Home > Database > Mysql Tutorial > How to Find the Oldest Person in Each Group Using SQL?

How to Find the Oldest Person in Each Group Using SQL?

Mary-Kate Olsen
Release: 2025-01-23 22:37:11
Original
515 people have browsed it

How to Find the Oldest Person in Each Group Using SQL?

Efficiently Identifying the Oldest Person in Each Group Using SQL

Database queries often require finding maximum values within specific groups. This example demonstrates how to pinpoint the oldest person in each group from a table with person, group, and age columns.

Solution:

This approach leverages a LEFT JOIN to compare each person's age within their group. Rows without a match represent the oldest person in that group.

Implementation:

<code class="language-sql">SELECT o.*
FROM Persons o
LEFT JOIN Persons b ON o.Group = b.Group AND o.Age < b.Age
WHERE b.Age IS NULL;</code>
Copy after login

Explanation:

  • Persons o: The Persons table is aliased as o (for "oldest").
  • LEFT JOIN Persons b: A LEFT JOIN is performed with the Persons table aliased as b (for "bigger age"). This joins each row in o with rows in the same group (o.Group = b.Group) that have a greater age (o.Age < b.Age).
  • WHERE b.Age IS NULL: This crucial clause filters the results. If a person in o has no match in b (meaning b.Age is NULL), it signifies that no one in their group is older. Therefore, only the oldest person from each group is selected.

Important Considerations:

  • Using an INNER JOIN would incorrectly omit the oldest person from groups with only one member.
  • This method is a highly efficient and widely recommended technique for this type of query. Similar approaches are discussed in "SQL Antipatterns Volume 1: Avoiding the Pitfalls of Database Programming".

The above is the detailed content of How to Find the Oldest Person in Each Group Using SQL?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template