Off the
Clock

The printouts on the floor are yours. I emptied the folder onto the rug. The table had dinner on it and the dinner was staying where it was. The robe is not a statement. The flat is cold. Page one is the LEFT JOIN where you put the filter in the WHERE. You asked for every customer and their shipped orders and you got two customers. Bruno and Dmitri are still customers. They have been customers this whole time. I have marked it. I am not sending it tonight. You would only fix it badly at midnight.

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

I did not bring the folder. The bag it lives in came, and it was inside the bag. Page two. NOT IN, with a NULL sitting in the subquery. Zero rows came back and you read zero rows as good news. Aiko and Bruno are in that subquery, so for them NOT IN is false. Carla and Dmitri are not, but order 204 is a guest checkout with a NULL customer id, so their last comparison comes back unknown. WHERE keeps only what is true. NOT EXISTS returns the two. I am going back in the water. Read the note when I am not looking at you.

SELECT c.customer_id, c.name
FROM customers c
WHERE c.customer_id NOT IN (SELECT o.customer_id FROM orders o)
ORDER BY c.customer_id;

The book in my lap is not work and the chair is not mine. I was going to give the evening to both of them. Then I remembered the revenue number you sent finance on Wednesday and I have been sitting here with the folder open since. You joined orders to line items and summed the order total. 455.00. The true figure is 245.00. Order 101 is worth 120.00 and got counted twice. Somebody approved that. I would like the name and the approval timestamp. I will be up anyway.

SELECT SUM(o.order_total) AS revenue
FROM orders o
JOIN line_items li ON li.order_id = o.order_id;

In uniform, in front of the room. I put your count on the board this morning with your name cut off the top. COUNT(*) says Aiko has five. COUNT(DISTINCT o.order_id) says two. Both are right. They count different things. Five is line items. Two is orders. You picked five because five is bigger. Carla has no orders at all and still reads one under COUNT(*). The folder went back in my bag at the bell. You worked it out from the board before I finished and did not say so. Good.

SELECT c.name,
       COUNT(*)                   AS count_star,
       COUNT(o.order_id)          AS count_order_id,
       COUNT(DISTINCT o.order_id) AS count_distinct_orders
FROM customers c
LEFT JOIN orders     o  ON o.customer_id = c.customer_id
LEFT JOIN line_items li ON li.order_id   = o.order_id
GROUP BY c.name;

Two hundred metres down. No signal, and nobody to ask me anything. I brought a novel. The folder was under the novel. Page five is your running total. Payments 2 and 3 both landed on 2026-03-02, which makes them peers. Under a bare ORDER BY the frame defaults to RANGE, and RANGE gives a row its peers as well, so payment 2 at 50.00 reports 225.00. Row order on a screen is not a clock. ROWS and a tiebreaker give it 150.00. Write the frame out. The jellyfish are on a schedule. I had twenty minutes. I spent nine of them writing the frame clause out in longhand for you.

SELECT payment_id, paid_on, amount,
       SUM(amount) OVER (ORDER BY paid_on) AS running_total
FROM payments
ORDER BY payment_id;

Sleeper train, 23:00, compartment to myself until you looked. The December close is on the table because it would not fit in my bag with the folder, so the folder is on the bunk. Your BETWEEN found three orders and 245.00. December has four and 745.00. Order 104 was placed at 2026-12-31 23:45:10, and BETWEEN’s upper bound is midnight at the start of the 31st, so a 500.00 order fell off the end of the year. You noticed the total looked light. You were right and you shipped it anyway. Next time the total looks light, count December by hand before you send it.

SELECT COUNT(*) AS orders, SUM(order_total) AS revenue
FROM orders
WHERE placed_at BETWEEN DATE '2026-12-01' AND DATE '2026-12-31';

Last page. I lost a bet, the bet had witnesses, and the costume department keeps records. You found the fanout and you tried to fix it with SUM(DISTINCT). 450.00 was obviously wrong, so somebody would have caught it. 165.00 looks like money and gets filed. DISTINCT inside an aggregate deduplicates values. It does not deduplicate rows. Orders 101 and 102 are both 120.00, so one of them is simply gone. 285.00 is the number. The folder is on your desk. Page seven is on top, face up, marked. Fix it before I get in.

SELECT SUM(DISTINCT o.order_total) AS revenue
FROM orders o
JOIN line_items li ON li.order_id = o.order_id;