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我正在尝试从我的主查询中加入 users_groups，但我似乎无法在别名表中加入它们。请帮我。它导致错误：“on 子句”中的未知列“node.id”

SELECT node.id, node.first_name, node.last_name, (COUNT( parent.id ) - ( sub_tree.depth +1 ) ) AS depth
FROM users AS node, users AS parent, users AS sub_parent, (
    SELECT node.id, node.first_name, node.last_name, (COUNT( parent.id ) -1) AS depth
    FROM users AS node, users AS parent
    WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.id =1
    GROUP BY node.id
    ORDER BY node.lft
) AS sub_tree
JOIN users_groups ON users_groups.user_id = node.id
WHERE node.lft BETWEEN parent.lft AND parent.rgt
    AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
    AND sub_parent.id = sub_tree.id
GROUP BY node.id
ORDER BY node.lft
LIMIT 0 , 30
4

2 回答 2

3

格式化你SQL的总是有帮助的

SELECT 
  node.id, 
  node.first_name, 
  node.last_name, 
  ( COUNT( parent.id ) - ( sub_tree.depth +1 ) ) AS depth
FROM 
  users AS node, 
  users AS parent, 
  users AS sub_parent, 
  (
    SELECT
      node.id, 
      node.first_name, 
      node.last_name, 
      ( COUNT( parent.id ) -1 ) AS depth
    FROM 
      users AS node, 
      users AS parent
    WHERE 
      node.lft BETWEEN parent.lft AND parent.rgt
        AND node.id =1
    GROUP BY 
      node.id
    ORDER BY 
      node.lft
  ) AS sub_tree
  JOIN users_groups ON users_groups.user_id = node.id
                                              ^^ should be sub_tree.id
WHERE 
  node.lft BETWEEN parent.lft AND parent.rgt
    AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
    AND sub_parent.id = sub_tree.id
GROUP BY 
  node.id
ORDER BY 
  node.lft
LIMIT 0 , 30
于 2013-04-19T18:30:29.987 回答
2

你必须使用你给它的别名..

JOIN users_groups ON users_groups.user_id = sub_tree.id

于 2013-04-19T18:22:30.520 回答