【问题标题】:Subquery in SQLite query not workingSQLite 查询中的子查询不起作用
【发布时间】:2009-05-09 16:35:40
【问题描述】:

我正在使用嵌套集模型将大量数据层次结构存储在 iPhone 上的本地 SQLite 数据库中。我从他们的网站上阅读了MySQL tech article 关于如何执行此操作的信息,但他们建议的查询之一(以及我需要的)似乎不适用于 SQLite,我不知道如何解决它。

SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
FROM line_items AS node,
    line_items AS parent,
    line_items AS sub_parent,
    (SELECT node.name, (COUNT(parent.name) - 1) AS depth
        FROM line_items AS node,
        line_items AS parent
        WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.name = 'Power Up'
        GROUP BY node.name
        ORDER BY node.lft
    ) AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
    AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
    AND sub_parent.name = sub_tree.name
GROUP BY node.name
HAVING depth <= 1
ORDER BY node.lft;

SQLite 报告 sub_tree.name 不是列,我认为这是因为它的子查询实现不完整。有没有人知道如何绕过这个限制?

查询的目的是获取给定父节点的所有直接子节点。

【问题讨论】:

    标签: iphone sql database sqlite


    【解决方案1】:

    尝试在子查询中使用“node.name AS name”,即

    SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
    FROM line_items AS node,
        line_items AS parent,
        line_items AS sub_parent,
        (SELECT node.name AS name, (COUNT(parent.name) - 1) AS depth
            FROM line_items AS node,
            line_items AS parent
            WHERE node.lft BETWEEN parent.lft AND parent.rgt
            AND node.name = 'Power Up'
            GROUP BY node.name
            ORDER BY node.lft
        ) AS sub_tree
    WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
        AND sub_parent.name = sub_tree.name
    GROUP BY node.name
    HAVING depth <= 1
    ORDER BY node.lft;
    

    至少似乎摆脱了错误。

    【讨论】:

    • 几乎完全修复了它。由于某种原因, HAVING DEPTH
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