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refactor(lib): remove keys and just get a slug from the label

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Lewis Wynne 2026-03-23 17:02:46 +00:00
parent 7dc4bd8a11
commit 456e7e8feb
5 changed files with 26 additions and 30 deletions
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6
src/lib/data/parse.ts
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@ -1,6 +1,7 @@
import { XMLParser } from 'fast-xml-parser';
import type { SpeciesData, CitizenshipData, LanguageData } from './types';
import type { Template, RecordDef, FieldDef, SelectOption } from '../types';
import { slugify } from '../utils/slugify';
const parser = new XMLParser({
ignoreAttributes: false,
@ -51,10 +52,6 @@ export function parseLanguages(xml: string): LanguageData[] {
}));
}
function slugify(label: string): string {
return label.toLowerCase().replace(/[^a-z0-9]+/g, '-').replace(/(^-|-$)/g, '');
}
function parseOptions(field: any): SelectOption[] {
if (!field.option) return [];
return field.option.map((o: any) => ({
@ -65,7 +62,6 @@ function parseOptions(field: any): SelectOption[] {
function parseField(raw: any): FieldDef {
const base = {
key: slugify(raw['@_label']),
label: raw['@_label']
};
const type = raw['@_type'];