgressListener() {

                            @Override
                            public void transferred(long num) {
                                publishProgress((int) ((num / (float) totalSize) * 100));
                            }
                        });

                File sourceFile = new File(filePath);

                // Adding file data to http body
                entity.addPart("image", new FileBody(sourceFile));



                totalSize = entity.getContentLength();
                httppost.setEntity(entity);

                // Making server call
                HttpResponse response = httpclient.execute(httppost);
                HttpEntity r_entity = response.getEntity();

                int statusCode = response.getStatusLine().getStatusCode();
                if (statusCode == 200) {
                    // Server response
                    responseString = EntityUtils.toString(r_entity);
                } else {
                    responseString = "Error occurred! Http Status Code: "
                            + statusCode;
                }

            } catch (ClientProtocolException e) {
                responseString = e.toString();
            } catch (IOException e) {
                responseString = e.toString();
            }

            return responseString;

        }

        @Override
        protected void onPostExecute(String result) {
            Log.e(TAG, "Response from server: " + result);

            // showing the server response in an alert dialog
            showAlert(result);

            super.onPostExecute(result);
        }

    }

    /**
     * Method to show alert dialog
     * */
    private void showAlert(String message) {
        AlertDialog.Builder builder = new AlertDialog.Builder(this);
        builder.setMessage(message).setTitle("Response from Servers")
                .setCancelable(false)
                .setPositiveButton("OK", new DialogInterface.OnClickListener() {
                    public void onClick(DialogInterface dialog, int id) {
                        // do nothing
                    }
                });
        AlertDialog alert = builder.create();
        alert.show();
    }

    }

我的PHP代码是这样的：

<?php

// Path to move uploaded files
$target_path = "uploads/";

// array for final json respone
$response = array();

// getting server ip address
$server_ip = gethostbyname(gethostname());

// final file url that is being uploaded
$file_upload_url = 'http://' . $server_ip . '/' . 'AndroidFileUpload' . '/' . $target_path;


if (isset($_FILES['image']['name'])) {
    $target_path = $target_path . basename($_FILES['image']['name']);



    $response['file_name'] = basename($_FILES['image']['name']);


    try {
        // Throws exception incase file is not being moved
        if (!move_uploaded_file($_FILES['image']['tmp_name'], $target_path)) {
            // make error flag true
            $response['error'] = true;
            $response['message'] = 'Could not move the file!';
        }

        // File successfully uploaded
        $response['message'] = 'File uploaded successfully!';
        $response['error'] = false;
        $response['file_path'] = $file_upload_url . basename($_FILES['image']['name']);
    } catch (Exception $e) {
        // Exception occurred. Make error flag true
        $response['error'] = true;
        $response['message'] = $e->getMessage();
    }
} else {
    // File parameter is missing
    $response['error'] = true;
    $response['message'] = 'Not received any file!F';
}

// Echo final json response to client
echo json_encode($response);
?>

Answer 1

来自此{"file_name":"IMG_20161014_110734.jpg","message":"File uploaded successfully!","error":false,"file_path":"http:\/\/197.28.3.‌209\/AndroidFileUplo‌ad\/uploads\/IMG_201‌61014_110734.jpg"}

您可以使用内置的json解析库

来解析文件

JsonObject obj = new JsonObject(reult);
String file_path = obj.getString("file_path");
Log.e("FilePath",file_path);

如何使用android将其上传到服务器后获取图像路径？

1 个答案: