php - 查询加入同一张表3次

Question

我创建了一个在 phpMyAdmin 中完美运行的查询，但是当我尝试在 .php 文件中调用它时，出现以下错误。

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

我的代码是：

$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;

我收到此错误是因为表“a”、“b”和“c”以及字段“mother_id”和“father_id”在通过 mysql_query($select_query) 函数调用 SQL 之前不存在。

第 56 行之前的代码按应有的方式查找、返回并显示结果。

Answer 1

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

Answer 2

如果您从 java 样式的断行中退后一步并因此攻击它，您可能会发现这一切都容易得多。PHP 中不需要拆分，多余的标点符号更容易出错。

$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;

这将让您使用非常方便的调试工具：

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

[ & ] 实际上是 < & >。对不起，SO新手。

然后，您将能够将实际查询从浏览器窗口复制/粘贴到 phpMyAdmin 中，然后运行该查询并查看它们是否实际上相同。