Every customer, even those without orders: every AI model's reply, tested

[5 lines every SQL prompt of ours shares, word for word: the whole prompt, on the methods page]
For every customer, show their name, the date of their first order that wasn't cancelled, and how many orders that weren't cancelled they have. Include customers who have no such orders, with NULL for the date and 0 for the count. Sort by name.
Write one SQLite query and reply with it in a ```sql code block.

SELECT c.name, MIN(o.ordered_on) AS first_order, COUNT(o.id) AS orders
FROM customers c
LEFT JOIN orders o ON o.customer_id = c.id AND o.status <> 'cancelled'
GROUP BY c.id, c.name
ORDER BY c.name

Exactly what this prompt's replies are checked against, with every other prompt of our test runs.

1. GPT-6 Luna

   Passed: Returned the right 8 rows.

   ```sql
   SELECT
     c.name,
     MIN(CASE WHEN o.status <> 'cancelled' THEN o.ordered_on END) AS first_order_date,
     COUNT(CASE WHEN o.status <> 'cancelled' THEN o.id END) AS order_count
   FROM customers AS c
   LEFT JOIN orders AS o
     ON o.customer_id = c.id
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   522 tokens in, 92 out · 2.4 s · $0.0001 · 1 message on Pro · answered by openai/gpt-6-luna via OpenAI · September 27, 2026

2. GLM 5.3 Flash

   Passed: Returned the right 8 rows.

   ```sql
   SELECT
     c.name,
     MIN(o.ordered_on) AS first_order_date,
     COUNT(o.id) AS order_count
   FROM customers c
   LEFT JOIN orders o
     ON o.customer_id = c.id AND o.status != 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   539 tokens in, 70 out · 1.0 s · $0.0001 · 1 message on Pro · answered by z-ai/glm-5.3-flash via Together · September 27, 2026

3. DeepSeek V4.1 Flash

   Passed: Returned the right 8 rows.

   ```sql
   SELECT
     c.name,
     MIN(o.ordered_on) AS first_order_date,
     COUNT(o.id) AS order_count
   FROM customers c
   LEFT JOIN orders o
     ON o.customer_id = c.id
    AND o.status <> 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```
   
   The `status <> 'cancelled'` condition lives in the `ON` clause (not `WHERE`) so customers with no qualifying orders are kept, yielding `NULL` and `0`.

   565 tokens in, 168 out (55 of them reasoning) · 1.7 s · $0.0002 · 1 message on Pro · answered by deepseek/deepseek-v4.1-flash via Novita · September 27, 2026

4. GLM 5.3

   Passed: Returned the right 8 rows.

   ```sql
   SELECT c.name,
          MIN(o.ordered_on) AS first_order_date,
          COUNT(o.id) AS order_count
   FROM customers c
   LEFT JOIN orders o
     ON o.customer_id = c.id AND o.status != 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   562 tokens in, 68 out (1 of them reasoning) · 1.0 s · $0.0007 · 1 message on Pro · answered by z-ai/glm-5.3 via Wafer · September 27, 2026

5. Gemini 3.8 Flash

   Passed: Returned the right 8 rows.

   ```sql
   SELECT 
       c.name,
       MIN(o.ordered_on) AS first_order_date,
       COUNT(o.id) AS non_cancelled_orders_count
   FROM customers c
   LEFT JOIN orders o 
       ON c.id = o.customer_id 
       AND o.status != 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   566 tokens in, 96 out · 3.7 s · $0.0008 · 1 message on Pro · answered by google/gemini-3.8-flash via Google · September 27, 2026

6. Claude Haiku 4.5

   Passed: Returned the right 8 rows.

   ```sql
   SELECT 
     c.name,
     MIN(o.ordered_on) AS first_order_date,
     COUNT(o.id) AS order_count
   FROM customers c
   LEFT JOIN orders o ON c.id = o.customer_id AND o.status != 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   569 tokens in, 90 out · 2.1 s · $0.0010 · 1 message on Pro · answered by anthropic/claude-haiku-4.5 via Amazon Bedrock · September 27, 2026

7. GPT-6 Sol

   Passed: Returned the right 8 rows.

   ```sql
   SELECT
       c.name,
       MIN(o.ordered_on) AS first_order_date,
       COUNT(o.id) AS non_cancelled_order_count
   FROM customers AS c
   LEFT JOIN orders AS o
       ON o.customer_id = c.id
      AND o.status <> 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   522 tokens in, 80 out · 1.4 s · $0.0018 · 1 message on Pro · answered by openai/gpt-6-sol via OpenAI · September 27, 2026

8. DeepSeek V4 Pro

   Passed: Returned the right 8 rows.

   ```sql
   SELECT c.name,
          MIN(o.ordered_on) AS first_order_date,
          COUNT(o.id) AS non_cancelled_orders
   FROM customers c
   LEFT JOIN orders o
     ON o.customer_id = c.id
    AND o.status <> 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   537 tokens in, 385 out (307 of them reasoning) · 4.5 s · $0.0028 · 1 message on Pro · answered by deepseek/deepseek-v4-pro-0813 via Venice · September 27, 2026

9. Claude Sonnet 5

   Passed: Returned the right 8 rows.

   ```sql
   SELECT
       c.name,
       MIN(o.ordered_on) AS first_order_date,
       COUNT(o.id) AS order_count
   FROM customers c
   LEFT JOIN orders o
       ON o.customer_id = c.id
       AND o.status != 'cancelled'
   GROUP BY c.id, c.name
   ORDER BY c.name;
   ```

   859 tokens in, 129 out · 2.2 s · $0.0030 · 1 message on Pro · answered by anthropic/claude-sonnet-5 via Claude Platform on AWS · September 27, 2026

10. Kimi K3

    Passed: Returned the right 8 rows.

    ```sql
    SELECT
        c.name,
        MIN(CASE WHEN o.status != 'cancelled' THEN o.ordered_on END) AS first_order_date,
        COUNT(CASE WHEN o.status != 'cancelled' THEN o.id END) AS order_count
    FROM customers c
    LEFT JOIN orders o ON o.customer_id = c.id
    GROUP BY c.id, c.name
    ORDER BY c.name;
    ```

    609 tokens in, 102 out (6 of them reasoning) · 2.5 s · $0.0031 · 1 message on Pro · answered by moonshotai/kimi-k3 via Together · September 27, 2026

11. Grok 4.7

    Passed: Returned the right 8 rows.

    ```sql
    SELECT
      c.name,
      MIN(o.ordered_on) AS first_order_date,
      COUNT(o.id) AS order_count
    FROM customers c
    LEFT JOIN orders o
      ON o.customer_id = c.id
     AND o.status != 'cancelled'
    GROUP BY c.id
    ORDER BY c.name;
    ```

    1,768 tokens in, 472 out (398 of them reasoning) · 5.3 s · $0.0037 · 1 message on Pro · answered by x-ai/grok-4.7 via xAI · September 27, 2026

12. Claude Sonnet 5.5

    Passed: Returned the right 8 rows.

    ```sql
    SELECT c.name,
           MIN(o.ordered_on) AS first_order_date,
           COUNT(o.id) AS order_count
    FROM customers c
    LEFT JOIN orders o
           ON o.customer_id = c.id
          AND o.status <> 'cancelled'
    GROUP BY c.id, c.name
    ORDER BY c.name;
    ```
    
    The status filter is in the `ON` clause so customers with no qualifying orders are kept, giving NULL for the date and 0 for the count. This assumes cancelled orders have the status `'cancelled'`, so adjust the string if your data uses different casing or wording.

    863 tokens in, 213 out · 2.3 s · $0.0043 · 1 message on Pro · answered by anthropic/claude-sonnet-5.5 via Anthropic · September 28, 2026

13. Gemini 3.1 Pro

    Passed: Returned the right 8 rows.

    ```sql
    SELECT 
        c.name, 
        MIN(o.ordered_on) AS first_order_date, 
        COUNT(o.id) AS order_count
    FROM customers c
    LEFT JOIN orders o 
        ON c.id = o.customer_id AND o.status != 'cancelled'
    GROUP BY c.id
    ORDER BY c.name;
    ```

    566 tokens in, 645 out (558 of them reasoning) · 7.9 s · $0.0089 · 1 message on Pro · answered by google/gemini-3.1-pro-preview via Google · September 27, 2026

14. GPT-6 Astra

    Passed: Returned the right 8 rows.

    ```sql
    SELECT
        c.name,
        MIN(o.ordered_on) AS first_order_date,
        COUNT(o.id) AS order_count
    FROM customers AS c
    LEFT JOIN orders AS o
        ON o.customer_id = c.id
        AND o.status <> 'cancelled'
    GROUP BY c.id, c.name
    ORDER BY c.name;
    ```

    522 tokens in, 77 out · 2.9 s · $0.0091 · 1 message on Pro · answered by openai/gpt-6-astra via OpenAI · September 27, 2026

15. Claude Opus 5.5

    Passed: Returned the right 8 rows.

    ```sql
    SELECT
        c.name,
        MIN(o.ordered_on) AS first_order_date,
        COUNT(o.id)       AS order_count
    FROM customers AS c
    LEFT JOIN orders AS o
        ON o.customer_id = c.id
       AND o.status <> 'cancelled'
    GROUP BY c.id, c.name
    ORDER BY c.name;
    ```
    
    This assumes cancelled orders have the status `'cancelled'` and that `ordered_on` is stored in ISO format (YYYY-MM-DD), so the earliest date is also the smallest text value.

    861 tokens in, 260 out (59 of them reasoning) · 5.0 s · $0.0094 · 1 message on Pro · answered by anthropic/claude-opus-5.5 via Claude Platform on AWS · September 27, 2026

16. Claude Fable 5.1

    Passed: Returned the right 8 rows.

    ```sql
    SELECT c.name,
           MIN(o.ordered_on) AS first_order_date,
           COUNT(o.id)       AS order_count
    FROM customers c
    LEFT JOIN orders o
           ON o.customer_id = c.id
          AND o.status <> 'cancelled'
    GROUP BY c.id, c.name
    ORDER BY c.name;
    ```
    
    The status filter is in the `ON` clause (not `WHERE`) so customers with only cancelled orders or no orders are still kept, yielding `NULL` and `0`.

    861 tokens in, 185 out · 5.2 s · $0.0199 · 1 message on Pro · answered by anthropic/claude-fable-5.1 via Anthropic · September 27, 2026

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