Get thumbnails of files from shared folders

[nadeemphpdots]

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" api call to get thumbnails. 

"2/files/get_thumbnail_v2" api return "path/not_found/." error invalid path error.

Is there any other way to achieve this ? 

I want thumbnails of file from shared folder.

[Greg-DB]

You should be able to use /2/files/get_thumbnail_v2 to get a thumbnail of a file from a shared link for a folder. I just tried it out and it worked for me. Here's an example of what the call might look like for curl, for instance:

curl -X POST https://content.dropboxapi.com/2/files/get_thumbnail_v2 \
  --header 'Authorization: Bearer <ACCESS_TOKEN>' \
  --header 'Dropbox-API-Arg: {"resource":{".tag":"link","path":"/test.jpg","url":"<SHARED_LINK>"}}' \
  -o out.jpg

In this example, "/test.jpg" is the relative path of the file for which I want the thumbnail in the linked folder.

If something's not working as expected for you, please show us the steps and code you're using that result in the issue so we can take a look. Make sure to redact any access token though.

[Здравко]

@nadeemphpdots, In addition to what Greg said, you can use those calls also with Application Authentication. In other words, you don't need to be logged in since shared resource isn't something private (by assumption).

@Greg-DB, I thought to post the example in App Authentication section, but there is a mistake. Can you use path in anonymous access, alone?! 😉 Maybe some documentation corrections are need.

[Greg-DB]

Yes, this works with app authentication as well. In that case it would look like:

curl -X POST https://content.dropboxapi.com/2/files/get_thumbnail_v2 \
  --header "Authorization: Basic <ENCODED_APP_KEY_AND_SECRET>" \
  --header 'Dropbox-API-Arg: {"resource":{".tag":"link","path":"/test.jpg","url":"<SHARED_LINK>"}}' \
  -o out.jpg

[nadeemphpdots]

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);

[nadeemphpdots]

Hi @Greg-DB,
I updated code and getting  result like 

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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.

[Здравко]

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.

[nadeemphpdots]

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&colon;image/jpg;charset=utf8;base64,'.$data.' />';

How can I get actual url of image Actual Image ?

[Здравко]

😁 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.

[nadeemphpdots]

@Здравко 
I am able to display thumbnails, But Now I want to open actual image (not thumbnails) on click of particular thumbnail image.
Is there any api to get both thumbnail and actual image?