PHP - 在mysqli綁定參數中的相同值

-1

我可以優化此代碼嗎？
mysqli_stmt_bind_param($statement, "iiiiiii", $user, $user, $user, $user, $user, $user, $user);PHP - 在mysqli綁定參數中的相同值

用於此請求：

$statement = mysqli_prepare($conn, "SELECT u_id, u_lastname, u_firstname, 
    (SELECT f_id from friend where (f_friend_one = u_id AND f_friend_two = ?) OR (f_friend_one = ? AND f_friend_two = u_id)), 
    (SELECT f_type from friend where (f_friend_one = u_id AND f_friend_two = ?) OR (f_friend_one = ? AND f_friend_two = u_id)) 
FROM user 
WHERE ((u_id = (SELECT f_friend_one FROM friend WHERE f_friend_one = u_id AND f_friend_two = ? AND (f_type = 1 OR f_type = 4))) 
OR  (u_id = (SELECT f_friend_two FROM friend WHERE f_friend_one = ? AND f_friend_two = u_id AND (f_type = 1 OR f_type = 3)))) 
AND  u_id != ? ") 
    or die(mysqli_error($conn));

來源

2015-10-15 Okn

可能不會使運行時間有差別，但我敢肯定，你會縮短查詢，但是，因爲它沒有包括... – Devon

查詢很可能被簡化成'JOIN'所以您不必重複所有這些查詢。我試圖弄清楚它在做什麼。 – Barmar

查詢獲取朋友信息，如果他們是朋友：我有一個有三列的朋友表：id，f_friend_one（發送邀請的用戶），f_friend_two。 – Okn

我想你可以把它簡化爲這樣：

SELECT u_id, u_lastname, f_id, f_type 
FROM user 
JOIN friend 
    ON (u_id = f_friend_one AND f_friend_two = ? AND f_type IN (1, 4)) 
     OR 
     (u_id = f_friend_two AND f_friend_one = ? AND f_type IN (1, 3)) 
WHERE u_id != ?

這有可能是不需要在年底WHERE條款，因爲有人不能向朋友發送好友請求。

OR可能會讓MySQL難以優化，所以最好將它分成兩個子查詢連接UNION。

SELECT u_id, u_lastname, f_id, f_type 
FROM user 
JOIN friend ON u_id = f_friend_one 
WHERE f_friend_two = ? AND f_type IN (1, 4)) 
    AND u_id != ? 

UNION 

SELECT u_id, u_lastname, f_id, f_type 
FROM user 
JOIN friend ON u_id = f_friend_two 
WHERE f_friend_one = ? AND f_type IN (1, 3)) 
    AND u_id != ?

來源

2015-10-15 23:16:47 Barmar

謝謝，工作很好！ – Okn

PHP - 在mysqli綁定參數中的相同值

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