Danbooru

Taken down posts.

Posted under Bugs & Features

John Ress

Hi, sometimes I'm searching images in danbooru, but there is no posts with it. Then I search for source and danbooru NTISAS and it shows that there is danbooru post with this image, but it's taken down (for example post #6800802). And sometimes I just use NTISAS, open the danbooru post in new tab, then after ~40 images I see this: "This page has been removed because of a takedown request.".
I'm feeling kind of sad, when I see this. Is there any way to take the source from post id or say to devs to mark the source?

Reply
Vasi1

Valiran9 said:

Would it be possible to leave the source it was uploaded from (Pixiv, DeviantArt, etc.) so we can see the art there?

That would be a sensible thing to do, I support this. On several occasions I just wanted to get to the full image source, because it would not show up in search results.

Reply
Minitajfun

Vasi1 said:

That would be a sensible thing to do, I support this. On several occasions I just wanted to get to the full image source, because it would not show up in search results.

I second that, this option would be very useful.

In the meantime, I leave a simple userscript that does just that below and hope that, in the future, it will become unnecessary.

Script
// ==UserScript==
// @name        Danbooru banned posts sources
// @match       https://danbooru.donmai.us/posts/*
// @grant       none
// @version     1.0
// @author      Minitajfun
// @description Simple script to display source on banned posts
// @run-at      document-end
// ==/UserScript==

xhttp = new XMLHttpRequest();
xhttp.onload = function() {
    res = JSON.parse(this.responseText);
    console.log(res);
    if (res.is_banned) {
        srcEl = document.createElement('p');
        srcElLink = document.createElement('a');
        srcEl.innerText = 'Original source: ';
        srcElLink.innerText = res.source;
        srcElLink.href = res.source;
        srcEl.appendChild(srcElLink);
        document.getElementById('page').appendChild(srcEl);
    }
}
xhttp.open("GET", window.location.href + '.json', true);
xhttp.send();
Reply
1