選擇在嵌套集模型的節點通過未唯一的名稱

Question

我試圖來管理節點的檢索在nested set model表，而不是通過唯一的ID，但通過名稱（串），和不同父母下的樹內的其他節點可以被調用相同的方式。

至於現在，我使用的唯一的ID來獲得嵌套組內的節點：

SELECT 
    node.name, node.lft, node.rgt 
FROM tbl AS parent, tbl AS node 
WHERE node.lft BETWEEN parent.lft AND parent.rgt 
AND node.id = '{$node_id}' 
GROUP BY node.id

嘗試此方法擴展到更一般的方法來檢索通過其名稱的節點，我想出了含有多達HAVING子句爲節點的深度查詢檢索，檢查節點名稱和它的深度：

SELECT 
    node.name, node.lft, node.rgt, COUNT(node.id) AS depth 
FROM tbl AS parent, tbl AS node 
WHERE node.lft BETWEEN parent.lft AND parent.rgt 
GROUP BY node.id 
HAVING 
    (node.name = 'myParentName' AND depth = 1) 
    OR 
    (node.name = 'myParent2Name' AND depth = 2) 
    OR 
    ... 
# and so on 

但它並非十全十美：具有相同的名稱和相同深度的兩個節點，但是在不同的父母身上，無論是否等級高，都可以被檢索出來他們屬於。

實施例：

 
ARTICLES 
    | 
    +--PHP 
    | +--the-origins 
    | +--syntax 
    +--JS 
     +--history 
     +--syntax 

在這種情況下，上面的查詢將返回任一文章/ PHP /語法或製品/ JS /語法：具有深度3，事實上一「語法」節點，或者是下PHP節點或JS節點下。 有沒有一條有效的路徑來解決這個問題？

Answer 1

我不太確定你要在這裏做什麼。您是否試圖訪問路徑名爲ARTICLES/PHP/syntax的節點？如果是這樣你需要做的將是每個父母級別自聯接：

SELECT n0.* 
FROM tbl AS n0 
JOIN tbl AS n1 ON n0.lft BETWEEN n1.lft+1 AND n1.rgt 
JOIN tbl AS n2 ON n1.lft BETWEEN n2.lft+1 AND n2.rgt 
WHERE n0.name='syntax' AND n1.name='PHP' AND n2.name='ARTICLES' 

ETA重新註釋：做直子在嵌套組比賽是不是很有趣。您必須嘗試在每個連接的行之間選擇一箇中間父行。這是你不想存在的行，所以你然後用空的左連接反轉這個條件。例如：

SELECT n0.* 
FROM tbl AS n0 
JOIN tbl AS n1 ON n0.lft BETWEEN n1.lft+1 AND n1.rgt 
JOIN tbl AS n2 ON n1.lft BETWEEN n2.lft+1 AND n2.rgt 
LEFT JOIN tbl AS nn1 ON nn1.lft BETWEEN n1.lft+1 AND n0.lft-1 
LEFT JOIN tbl AS nn2 ON nn2.lft BETWEEN n2.lft+1 AND n1.lft-1 
WHERE nn1.lft IS NULL AND nn2.lft IS NULL 
AND n0.name='syntax' AND n1.name='PHP' AND n2.name='ARTICLES'