使用MySQL使用LEFT JOIN返回結果，其中一個表不包含記錄

Question

這是一個隨機問題，我一直在使用MySQL中的多個查詢和一小部分PHP進行此操作。

我有一個MySQL查詢：

SELECT * 
FROM (reviews LEFT JOIN orders ON reviews.orderid = orders.orderid) 
LEFT JOIN customers ON orders.custid = customers.id 
WHERE customers.email = '$email' 

返回從特定客戶的所有產品評價。

我想知道構造一個單獨的查詢，它將返回尚未審查的所有客戶訂單。也就是說，在評論表中沒有特定orderid的記錄。

所以......

SELECT * 
FROM orders LEFT JOIN customers ON orders.custid = customers.id 
WHERE customers.email = '$email' 

...這將返回所有客戶的訂單，但後來我想也許是使用其他WHERE子句和LEFT JOIN，因此現在只有訂單ID的，沒有相應的審查記錄回。我一直在試圖找到一些能夠做到這一點的東西，現在沒有任何運氣。似乎沒有任何工作。如上所述，我已經設法使用PHP/MySQL的組合，但它不是非常高效，所以我想知道是否有人有任何建議？

在此先感謝。

瑞安

Answer 1

SELECT * 
FROM customers 
    JOIN orders 
    ON orders.custid = customers.id 
WHERE customers.email = '$email' 
    AND orders.orderid NOT IN 
     (SELECT orderid FROM reviews) 

...也許

Answer 2

你嘗試呢？

SELECT orders.* FROM orders 
LEFT JOIN customers ON orders.custid = customers.id 
LEFT JOIN reviews ON reviews.orderid=orders.id 
WHERE customers.email = '$email' AND reviews.orderid IS NULL 

，應返回所有未查看訂單郵件$電子郵件的客戶

Answer 3

試試這個：

SELECT * 
FROM orders a LEFT JOIN customers b ON a.custid = b.id 
WHERE b.email = '$email' AND 
     a.orderid NOT IN (SELECT orderid FROM reviews) 

Answer 4

您可以LEFT JOIN客戶/訂單到reviews，並沒有相應的review選擇行：

SELECT * 
FROM orders LEFT JOIN customers ON orders.custid = customers.id 
LEFT JOIN reviews ON reviews.orderid = orders.orderid 
WHERE customers.email = '$email' 
AND reviews.orderid IS NULL 

Answer 5

這將工作，但可能會很慢。

SELECT * 
FROM orders 
LEFT JOIN customers 
    ON orders.custid = customers.id 
WHERE customers.email = '$email' 
AND orders.orderid NOT IN (SELECT orderid FROM reviews) 

這會更快。

SELECT orders.orderid 
FROM orders 
LEFT JOIN customers 
    ON orders.custid = customers.id 
LEFT JOIN reviews 
    ON reviews.orderid = orders.orderid 
WHERE customers.email = '$email' 
GROUP BY orders.orderid 
HAVING count(reviews) = 0