Generatore di Query SQL

Input

Cosa dovrebbe restituire la query?

Monthly revenue per country for 2025, counting only customers who placed more than 3 orders in 2025. Sort by month, then revenue descending.

Tabelle e colonne (opzionale)

orders(id, customer_id, total_cents integer, created_at timestamp) customers(id, country text, signup_at timestamp)

Dialetto SQL

PostgreSQL

Stile della query

Leggibile (CTE, una clausola per riga)

Spiegazione

Breve

Output (estratto)

```sql
WITH active_customers AS (
  SELECT o.customer_id
  FROM orders AS o
  WHERE o.created_at >= DATE '2025-01-01'
    AND o.created_at <  DATE '2026-01-01'
  GROUP BY o.customer_id
  HAVING COUNT(*) > 3
)
SELECT
  DATE_TRUNC('month', o.created_at)::date AS month,
  c.country,
  SUM(o.total_cents) / 100.0            AS revenue
FROM orders AS o
JOIN active_customers AS a ON a.customer_id = o.customer_id
JOIN customers        AS c ON c.id = o.customer_id
WHERE o.created_at >= DATE '2025-01-01'
  AND o.created_at <  DATE '2026-01-01'
GROUP BY 1, 2
ORDER BY month, revenue DESC;
```

**Schema usato così com'è.**

Il CTE individua i clienti con più di tre ordini nel 2025; la query principale conserva solo i loro ordini, tronca ogni ordine al mese e somma il ricavo per mese e paese. L'intervallo di date è semi-aperto, quindi il 31 dicembre viene contato una sola volta…

Input

Cosa dovrebbe restituire la query?

For every customer, their most recent order with its total and date.

Tabelle e colonne (opzionale)

orders(id, customer_id, total_cents integer, created_at timestamp) customers(id, email text)

Dialetto SQL

MySQL / MariaDB

Stile della query

Analitico (funzioni finestra dove utili)

Spiegazione

Dettagliata (clausola per clausola)

Output (estratto)

```sql
WITH ranked AS (
  SELECT
    o.customer_id,
    o.id          AS order_id,
    o.total_cents,
    o.created_at,
    ROW_NUMBER() OVER (PARTITION BY o.customer_id ORDER BY o.created_at DESC, o.id DESC) AS rn
  FROM orders AS o
)
SELECT c.id AS customer_id, c.email, r.order_id, r.total_cents / 100.0 AS total, r.created_at
FROM customers AS c
JOIN ranked AS r ON r.customer_id = c.id AND r.rn = 1
ORDER BY r.created_at DESC;
```

**Schema usato così com'è.**

| Clausola | Cosa fa |
|---|---|
| `ROW_NUMBER() OVER (PARTITION BY … ORDER BY created_at DESC, id DESC)` | numera gli ordini di ogni cliente dal più recente; il tie‑break sull'id rende il risultato deterministico |
| `JOIN ranked … AND r.rn = 1` | mantiene esattamente una riga per cliente |…