Fix skill list namespace layout to stack vertically

The namespace was appearing side-by-side with the skill name because
they were in the same flex row. Restructured the layout so the
namespace appears on its own line above, with the icon and skill name
on the row below it.

Co-Authored-By: Paperclip <noreply@paperclip.ing>
This commit is contained in:
Dotta
2026-03-16 18:59:36 -05:00
parent 9e843c4dec
commit b339f923d6

View File

@@ -422,23 +422,25 @@ function SkillList({
className="flex min-w-0 items-center self-stretch pr-2 text-left no-underline"
onClick={() => onSelectSkill(skill.id)}
>
<span className="flex min-w-0 items-center gap-2 self-center">
<Tooltip>
<TooltipTrigger asChild>
<span className="flex h-4 w-4 shrink-0 items-center justify-center text-muted-foreground opacity-75 transition-opacity group-hover:opacity-100">
<SourceIcon className="h-3.5 w-3.5" />
<span className="sr-only">{source.managedLabel}</span>
</span>
</TooltipTrigger>
<TooltipContent side="top">{source.managedLabel}</TooltipContent>
</Tooltip>
<span className="min-w-0 self-center">
{skill.key.includes("/") && (
<span className="block truncate font-mono text-[11px] leading-4 text-muted-foreground">
{skill.key.split("/").slice(0, -1).join("/")}
</span>
)}
<span className="block min-w-0 overflow-hidden text-[13px] font-medium leading-5 [display:-webkit-box] [-webkit-box-orient:vertical] [-webkit-line-clamp:3]">
{skill.name}
<span className="flex min-w-0 items-center gap-2">
<Tooltip>
<TooltipTrigger asChild>
<span className="flex h-4 w-4 shrink-0 items-center justify-center text-muted-foreground opacity-75 transition-opacity group-hover:opacity-100">
<SourceIcon className="h-3.5 w-3.5" />
<span className="sr-only">{source.managedLabel}</span>
</span>
</TooltipTrigger>
<TooltipContent side="top">{source.managedLabel}</TooltipContent>
</Tooltip>
<span className="min-w-0 overflow-hidden text-[13px] font-medium leading-5 [display:-webkit-box] [-webkit-box-orient:vertical] [-webkit-line-clamp:3]">
{skill.name}
</span>
</span>
</span>
</Link>