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Forum Discussion
Get thumbnails of files from shared folders
Hi there,
I get list of files inside shared folder using "2/files/list_folder" with "shared_link" parameter and I am using response from "2/files/list_folder" into "2/files/get_thumbnail_v2"...
Hi @Greg-DB, Здравко Thanks for reply,
I use below code and getting error
"Error in call to API function "files/get_thumbnail:2": HTTP header "Dropbox-API-Arg": could not decode input as JSON"
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, 'https://content.dropboxapi.com/2/files/get_thumbnail_v2');
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_POST, 1);
$headers = array();
$headers[] = 'Authorization: Bearer sl.BIgvvyUur9m--oAuUapVgE__HaC8DnOZqdWuE9Q0Cs6QfpGbsYtWgAzQI6jGfW_seiBVA8gIeE0FFb0oUBM__kr3Br08XnoMD2EBSv_3g-MqrKibUT4_e6JlwJr_eCDyTMcwkZMm8RWV';
$headers[] = 'Dropbox-Api-Arg: {\"resource\":{\".tag\":\"link\",\"url\":\"https://www.dropbox.com/sh/ns2y8e6lqyh5avz/AAAHX5tPck8CHwpcSKMwV7cba?dl=0\",\"path\":\"/1515 Lochstone Drive 8.jpg\"}}';
$headers[] = 'Content-Type: application/octet-stream; charset=utf-8';
curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
$result = curl_exec($ch);
dd($result);
Hi @Greg-DB,
I updated code and getting result like
����JFIF��C % # , #&')*) -0-(0%()(��C (((((((((((((((((((((((((((((((((((((((((((((((((((��+@"�� ���}!1AQa"q2���#B��R��$3br� %&'()*456789:CDEFGHIJSTUVWXYZcdefghijstuvwxyz��������������������������������������������������������������������������� ���w!1AQaq"2�B���� #3R�br� $4�%�&'()*56789:CDEFGHIJSTUVWXYZcdefghijstuvwxyz��������������������������������������������������������������������������?��S���v���0GpjQ& ��-�;���v5Q���T�P�4�Z6�dp3XWZ��qG��r�J G`�29���Gr�-��\/�WipMAg���L[�ˡ����0ʞ�}��95 ���%G#/?Ni{H����� #��u�N��p��(G�9���M[M{�g �n� ‰*�O��p9��Z�u������r;Ckh�<�r���; ����J��*1��_��]ni,4�6 -W,�!�O�8����Ct�j�)`�c ��VU���.��4Km� �Fs���6—W���`�#a����(�z�q�k�:OY�~#��쮰C m�� �q]�e{Kx���ъ�; �G��u>��hZ`�N�H��f�`t�?T��u�t��:ItA�xQ��X)ʢ7Q�7���g ��Գ�Hm�*��N��wN �t�;C����Ծk/�cm�rh'��خ^ K[�E��w��y�V#��#h��֯�WH��l��x̦Y���#=q�[R��&gR���H�� �/�{!��2�� v�1��&��y����=���L�r� 7/=qבUmt[��L��� ���C��� s��+��-l������l���i\�x �<�E� R�m���Ѽ'q�d��J0�D [k�=���u� ��Z\�#��9������I� V~���jK�~צ�gf=�t m ��L��_z�lX�ߞ���y`�� >�XnL��9��(=kzS�wFSM�xԷ)q�F��T&ߝ��y5ne"M�?�_) \s�� b��V�Z��4V �0�;Q��n��Z�ic>���I am trying to display all thumbnails from shared folder in site and User can select particular thumbnail from list and I want to url of that selected thumbnail.
Hi nadeemphpdots,
Seems you have found out your PHP string error. 🙂.... And... What's wrong here? 🤔 Everything seems normal. Your jpeg, you opened as text, might confuse you. Isn't so? 😉
Hope this helps.
Hi Здравко
I am trying to display all thumbnails from shared folder in site and User can select particular thumbnail from list and I want to url of that selected thumbnail.Now I will display Image by
echo '<img src="data:image/jpg;charset=utf8;base64,'.$data.' />';How can I get actual url of image Actual Image ?
😁 I'm not sure you understand, but you are trying to put image file binary content as a source URL of an image. I thought my post above was clear, but seems it's not enough. Think of your "$data" as a content of an image read from file available on your server (not as a URL to there). What will be the result? 🧐 🙂 You have to pass the image data as such not as a URL! There are many correct way this to be done. 😉 Only your imagination ability can restrict you. Of course, there are many wrong ways (like mixing URL and data).
Hope it's a bit more clear now.
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