SQL檢索數據關係

$sql = "SELECT distinct `customer_id` FROM(SELECT * FROM `booking` WHERE `confirmed` = 1 ORDER BY `reserve_date` DESC) as inv"; 
$query = $this->db->query($sql);

完美運行在笨文件的模型PHP CMS的SQL檢索數據關係

有一列名爲「客戶」表命名爲「名」想訂購由$query獲得的data(customer_id)場名爲「客戶」表中的「名」，並將其存儲在$result 那麼，做什麼都我$result

想怎麼辦呢 - 請幫我 - 是新來的笨和PHP。非常感謝。

來源

2015-07-20 WebSuccess

你的問題是模糊的，請寫正確。 –

-什麼不明白 - – WebSuccess

- 請告訴我 - 其他人瞭解 - – WebSuccess

加入客戶表並按名稱排序。

SELECT DISTINCT 
    customer_id 
FROM 
    (SELECT 
     * 
    FROM 
     booking 
    WHERE 
     confirmed = 1 
    ORDER BY reserve_date DESC) AS inv 
     INNER JOIN 
    customer AS cu ON cu.id = inv.customer_id 
ORDER BY cu.name

如果你永遠也不會從「預訂」 subquerying使用限制你可以通過

ORDER BY null

更換

ORDER BY reserve_date DESC

優化它，然後你甚至可以重寫查詢成爲

SELECT DISTINCT 
    inv.customer_id 
FROM 
    booking AS inv 
     INNER JOIN 
    customer AS cu ON cu.id = inv.customer_id 
WHERE 
    inv.confirmed = 1 
ORDER BY cu.name;

不要忘記在id, customer_id, confirmed and name列上建立必要的索引。

來源

2015-07-20 06:09:17

如何索引亞歷山大 - 謝謝 – WebSuccess

http://www.databasejournal.com/features/mysql/article.php/1382791/Optimizing-MySQL-Queries-and-Indexes.htm 接受答案我回答了你的問題:) –

@websuccess這看起來像正確的答案 – Orangepill

$ SQL = SELECT DISTINCT booking.customer_id 從預訂 INNER JOIN客戶ON booking.customer_id = customer.customer_id WHERE confirmed = 1個 ORDER BY customer.name

來源

2015-07-21 10:06:23 WebSuccess

SQL檢索數據關係

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