Question

我试图返回一个表格，其中包含使用嵌套集合模型表示的层次结构中的节点深度，我遵循本教程，但查询中使用的查询＆＃39;查找节点的深度＆＃39;对我不起作用：http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/

SELECT node.GroupName, (COUNT(parent.GroupName) - 1) AS depth
FROM CompanyGroup AS node,
        CompanyGroup AS parent
WHERE node.LeftID BETWEEN parent.LeftID AND parent.RightID
GROUP BY node.GroupName
ORDER BY node.LeftID;

运行此查询我收到错误＆＃34; Column＆＃39; CompanyGroup.GroupName＆＃39;在选择列表中无效，因为它不包含在聚合函数或GROUP BY子句中。＆＃34;

有人可以解释为什么吗？

编辑：错误消息中的列错误，我的道歉错误是：＆＃34; Column＆＃34; CompanyGroup.LeftID＆＃34;无效...... ＆＃34;

Answer 1

试试这个 -

SELECT 
      node.GroupName
    , depth = COUNT(parent.GroupName) - 1
FROM CompanyGroup node
JOIN CompanyGroup parent ON node.LeftID BETWEEN parent.LeftID AND parent.RightID
GROUP BY node.GroupName
ORDER BY MIN(node.LeftID) --<--

或试试这个 -

SELECT 
      node.GroupName
    , depth = COUNT(parent.GroupName) - 1
FROM CompanyGroup node
JOIN CompanyGroup parent ON node.LeftID BETWEEN parent.LeftID AND parent.RightID
GROUP BY node.GroupName, node.LeftID
ORDER BY node.LeftID

Answer 2

您是否正在运行其他查询？您不应该收到该特定错误，而是"Column "CompanyGroup.LeftID" is invalid in the ORDER BY..."

此查询应该有效：

SELECT node.GroupName, (COUNT(parent.GroupName) - 1) AS depth
FROM CompanyGroup AS node,
        CompanyGroup AS parent
WHERE node.LeftID BETWEEN parent.LeftID AND parent.RightID
GROUP BY node.GroupName
;

SQL Fiddle Demo

Answer 3

我不认为这是你从你发布的查询中收到的错误 - 你确定你没有弄乱它吗？

无论如何，这里还有其他问题：你正在使用node.LeftID订购结果，但是你不能这样做。您应该收到此消息：

Column "node.LeftID" is invalid in the ORDER BY clause because it is not contained in either an aggregate function or the GROUP BY clause.

尝试删除ORDER BY并再次运行。

列在选择列表中无效，因为它不包含在聚合函数或GROUP BY子句中

3 个答案: