小编Siv*_*r S的帖子

从中选择图像的第一个片段

           iv.setImageURI(Uri.fromFile(pictureFile));
                String stringUri;
                stringUri = pictureFile.toString();

                FreeFragment ldf = new FreeFragment ();
                Bundle args = new Bundle();
                args.putString("Image", stringUri);
                ldf.setArguments(args);
                Log.d("Passing image", String.valueOf(args));
                getFragmentManager().beginTransaction().add(R.id.container, ldf).commit();

第二个片段接收图像并显示它

 String bbb = getArguments().getString("Image");
    Bitmap bitmap = BitmapFactory.decodeFile(bbb);
    iv.setImageBitmap(bitmap);

http://example/map.php？cust_id = 1&store_id = 2,在onclick链接期间从url获取cust_id和store_id并将其保存在另一个共享首选项中.我从其他类调用这两个id并希望将这两个id发送到其他url但是id的值没有上传到该url,如果仅传递cust_id意味着一切正常,而传递两个值则无效.这个概念的新手请帮助我.

    @Override
    public boolean shouldOverrideUrlLoading(WebView view, String url) {
        view.loadUrl(url);
        CharSequence urlstring1="cust_id";
        CharSequence urlstring2="store_id";
        String[] details = url.split("[?]");
        String[] strdetails =details[1].split("&");
        String strcust_id = strdetails[1];
        String strstore_id= strdetails[1];

        if (url.contains(urlstring1)& url.contains(urlstring2)){
            SharedPreferences settings = getActivity().getSharedPreferences(PREFS_NAME, Context.MODE_PRIVATE);
            SharedPreferences.Editor editor = settings.edit();
            editor.putString("Customer", strcust_id).putString("Store", strstore_id);
            editor.commit();

            AboutusFragment fragment2 = new AboutusFragment();
            FragmentManager fragmentManager = getFragmentManager();
            android.support.v4.app.FragmentTransaction fragmentTransaction = fragmentManager.beginTransaction();
            fragmentTransaction.replace(R.id.container, fragment2);
            fragmentTransaction.commit();
            return false;
        }else{
            return super.shouldOverrideUrlLoading(view, url);
        }
    }

在下一堂课

if(CheckNetwork.isInternetAvailable(getContext().getApplicationContext())) //returns true if internet available …