Article “Wolfgang Amberger” in German Wikipedia has 11.8 points for quality (as of July 1, 2025). The article contains 1 references and 7 sections.
In June 2025 the article “Wolfgang Amberger” was edited by 1 authors in German Wikipedia and written by 1 authors in all languages.
Since the creation of article “Wolfgang Amberger”, its content was written by 6 registered users of German Wikipedia and edited by 6 registered Wikipedia users in all languages.
The article is cited 9 times in German Wikipedia and cited 9 times in all languages.
The highest Authors Interest rank from 2001:
The highest popularity rank from 2008:
There is 1 language version for this article in the WikiRank database (of the considered 55 Wikipedia language editions).
The quality and popularity assessment was based on Wikipédia dumps from July 1, 2025 (including revision history and pageviews for previous years).
| Best Rank | German: #779858 02.2021 |
Global: #4635732 01.2008 |
| Best Rank | German: #15564 02.2021 |
Global: #272427 02.2021 |
On 16 March 2026 in multilingual Wikipedia, Internet users most often read articles on the following topics: 98th Academy Awards, Michael B. Jordan, Sinners, Jessie Buckley, Paul Thomas Anderson, Sean Penn, One Battle After Another, Amy Madigan, Hamnet, Timothée Chalamet.
In German Wikipedia the most popular articles on that day were: Oscarverleihung 2026, One Battle After Another, Blood & Sinners, Michael B. Jordan, Jessie Buckley, Jürgen Habermas, Sean Penn, Banksy, Liste der größten Auslegerbrücken, Amy Madigan.
The WikiRank project is intended for automatic relative evaluation of the articles in the various language versions of Wikipedia. At the moment the service allows to compare over 44 million Wikipedia articles in 55 languages. Quality scores of articles are based on Wikipedia dumps from July, 2025. When calculating current popularity and AI of articles data from June 2025 was taken into account. For historical values of popularity and AI WikiRank used data from 2001 to 2025... More information
$$Quality=1/c∑↙{i=1}↖c nm_i-RS=1/5(9.32+1.43+0+33.33+15.13)$$
where:
The results of the calculations may be slightly different because of rounded values of normalized metrics in formulas
