Re: License for libdrm packages

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I have read that article in ArchWiki. I understand that point that MIT licences are all custom because of individual copyright line. But then I do not understand when should I use license=('MIT') instead of license=('custom')?
I have read that MIT is a set of licenses, but it is kinda unclear. I guess that if there is clear text that it is a MIT license, then I use MIT, otherwise for MIT-style licence I just use custom. Am I correct?

23.05.2019, 20:06, "mpan" <archml-y1vf3axu@xxxxxxx>:
>>  Hello. I was repacking amdgpu-pro deb files and when I started converting licences, I have noticed that libdrm* packages have a MIT Licence text in copyright file. I decided to check if AUR/libdrm-git and Extra/libdrm uses MIT licence, but they don't. I contacted Lone_Wolf (maintainer of libdrm-git) and he said that he used a licence from Extra/libdrm.
>>  Should not it be changed to MIT instead of custom?
>
>   “MIT-style license” is a class of licenses, not a specific one. Each
> software using MIT-style licensing is having its own, independent
> license text. While in practice they may be nearly identical (modulo
> copyright line), they’re in fact separate licenses. The topic is
> discussed on the wiki:
> <https://wiki.archlinux.org/index.php/Pkgbuild#license>.
>
> In particular the terms include:
>  ------
>   The above copyright notice and this permission notice (including the
>   next paragraph) shall be included in all copies or substantial
>   portions of the Software.
>  ------
> Which means that if multiple MIT-licensed pieces of software are
> combined, the complete notice of each of them has to be included in the
> final work. And this is exactly what happens in case of libdrm:
> <https://git.archlinux.org/svntogit/packages.git/tree/trunk/COPYING?h=packages/libdrm&id=b080357775c306e74a4257099ab4197604c4f57b>.



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