树形菜单的json字符串的拼接

最近在学习权限管理, 要用到树形按钮, 但是字符串的拼接是一个难理解的问题, 然后从网上找了一个从前台用js来遍历组成这个json字符串, 很好! 但是没看懂...

    var data = [
        {"id":1,"parendId":0,"name":"Foods"},
        {"id":2,"parentId":1,"name":"Fruits"},
        {"id":3,"parentId":1,"name":"Vegetables"},
        {"id":4,"parentId":2,"name":"apple"},
        {"id":5,"parentId":2,"name":"orange"},
        {"id":6,"parentId":3,"name":"tomato"},
        {"id":7,"parentId":3,"name":"carrot"},
        {"id":8,"parentId":3,"name":"cabbage"},
        {"id":9,"parentId":3,"name":"potato"},
        {"id":10,"parentId":3,"name":"lettuce"}
    ];


    function exists(rows, parentId){
        for(var i=0; i<rows.length; i++){
            if (rows[i].id == parentId) {
                return true;
            }
        }
        return false;
    }

    function convert(rows){
        var nodes = [];
        //取到第一级菜单
        for(var i=0; i<rows.length; i++){
            var row = rows[i];
            if (!exists(rows, row.parentId)){
                nodes.push({
                    id:row.id,
                    text:row.name
                });
            }
        }

        var toDo = [];
        for(var i=0; i<nodes.length; i++){
            toDo.push(nodes[i]);
        }
        while(toDo.length){
            var node = toDo.shift();    // the parent node
            //取得子菜单
            for(var i=0; i<rows.length; i++){
                var row = rows[i];
                if (row.parentId == node.id){
                    var child = {id:row.id,text:row.name};
                    if (node.children){
                        node.children.push(child);
                    } else {
                        node.children = [child];
                    }
                    toDo.push(child);
                }
            }
        }
        return nodes;
    }

    var data1 = convert(data);

    $('#tt').tree({
        data:data1
    });

下面是从后台拼接字符串

贴在这里从后台实现的方法不是一种可取的办法, 是利用多次从数据库取出数据进行拼接, 最后也能实现效果, 只是这个数据量太小了, 如果是企业里比较大型的项目, 这样会占用不少电脑的性能, 从网上找的那些基本没看懂

import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.Statement;
import java.util.ArrayList;
import java.sql.Connection;


/**
 * @author 25673
 * @数据库连接类
 */
public class ConnectionUtil {
    public static Connection getConnection() {
        Connection conn = null;
        try {
            Class.forName("com.mysql.jdbc.Driver");
            //"jdbc:mysql://数据库地址:3306/数据库名","用户名","用户密码"
            conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/quanxian","root","");
        } catch (Exception e) {
            e.printStackTrace();
        }
        return conn;
    }
    
    public static ArrayList<Menu> findMenu(String parentid) {
        String sql = "select * from sys_menu where parent_id = '"+parentid+"'";
        Connection c = null;
        Statement s = null;
        ResultSet r = null;
        ArrayList<Menu> arrList = null;
        
        c = getConnection();
        try {
            s = c.createStatement();
            r = s.executeQuery(sql);
            arrList = new ArrayList<Menu>();
            while(r.next()) {
                Menu menu = new Menu();
                menu.setId(r.getString("id"));
                menu.setParentid(r.getString("parent_id"));
                menu.setMenuName(r.getString("menu_name"));
                arrList.add(menu);
            }
            
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            try {
                c.close();
                s.close();
                r.close();
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
        return arrList;
    }
    
    /**
     * @param arrList 查询出来的所有符合条件的结果
     * @param parentid 第一级菜单的parentid
     * @return
     */
    public static String strJson(String parentid) {
        /*
          每调用一次这个方法, 就要去连接一次数据库, 总觉这样会增加很多工作量, 因为是进行了多次查询
          这里使用了递归的方式
         */
        ArrayList<Menu> arrList = findMenu(parentid);
        StringBuilder sb = new StringBuilder();
        for(int n = 0;n < arrList.size();n++) {
            sb.append("{");
            sb.append(""id":""+arrList.get(n).getId()+"",");
            sb.append(""text":""+arrList.get(n).getMenuName()+""");
            String str = strJson(arrList.get(n).getId());
            if(!str.equals("")) {
                sb.append(","state":"closed"");
                sb.append(","children":["+str+"]");
            }
            sb.append("}");
            if(n < (arrList.size() - 1)) {
                sb.append(",");
            }
        }
        return sb.toString();
    }
}

还在继续思考怎么从后台一次性查出所有的信息, 直接用拼接的方式组成这个字符串, 希望能有人指点一下, 谢谢