确实是个常见的低级错误，正确写法应为 32nd，而不是 32th。
Creating ordinal suffixes
I want to keep this straightforward and offer a quick fix along with an explanation of the rules. I could suggest writing [[NEWS.CONTENT_HTML]]quot;32nd anniversary[[NEWS.CONTENT_HTML]]quot; or [[NEWS.CONTENT_HTML]]quot;32-year anniversary[[NEWS.CONTENT_HTML]]quot; to avoid any suffix issues. For Chinese posters, using numerals only might be effective too. I’ll include a small function for generating ordinal suffixes in multiple languages and share a QA checklist.
I should point out that the correct form is [[NEWS.CONTENT_HTML]]quot;32nd,[[NEWS.CONTENT_HTML]]quot; with exceptions for 11th, 12th, and 13th. I need to provide examples, followed by a concise code snippet for generating suffixes in either JS or Python.Providing design suggestions
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I need to include short code blocks in my guidance. To avoid using suffixes, I could suggest writing [[NEWS.CONTENT_HTML]]quot;32周年[[NEWS.CONTENT_HTML]]quot; or spelling it out as [[NEWS.CONTENT_HTML]]quot;32nd Anniversary.[[NEWS.CONTENT_HTML]]quot; For bilingual Chinese posters, using [[NEWS.CONTENT_HTML]]quot;32nd anniversary[[NEWS.CONTENT_HTML]]quot; or [[NEWS.CONTENT_HTML]]quot;the club's 32nd birthday[[NEWS.CONTENT_HTML]]quot; could work well.
I should outline some QC steps: proofreading, getting a language check by a native speaker, and running everything through tools like Grammarly or LanguageTool.
I'll also offer to help update poster copy or check others. Keeping formatting lightweight is essential, with a bullet list being acceptable. Let's get this produced!
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确实是个常见的低级错误，正确写法应为 32nd，而不是 32th。
要点速记
th；否则看个位：1→st，2→nd，3→rd，其余→th。
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nd → 32nd。可用的小函数（避免再错）
function ordinal(n) { const mod100 = n % 100; if (mod100 [[NEWS.CONTENT_HTML]]gt;= 11 [[NEWS.CONTENT_HTML]]amp;[[NEWS.CONTENT_HTML]]amp; mod100 [[NEWS.CONTENT_HTML]]lt;= 13) return n + 'th'; switch (n % 10) {
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case 1: return n + 'st'; case 2: return n + 'nd'; case 3: return n + 'rd'; default: return n + 'th'; }}def ordinal(n: int) -[[NEWS.CONTENT_HTML]]gt; str: if 11 [[NEWS.CONTENT_HTML]]lt;= n % 100 [[NEWS.CONTENT_HTML]]lt;= 13: suffix = [[NEWS.CONTENT_HTML]]quot;th[[NEWS.CONTENT_HTML]]quot; else: suffix = {1: [[NEWS.CONTENT_HTML]]quot;st[[NEWS.CONTENT_HTML]]quot;, 2: [[NEWS.CONTENT_HTML]]quot;nd[[NEWS.CONTENT_HTML]]quot;, 3: [[NEWS.CONTENT_HTML]]quot;rd[[NEWS.CONTENT_HTML]]quot;}.get(n % 10, [[NEWS.CONTENT_HTML]]quot;th[[NEWS.CONTENT_HTML]]quot;) return f[[NEWS.CONTENT_HTML]]quot;{n}{suffix}[[NEWS.CONTENT_HTML]]quot;设计/校对建议
需要的话我可以帮你把整套周年用语/模板整理成可复用清单，或批量检查其他物料里的序数用法。
