Question

我有所有叶子都有索引的树，当树以数据库的形式递归时，数据库将通过索引对树进行排序。首先，它获取按索引排序的根节点，依此类推。现在我需要实现操作，用户可以通过按向上/向下箭头图标对这些索引进行排序。当用户按下时，索引应该采用它自己索引下的索引，当按下向上箭头时，反之亦然。我只是不知道实现这种功能的最佳方法是什么。

Answer 1

由于你的问题有点模糊，这个答案假设你知道你在做数据库时你在做什么（如果没有，我会推荐hibernate for java）以下代码是为了给你一些实现的想法你的解决方案。

//If I have understood your question, you want two nodes to swap position in the tree structure
public static swapNode(Node parent, Node child)
{
    Long superId = parent.getParentId();
    child.parentId(superId);
    parent.setParentId(child.getId());
    child.setId(parentId);
    //update children lists of parent and child
    //update parent ids of children lists

    //save changes to database
}

//create tree structure from database. Assumes nodes have been loaded from a database
//table where each row represents a node with a parent id column the root node which has parent id null)
//invoke this with all nodes and null for parentId argument
public static List<Node> createNodeTree(List<Node> allNodes, Long parentId)
{
    List<Node> treeList = new ArrayList<Node>();
    for(Node node : nodes)
    {
        if(parentIdMatches(node, parentId))
        {
            node.setChildren(createNodeTree(allNodes, node.getId()));
            treeList.add(node);
        }
    }
    return treeList;
}

private static boolean parentIdMatches(Node node, Long parentId)
{
    return (parentId != null && parentId.equals(node.getParentId())) 
        || (parentId == null && node.getParentId() == null);
}

//The objects loaded from the database should implement this interface
public interface Node
{
    void setParentId(Long id);
    Long getParentId();
    Long getId();
    List<Node> getChildren();
    void setChildren(List<Node> nodes);
}

递归树索引的顺序？

1 个答案: