You are so close! All you need to do is select BOTH the home and its max date time, then join back to the topten table on BOTH fields: SELECT tt.* FROM topten tt INNER JOIN (SELECT home, MAX(datetime) AS MaxDateTime FROM topten GROUP BY home) groupedtt ON tt.home = groupedtt.home AND tt.datetime = groupedtt.MaxDateTime
MySQL 8.0 now supports windowing functions, like almost all popular SQL implementations. With this standard syntax, we can write greatest-n-per-group queries: WITH ranked_messages AS ( SELECT m.*, ROW_NUMBER() OVER (PARTITION BY name ORDER BY id DESC) AS rn FROM messages AS m ) SELECT * FROM ranked_messages WHERE rn = 1; This and other approaches … Read more
At first glance… All you need is a GROUP BY clause with the MAX aggregate function: SELECT id, MAX(rev) FROM YourTable GROUP BY id It’s never that simple, is it? I just noticed you need the content column as well. This is a very common question in SQL: find the whole data for the row … Read more
DISTINCT ON is typically simplest and fastest for this in PostgreSQL. (For performance optimization for certain workloads see below.) SELECT DISTINCT ON (customer) id, customer, total FROM purchases ORDER BY customer, total DESC, id; Or shorter (if not as clear) with ordinal numbers of output columns: SELECT DISTINCT ON (2) id, customer, total FROM purchases … Read more