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Sort NavigationTree alphabetically #112

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merged 7 commits into from
Jun 21, 2023
+56 −0
Merged

Sort NavigationTree alphabetically #112

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2483e74
add tree sorting func
prestonbourne Jun 20, 2023
28f0d69
Merge branch 'main' into pres.sort-nav-tree
prestonbourne Jun 20, 2023
344f491
add accordion
prestonbourne Jun 20, 2023
5ea8f66
Tests for `getNavigationTree` (#110)
prestonbourne Jun 21, 2023
d62e01d
Merge branch 'main' into pres.sort-nav-tree
prestonbourne Jun 21, 2023
1bb7a01
rewrite sort
prestonbourne Jun 21, 2023
fce23a9
new line
prestonbourne Jun 21, 2023
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8 changes: 8 additions & 0 deletions examples/basic/components/accordion/docs.mdx
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Adding a simple accordion to show the sorting

Original file line number Diff line number Diff line change
@@ -0,0 +1,8 @@
---
title: 'Accordion'
description: 'UI component named after a funky instrument 🪗'
path: 'Content/Accordion'
---
import {Accordion} from '.'

<Accordion/>
15 changes: 15 additions & 0 deletions examples/basic/components/accordion/index.tsx
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Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
import { ReactElement } from 'react'

interface AccordionProps {
color?: string
children: ReactElement
}

export function Accordion({ children }: AccordionProps) {
return (
<details>
<summary>Hello 👋</summary>
Nice meeting you! How's everything going?
</details>
)
}
Comment on lines +8 to +15
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Using the built in HTML elements for this, I imagine down the line we'll use the components package Dylan is working on

25 changes: 25 additions & 0 deletions packages/swingset/src/get-nav-tree.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -4,6 +4,7 @@ import {
ComponentNode,
CategoryNode,
NavigationTree,
NavigationNode,
} from './types.js'

export function getNavigationTree(
Expand Down Expand Up @@ -83,5 +84,29 @@ export function getNavigationTree(
}

const tree = Array.from(categories.values())

tree.sort(compareTitleSort)

return tree
}

function compareTitleSort<T extends CategoryNode | NavigationNode>(
a: T,
b: T
): -1 | 0 | 1 {
const aHasChildren = 'children' in a && a.children.length > 0
const bHasChildren = 'children' in b && b.children.length > 0
if (aHasChildren) {
a.children.sort(compareTitleSort)
}
if (bHasChildren) {
b.children.sort(compareTitleSort)
}
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I'm thinking this might not be the proper place to call sort recursively, as each element might get passed to the compare function more than once, which would result in us sorting an element's children multiple times. We'll likely want to separate out traversing the tree from the actual sorting.

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Are you suggesting, something like:

  • Create a toBeSorted = Map()
  • Traverse the tree, while checking which nodes have children
  • If hasChildren == true. Store reference to the node in toBeSorted
  • Loop through the map and sort

if (a.title > b.title) {
return 1
}
if (b.title > a.title) {
return -1
}
return 0
}
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Made the compare function recursive per our discussion. @BRKalow

This way the tree will be sorted no matter how deep the structure gets as long as we keep using the children key to handle nesting