Android:上传照片

时间:2011-01-24 11:13:58

标签: android upload photo

问候,

我在从Android手机上传照片时遇到问题。这是我用来提示用户选择照片的代码:

public class Uploader{
        public void upload(){
            Log.i("EOH","hi...");
            Intent photoPickerIntent = new Intent(Intent.ACTION_GET_CONTENT);
            photoPickerIntent.setType("image/*");
            startActivityForResult(photoPickerIntent, 1);
        }
        protected void onActivityResult(int requestCode, int resultCode, Intent data)
        {
            if (resultCode == RESULT_OK)
            {
                Bundle extras = data.getExtras();
                Bitmap b = (Bitmap) extras.get("data");
                //what do do now with this Bitmap...
                    //how do I upload it?




            }
        }

    }

所以我有位图b,但我不知道下一步该做什么?

我有以下用于发送POST请求的代码:

List<NameValuePair> params = new ArrayList<NameValuePair>(2);  
params.add(new BasicNameValuePair("someValA", String.valueOf(lat)));  
params.add(new BasicNameValuePair("someValB", String.valueOf(lng)));  
new HttpConnection(handler).post("http://myurl.com/upload.php",params);

如何将Bitmap图像连接到此?我已经搜索谷歌多年了,找不到一个好方法。

我希望有人可以提供帮助。

非常感谢,


好的,我尝试过chirag shah的建议。这是我的代码:

protected void onActivityResult(int requestCode, int resultCode, Intent data)
        {
            if (resultCode == RESULT_OK)
            {
                Uri imageUri=data.getData();
                List<NameValuePair> params = new ArrayList<NameValuePair>(1);
                params.add(new BasicNameValuePair("image", imageUri.getPath()));
                post("http://www.myurl.com/ajax/uploadPhoto.php",params);
            }
        }

其中post函数(推荐)是:

public void post(String url, List<NameValuePair> nameValuePairs) {
            HttpClient httpClient = new DefaultHttpClient();
            HttpContext localContext = new BasicHttpContext();
            HttpPost httpPost = new HttpPost(url);

            try {
                MultipartEntity entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);

                for(int index=0; index < nameValuePairs.size(); index++) {
                    if(nameValuePairs.get(index).getName().equalsIgnoreCase("image")) {
                        // If the key equals to "image", we use FileBody to transfer the data
                        entity.addPart(nameValuePairs.get(index).getName(), new FileBody(new File (nameValuePairs.get(index).getValue())));
                    }else{
                        // Normal string data
                        entity.addPart(nameValuePairs.get(index).getName(), new StringBody(nameValuePairs.get(index).getValue()));
                    }
                }

                httpPost.setEntity(entity);

                HttpResponse response = httpClient.execute(httpPost, localContext);
                Log.i("EOH",response.toString());
            } catch (IOException e) {
                e.printStackTrace();
            }
    }

但是myurl.com/ajax/uploadPhoto.php什么也没得到。我已经创建了一个PHP日志来记录uploadPhoto.php上的任何活动,并且没有任何显示。值得注意的是,如果我只是将myurl.com/ajax/uploadPhoto.php直接输入网络浏览器,它就可以正常登录。

还有其他建议吗?

非常感谢,

1 个答案:

答案 0 :(得分:0)

如果你还没有。尝试将帖子包装到异步任务中的服务器并调用该任务。我发现我使用的api需要远离主线程的http posts / gets,以避免锁定主线程的可能问题。

将它设置为异步任务并开始实际工作相对容易。

有关异步任务的更多信息:http://developer.android.com/reference/android/os/AsyncTask.html

对于一个简单的例子,这是我的异步任务之一:

private class SomeAsyncTask extends AsyncTask<String, Void, JSONObject> {

    @Override
    protected JSONObject doInBackground(String... arg0) {

        UserFunctions uf = new UserFunctions();
        JSONObject res = uf.doPostToServer(arg0[0], arg0[1], arg0[2], getApplicationContext());

        return res;
    }

    @Override
    protected void onPostExecute(JSONObject result) {

        try {
            int success = result.getInt("success");
            if(success == 1) {

                Intent vcact = new Intent(getApplicationContext(), OtherActivity.class);
                vcact.putExtra("id", cid);
                startActivity(vcact);
                // close main screen
                finish();

            } else {
                Toast.makeText(SomeActivity.this, "Error saving to server. Try again.", Toast.LENGTH_SHORT).show();
            }
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }
}

并使用

调用onclick
new SomeAsyncTask().execute(cid, otherdata, moredata);
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