php - INNER JOIN : where 子句太模棱两可

Question

我收到错误消息"Integrity constraint violation: 1052 Column 'restaurant_id' in where clause is ambiguous' in"。

a) 如何在 JOIN 上使用 bindparam？这甚至是导致此错误的问题吗？

function restaurant(PDO $dbh, $username) {
   global $dbh;
   $stmt = $dbh->prepare("
      SELECT        *
      FROM          users u
      INNER JOIN    menues m
      ON            u.user_id = m.restaurant_id
      INNER JOIN    users_slider s
      ON            m.restaurant_id = s.restaurant_id
      WHERE         restaurant_id = :restaurant_id
   ");

   $stmt->bindParam(":restaurant_id", $_GET['r']);
   $stmt->execute();
   return $stmt->fetchAll();
}

Answer 1

改变

...
WHERE     restaurant_id = :restaurant_id

至

...
WHERE     m.restaurant_id = :restaurant_id
          ^^

Answer 2

请根据您的表格使用m.restaurant_id或s.restaurant_id在条件下使用它将解决您的问题。

Answer 3

始终使用别名以避免连接查询中的冲突。这里restaurant_id应该是m.restaurant_id。

试试这个查询： -

SELECT        *
      FROM          users u
      INNER JOIN    menues m
      ON            u.user_id = m.restaurant_id
      INNER JOIN    users_slider s
      ON            m.restaurant_id = s.restaurant_id
      WHERE         m.restaurant_id =:restaurant_id

Answer 4

将 where 条件放在 WHERE m.restaurant_id = :restaurant_id

function restaurant(PDO $dbh, $username) {
   global $dbh;
   $stmt = $dbh->prepare("
      SELECT        *
      FROM      users u
      INNER JOIN    menues m
      ON            u.user_id = m.restaurant_id
      INNER JOIN    users_slider s
      ON            m.restaurant_id = s.restaurant_id
      WHERE     m.restaurant_id = :restaurant_id
   ");

   $stmt->bindParam(":restaurant_id", $_GET['r']);
   $stmt->execute();
   return $stmt->fetchAll();
}