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Improve sort & filter of language options in search (#5709)
authored by
Paul Frazee
and committed by
GitHub
5ebb630c
9a769b9a
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src/view/screens/Search/Search.tsx
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-3
src/view/screens/Search/Search.tsx
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import AsyncStorage from '@react-native-async-storage/async-storage'
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import {useFocusEffect, useNavigation} from '@react-navigation/native'
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import {LANGUAGES} from '#/lib/../locale/languages'
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import {APP_LANGUAGES, LANGUAGES} from '#/lib/../locale/languages'
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import {createHitslop} from '#/lib/constants'
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import {HITSLOP_10} from '#/lib/constants'
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import {useNonReactiveCallback} from '#/lib/hooks/useNonReactiveCallback'
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key: '*',
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},
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].concat(
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LANGUAGES.filter(l => Boolean(l.code2))
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LANGUAGES.filter(
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(lang, index, self) =>
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Boolean(lang.code2) && // reduce to the code2 varieties
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index === self.findIndex(t => t.code2 === lang.code2), // remove dupes (which will happen)
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)
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.map(l => ({
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label: l.name,
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inputLabel: l.name,
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value: l.code2,
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key: l.code2 + l.code3,
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}))
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.sort(a => (contentLanguages.includes(a.value) ? -1 : 1)),
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.sort((a, b) => {
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// prioritize user's languages
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const aIsUser = contentLanguages.includes(a.value)
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const bIsUser = contentLanguages.includes(b.value)
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if (aIsUser && !bIsUser) return -1
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if (bIsUser && !aIsUser) return 1
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// prioritize "common" langs in the network
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const aIsCommon = !!APP_LANGUAGES.find(al => al.code2 === a.value)
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const bIsCommon = !!APP_LANGUAGES.find(al => al.code2 === b.value)
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if (aIsCommon && !bIsCommon) return -1
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if (bIsCommon && !aIsCommon) return 1
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// fall back to alphabetical
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return a.label.localeCompare(b.label)
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}),
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)
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}, [_, contentLanguages])
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