Question

任务是在node.js中递归构建树结构。也就是说，我使用一副幻灯片，可能还包括其他子甲板。我是函数式编程的新手，我不知道自己需要做什么。我知道，有更多高级方法来控制流程，但我想在开始时尝试以嵌套方式执行，如果可能的话。我现在拥有的：

server.js调用api并使用json对象进行响应：

router.get('/deck/tree/:rev_id', function(req, res) {       
    var Deck = require('./models/deck');
    var deck = new Deck(connection);
    deck.getTree(req.params.rev_id ,function (tree) {res.json(tree)});      
});

在甲板模块中我有几个功能： 2.a获得牌组标题的功能：

this.getTitle = function(rev_id, callback){
    var sql = "SELECT title FROM ?? WHERE ?? = ?";
    var inserts = ['deck_revision', 'id', rev_id];
    sql = mysql.format(sql, inserts);
    connection.query(sql, function(err, results) {
    if (err) throw err;

    callback(results[0].title);
    }); 
};

2.b获得＆＃34;直接＆＃34;的功能甲板上的孩子们：

this.getChildren = function(acc, callback){
    acc.children = [];
    var sql = "SELECT item_id AS id, item_type AS type FROM ?? WHERE ?? = ? ORDER BY position";
        var inserts = ['deck_content', 'deck_revision_id', acc.id];
        sql = mysql.format(sql, inserts);
        connection.query(sql, function(err, results) {
            if (err) throw err;
            results.forEach(function(res_item){
                acc.children.push({'id': res_item.id, 'type': res_item.type});
                if (acc.children.length === results.length){
                    callback(acc);
                }
            });
        });     
};

应该补充树的功能（暂时不起作用）：

var results = [];
this.getTree = function(id, callback) {
    var acc = {};   
    var deck = this;
    var new_slide = new Slide(connection);
    deck.getTitle(id, function (title_str) {
        acc.title = title_str;
        acc.id = id;
        acc.type = 'deck';
        deck.getChildren(acc, function(new_acc){
            new_acc.children.forEach(function(element){
                if (element.type === 'deck'){

      // I suppose the help is needed here!    

                    deck.getTree(element.id, function(new_element){
                        results.push(new_element);
                        if(results.length === new_acc.children.length) {
                            callback(results);
                        }
                    });

                }else{
                    new_slide.getTitle(element.id, function(title_str){
                        element.title = title_str;
                        results.push(element);
                        if(results.length === new_acc.children.length) {
                            callback(results);
                        }
                    });

                }

Answer 1

如果有人感兴趣，请点击此处我的工作原理：

 this.getChildren = function(id, callback){
    var sql = "SELECT item_id AS id, item_type AS type, position FROM ?? WHERE ?? = ? ORDER BY position";
    var inserts = ['deck_content', 'deck_revision_id', id];
    sql = mysql.format(sql, inserts);
    connection.query(sql, function(err, results) {
        if (err) throw err;
        callback(results);
    }); 
};

 this.getTree = function(id, acc, callback) {
    deck = new Deck(connection);
    var new_slide = new Slide(connection);
    deck.getTitle(id, function (title_str) {
        acc.title = title_str;
        acc.id = id;
        acc.type = 'deck';
        acc.children = [];
        deck.getChildren(acc.id, function(children){
            children.forEach(function(element){
                if (element.type === 'deck'){
                    deck.getTree(element.id, element, function(new_element){
                        acc.children[new_element.position - 1] = new_element;
                        var size = acc.children.filter(function(value) { return value !== null }).length;
                        if (size === children.length){
                            callback(acc);
                        }
                    });                            
                }
                else{
                    new_slide.getTitle(element.id, function(title_str){
                        element.title = title_str;
                        acc.children[element.position - 1] = element;
                        var size = acc.children.filter(function(value) { return value !== null }).length;
                        if(size === children.length) {
                            callback(acc);
                        }
                    });                            
                }
            });                    
        });
    });
};

node.js中树结构的递归

1 个答案: