Question

我有这个mysql表＆＃34;类别＆＃34;：

BigInteger bi = new BigInteger(1000);

我想得到这个结果：

id  category        parent
1   category1       NULL
2   subcategory1    1
3   category2       NULL
4   subcategory2    2

得到这个结果我使用这段代码，但速度很慢：

category1
    subcategory1
        subcategory2
category2

你有更快的解决方案吗？感谢

Answer 1

如果您知道树的深度（最大或期望），则可以将其全部放在SQL中，展平为表示路径的行：

SELECT c.category, c1.category AS sub1, c2.category AS sub2, c3.category AS sub3
FROM categories c
    LEFT JOIN 
        (categories c1 
             LEFT JOIN
                 (categories c2
                      LEFT JOIN categories c3 
                      ON c3.parent = c2.category)
                 ON c2.parent = c1.category)
         ON c1.parent = c.category
WHERE c.parent IS NULL;

有了这个，相应地在UI中呈现它并不是什么大不了的事。

Answer 2

private List<PolyRegion2D> Joiner(IEnumerable<Polygon2D> polygons) {
        // The resulting polygons are unconnected
        List<PolyRegion2D> res = new List<PolyRegion2D>();

        // Put every polygon in a region to do the unions.
        LinkedList<PolyRegion2D> polygonRegions = new LinkedList<PolyRegion2D>();
        foreach (Polygon2D polygon in polygons) {
            polygonRegions.AddLast(new PolyRegion2D(new Polygon2D[]{polygon}));
        }

        while (polygonRegions.Count > 0) {
            PolyRegion2D first = polygonRegions.First.Value;
            polygonRegions.RemoveFirst();
            PolyRegion2D union;
            LinkedListNode<PolyRegion2D> connected = FindConnected(first, polygonRegions, out union);
            if (connected == null) {
                // Unconnected polygon
                res.Add(first);
            } else {
                // Intersection found
                polygonRegions.Remove(connected);
                polygonRegions.AddFirst(union);
            }
        }

        return res;
    }

    private LinkedListNode<PolyRegion2D> FindConnected(PolyRegion2D poly, LinkedList<PolyRegion2D> polys, out PolyRegion2D union) {
        LinkedListNode<PolyRegion2D> node = polys.First;
        while(node != null){
            PolyRegion2D[] union_ = PolyRegion2D.Union(poly, node.Value);
            if (union_.Length == 1) {
                union = union_[0];
                return node;
            }
            node = node.Next;
        }
        union = null;
        return null;
    }

处理mysql / php中的父/子类别

2 个答案: