我如何使用WHERE加入左邊？

Question

我有一個MySQL的要求，那是工作的罰款：

$pdo = $db->query(' 
SELECT *, 
GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, 
monkey.id AS id, 
duck.id AS horse_id, 
bird.id AS elephant_id 
FROM monkey 
LEFT JOIN duck ON monkey.horse=duck.id 
LEFT JOIN bird ON monkey.elephant=bird.id 
LEFT JOIN cat ON monkey.id=cat.mouse_id 
GROUP BY monkey.id ASC;'); 

但現在我想選擇其中猴=「ID」僅值。

所以我試圖使這個解決方案：

$pdo = $db->query(' 
    SELECT *, 
    GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, 
    monkey.id AS id, 
    duck.id AS horse_id, 
    bird.id AS elephant_id 
    FROM monkey WHERE id = "'.$id.'" 
    LEFT JOIN duck ON monkey.horse=duck.id 
    LEFT JOIN bird ON monkey.elephant=bird.id 
    LEFT JOIN cat ON monkey.id=cat.mouse_id 
    GROUP BY monkey.id ASC;'); 

但我得到這個錯誤：

Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'LEFT JOIN duck ON monkey.chorse=duck.id LEFT JOIN bird ON monkey.bir' at line 7 in ..mypage.php:28 Stack trace: #0 ..mypage.php(28): PDO->query('\nSELECT *, \nGRO...') #1 {main} thrown in ...mypage.php on line 28

Answer 1

首先，你必須使joins，然後where，試試這個：

$pdo = $db->query(' 
    SELECT *, 
    GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, 
    monkey.id AS id, 
    duck.id AS horse_id, 
    bird.id AS elephant_id 
    FROM monkey 
    LEFT JOIN duck ON monkey.horse=duck.id 
    LEFT JOIN bird ON monkey.elephant=bird.id 
    LEFT JOIN cat ON monkey.id=cat.mouse_id 
    WHERE monkey.id = "'.$id.'" 
    GROUP BY monkey.id ASC;'); 

Answer 2

所以，你有幾個問題。首先，你不是在清理你的id（查看Bobby Tables）。其次，你把WHERE子句放在錯誤的地方，第三，你不告訴SQL使用哪個ID列。

關注這個作爲一個一般模式：

SELECT 
    <table1>.<column1>, 
    <table2>.<column2>, 
    (etc) 
FROM 
    <table1> 
    JOIN 
    <table> 
    ON (<table1>.<id-column> = <table2>.<id-column) 
WHERE 
    <table1>.<id-column> = ${id-value} 

在你的情況，

$query = 'SELECT *, 
    GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, 
    monkey.id AS id, 
    duck.id AS horse_id, 
    bird.id AS elephant_id 
FROM monkey 
    LEFT JOIN duck ON monkey.horse=duck.id 
    LEFT JOIN bird ON monkey.elephant=bird.id 
    LEFT JOIN cat ON monkey.id=cat.mouse_id 
WHERE monkey.id = "'.$db->escape($id).'" 
GROUP BY monkey.id ASC';