MySQL Relational Data Modeling

How it works

The workshop has one row per depot, one row per kind of part and one row per recorded job. A part belongs to one depot; a job refers to a part and a volunteer. Keys connect those records without copying a depot name into every job. A relational model lets the same facts support several questions while constraints protect permitted relationships. The example inspects counts independently so no join accidentally multiplies them.

Run in a disposable MySQL lab

Use Oracle MySQL 8.4 with InnoDB, not Bhylu’s SQLite playground. The examples were reviewed on MySQL 8.4.8. XAMPP may provide MariaDB, which has different feature and collation behaviour. Check your actual server before running the script. These lessons do not require a change to Bhylu’s PHP, Apache or normal database service.

1. On a separate lab instance, create the empty bhylu_mysql_lab schema. Stop if it already exists; do not overwrite an existing database.
2. Select that schema and import the complete repair-parts seed once. It sets the documented session mode, utf8mb4 collation, UTC and English date-name locale.
3. Run this lesson in a fresh copy of the seed, not after another lesson’s writes. Use the command-line client or a compatible editor. DELIMITER lines are client instructions for stored-program examples.

Download this complete lesson script. Its setup is the linked seed; review the dataset before changing it. Administrative and destructive examples are restricted to the specifically named disposable objects. Every script repeats the session settings because each new connection has its own session. No MySQL browser runner is currently available.

Example

SET NAMES utf8mb4 COLLATE utf8mb4_0900_ai_ci;
SET SESSION sql_mode = 'STRICT_TRANS_TABLES,ONLY_FULL_GROUP_BY,ERROR_FOR_DIVISION_BY_ZERO,NO_ENGINE_SUBSTITUTION';
SET SESSION time_zone = '+00:00';
SET SESSION lc_time_names = 'en_US';
SELECT (SELECT COUNT(*) FROM depots) AS depots,
       (SELECT COUNT(*) FROM parts) AS part_types,
       (SELECT COUNT(*) FROM volunteers) AS volunteers,
       (SELECT COUNT(*) FROM jobs) AS jobs;

Expected result

3 | 4 | 3 | 3

The three subqueries count their own tables and return one summary row. West is a depot without parts, and Cotton cord has zero stock; these deliberate cases help test missing relationships and zero values later. The jobs table contains both completed and open work. A job is not itself a stock movement in this seed, so do not subtract every job blindly from current quantity.