SQL總和，每

Question

天

用戶唯一我有一個Postgres的表看起來像這樣：

id | user_id | state | created_at 

狀態可以是下列任何一種：

new, paying, paid, completing, complete, payment_failed, completion_failed 

我需要一個返回與下面的報告語句：按日期

1. 款項支付所有國家的
2. 按日期之所有已完成的狀態
3. 和所有新的，支付，完成日期的狀態只有一個每用戶每天要算
4. 總和的所有payment_failed，按日期completion_failed與每個用戶每天只有一個要算

到目前爲止，我有這樣的：

SELECT 
    DATE(created_at) AS date, 
    SUM(CASE WHEN state = 'complete' THEN 1 ELSE 0 END) AS complete, 
    SUM(CASE WHEN state = 'paid' THEN 1 ELSE 0 END) AS paid 
FROM orders 
WHERE created_at BETWEEN ? AND ? 
GROUP BY DATE(created_at) 

的進展和失敗國家的總和爲加入這選擇很容易：

SUM(CASE WHEN state IN('new','paying','completing') THEN 1 ELSE 0 END) AS in_progress, 
SUM(CASE WHEN state IN('payment_failed','completion_failed') THEN 1 ELSE 0 END) AS failed 

但我無法弄清楚如何每天只有一個user_id in_progress和失敗的狀態進行計數。

我需要這個的原因是在我們的統計數據中操縱失敗率，因爲許多觸發失敗或不完整訂單的用戶繼續觸發更多事件，從而導致我們的失敗率上升。

感謝您提前。

Answer 1

試着這麼做：

SELECT 
    DATE(created_at) AS date, 
    SUM(CASE WHEN state = 'complete' THEN 1 ELSE 0 END) AS complete, 
    SUM(CASE WHEN state = 'paid' THEN 1 ELSE 0 END) AS paid, 
    COUNT(DISTINCT CASE WHEN state IN('new','paying','completing') THEN user_id ELSE NULL END) AS in_progress, 
    COUNT(DISTINCT CASE WHEN state IN('payment_failed','completion_failed') THEN user_id ELSE NULL END) AS failed 
FROM orders 
WHERE created_at BETWEEN ? AND ? 
GROUP BY DATE(created_at); 

主要思想 - COUNT (DISTINCT ...)將依靠獨特user_id並不會算NULL值。

詳情：aggregate functions，4.2.7. Aggregate Expressions

具有相同的風格計數整個查詢和簡化CASE WHEN ...：

SELECT 
    DATE(created_at) AS date, 
    COUNT(CASE WHEN state = 'complete' THEN 1 END) AS complete, 
    COUNT(CASE WHEN state = 'paid' THEN 1 END) AS paid, 
    COUNT(DISTINCT CASE WHEN state IN('new','paying','completing') THEN user_id END) AS in_progress, 
    COUNT(DISTINCT CASE WHEN state IN('payment_failed','completion_failed') THEN user_id END) AS failed 
FROM orders 
WHERE created_at BETWEEN ? AND ? 
GROUP BY DATE(created_at); 

Answer 2

SELECT created_at::date AS the_date 
     ,SUM(CASE WHEN state = 'complete' THEN 1 ELSE 0 END) AS complete 
     ,SUM(CASE WHEN state = 'paid' THEN 1 ELSE 0 END) AS paid 
     ,COUNT(DISTINCT CASE WHEN state IN('new','paying','completing') 
         THEN user_id ELSE NULL END) AS in_progress 
     ,COUNT(DISTINCT CASE WHEN state IN('payment_failed','completion_failed') 
         THEN user_id ELSE NULL END) AS failed 
FROM orders 
WHERE created_at BETWEEN ? AND ? 
GROUP BY created_at::date 

我用the_date作爲別名，因爲它是不明智的（同時允許）使用關鍵詞日期作爲標識符。

你可以使用complete和paid類似的技術，一個是作爲其他有好：

COUNT(CASE WHEN state = 'complete' THEN 1 ELSE NULL END) AS complete