查詢聯接同一表3次

Question

我創建了一個在phpMyAdmin中完美運行的查詢，但是當我嘗試在.php文件中調用它時，出現以下錯誤。

Undefined variable: mothers_name in C:\wamp\www\Family_Tree\showfamily.php on line 56

我的代碼是：

$select_query = "SELECT a.id, CONCAT( a.surname,  ', ', a.first_names ) AS child_name, " . 
"CONCAT( b.surname,  ', ', b.first_names ) AS mothers_name, " .
"CONCAT( c.surname, ', ', c.first_names ) AS fathers_name " .
"FROM family_members a " .
"INNER JOIN family_members b ON a.mother_id = b.id " .
"INNER JOIN family_members c ON a.father_id = c.id" .
"WHERE a.id = " . $user_id;

我收到此錯誤消息是因為在通過mysql_query（$ select_query）函數調用SQL之前，表“ a”，“ b”和“ c”以及字段“ mother_id”和“ father_id”並不存在。

第56行之前的代碼將查找，返回並按原樣顯示結果。

Answer 1

..._id = c.id" .               // <-- you forgot a space, results in c.idWHERE
"WHERE a.id = " . $user_id;

Answer 2

如果您退出Java樣式的換行並進行攻擊，那么可能會發現這一切容易得多。 PHP無需對其進行拆分，並且額外的標點符號使錯誤更容易實現。

$select_query = 
"SELECT 
     a.id, 
     CONCAT( a.surname, ', ', a.first_names ) AS child_name, 
     CONCAT( b.surname, ', ', b.first_names ) AS mothers_name,
     CONCAT( c.surname, ', ', c.first_names ) AS fathers_name 
FROM family_members a 
INNER JOIN family_members b ON a.mother_id = b.id 
INNER JOIN family_members c ON a.father_id = c.id
WHERE a.id = " . $user_id;

這將使您使用非常方便的調試工具：

echo "[pre]" . $query . "[/pre]";

[＆]實際上是<＆>。 抱歉，SO新手。

然后，您將能夠將實際查詢從瀏覽器窗口復制/粘貼到phpMyAdmin中，然后運行該查詢，看看它們是否完全相同。