我创建了一个在phpMyAdmin中运行良好的查询，但当我试图在.php文件中调用它时，我会收到以下错误。
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 = " . $user_id;
我收到这个错误是因为表"a"、"b"answers"c"以及字段"mother_id"answers"parent_id"在通过mysql_query（$select_query）函数调用SQL之前不存在。
第56行之前的代码查找、返回并显示结果。
..._id = c.id" . // <-- you forgot a space, results in c.idWHERE
"WHERE a.id = " . $user_id;
如果你从java风格的换行中退一步，这样攻击它，你可能会发现这一切都会容易得多。PHP中不需要将其拆分，而且额外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;
这将允许您使用非常方便的调试工具：
echo "[pre]" . $query . "[/pre]";
而[&]实际上是<amp；>。抱歉，SO 新手
然后，您可以将浏览器窗口中的实际查询复制/粘贴到phpMyAdmin中，然后运行该查询，看看它们是否真的相同。