Searching for an artist in Listenbrainz

ListenBrainz
1

Hi, is it possible to search for an artist in Listenbrainz?
I know if you see an artist listed in your listens (or someone else’s) you can click on the name and get to a nice page which lists all their albums, top listeners etc… but a search, or even a link from Musicbrainz would be good.
Thanks.

2 Likes
2

Hi!

Sadly it isn’t possible yet, but we’re very aware that this feature is missing. We hope to work on it in the short term, but don’t have a date in mind yet for when this might come.

7 Likes
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Thanks for the reply, glad it is on the cards at some point - brilliant! That is the sort of thing that would help LB with replacing Last.FM. Artist pages are relatively new, right?

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Yes and the second we released those, we realized. Shit, now we need search. lol.

At least we can use the MB search backend for this feature, so its not very hard.

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Good point, I was missing that too. It would be great to have that on LB.

A similar thing I have noticed is in the other direction: When viewing an artist on MB, it’s impossible to get to the artist page on LB. Releases have a “Listen on LB” button, but on the artist itself if would be nice to have a similar button to jump to the artist page.

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there’s a ticket for this, actually~

speaking of tickets, I’m hoping that in the process we finish adding a unified search to MusicBrainz as well, and primarily use that for ListenBrainz (but still of course allow specifying an entity type)

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In the mean time, what you can do is look up an artist on MB and then change the url from https://musicbrainz.org/artist/… into https://listenbrainz.org/artist/…

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Thanks for the hint, that works well. Not ideal of course, but it’s possible to access the artist page this way :+1:

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Repurposed an old bookmarklet:

javascript:switch(!0)%7Bcase%20-1%3Clocation.toString().indexOf(%22musicbrainz.org%2Fartist%2F%22)%3Aopen(%22https%3A%2F%2Flistenbrainz.org%2Fartist%2F%22%2Blocation.pathname.toString().slice(8%2C44))%3Bbreak%3Bcase%20-1%3Clocation.toString().indexOf(%22musicbrainz.org%2Frelease-group%2F%22)%3Aopen(%22https%3A%2F%2Flistenbrainz.org%2Falbum%2F%22%2Blocation.pathname.toString().slice(15%2C51))%7D%3Bvoid+0
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AMAZING! Thank you! :beers: