Template talk:Label
Would it be possible to implement the language code?
 | This template does not require a rating on Wikipedia's content assessment scale. It is of interest to the following WikiProjects: | |||||||
| ||||||||
Language code
Would it be possible to implement the language code?
We would be interested in using it to display the label in a secondary language. Thanks! Syced (talk) 02:38, 21 May 2015 (UTC)
- Another use waiting on this functionality. Bazj (talk) 08:54, 19 July 2015 (UTC)
- Another example would be for British people where if my display language is en-gb for Q7540 it would show "Aubergine" instead of "Eggplant" Back ache (talk) 10:50, 7 December 2021 (UTC)
LinkedLabel variant
ResolvedSolution:
{{wikidata|label|linked|Q1121807}}- mainsail
Solution 2:
{{LinkedLabel|Q1121807}}- mainsail
Original request below:
{{wikidata:Template:LinkedLabel|Q1121807}}- {{wikidata:Template:LinkedLabel|Q1121807}}
Don't need to show Q ids here: Template:Recreational yacht D1gggg (talk) 17:10, 22 October 2017 (UTC)
- Links should lead to wikipedia sitelinks, not wikidata D1gggg (talk) 17:11, 22 October 2017 (UTC)
Something should be returned when the label is not found
See Template:Label/testcases#Invalid Q number. The template silently returns nothing instead of an error message and a tracking category. What should be returned instead of nothing? – Jonesey95 (talk) 01:37, 30 December 2025 (UTC)
Content Disclaimer
Informasi ini disarikan dari Wikipedia dan disajikan kembali untuk tujuan edukasi. Konten tersedia di bawah lisensi CC BY-SA 3.0. Kami tidak bertanggung jawab atas ketidakakuratan data yang bersumber dari kontribusi publik tersebut.
- The information displayed on this website is sourced in part or in whole from Wikipedia and has been adapted for the purpose of restating it. We strive to provide accurate and relevant information, however:
- There is no guarantee of absolute accuracy. Wikipedia is an open, collaborative project that can be edited by anyone, so information is subject to change.
- It is not intended to constitute professional advice. The content displayed is for informational and educational purposes only. For important decisions (e.g., medical, legal, or financial), please consult a professional.
- Content copyright. Wikipedia is licensed under the Creative Commons Attribution-ShareAlike License (CC BY-SA). This means that content may be reused with appropriate attribution and shared under a similar license.
- Responsible use. Any risk arising from the use of information from this website is entirely the responsibility of the user.