功能点分析：商品类目表

• 前言

　　　　作为电商网站，必然要有商品类目表，以便商品分类检索。而设计商品类目表，又是一件特别繁杂的事情。一件商品可能有多个类目来检索出来，比如苹果手机，可以从品牌检索也可以从手机检索。一个类目对应多个商品，比如手机下对应了多款属于手机类的商品。而类目是一个多叉树结构，类似于文件夹的结构。通常大电商都是把类目整理好了放在cache上，就不用频繁访问数据库和整理排序了。

• 个人类目标的结构设计

　　　　参考了一下网上无限级分类的表设计，简化模型为一个表内存在父子级从属关系。

　　　　表中的isRoot可以去掉，只要parent为0就认为是根节点就好了。而为什么不存子节点而是存父节点，因为子节点的映射类，生成整理的时候比存父节点要复杂一些，当然程序写的不复杂而是这个问题需要花费的时间就会多一些。

　　　　表设计好了之后，就需要查询数据并且整理出他们的从属关系。为了减少数据库的查询次数，我采用了全部查询，在内存中生成多叉树的形式。节点类的定义想法非常直接，它存在唯一识别：id，真实信息：name，以及它的子树：List or Map。

/*
 * 树型存储
 */
class CategoryTree{
    private long id;
    private String name;
    private List<CategoryTree> children;
    public CategoryTree(){
        children = new ArrayList<>();
    }
    public long getId() {
        return id;
    }
    public void setId(long id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public List<CategoryTree> getChildren() {
        return children;
    }
    public void setChildren(List<CategoryTree> children) {
        this.children = children;
    }
    
}
/*
 * 哈希表存储
 */
class CategoryMap{
    private long id;
    private String name;
    private Map<Long,CategoryMap> children;
    public CategoryMap(){
        children = new HashMap<>();
    }
    public long getId() {
        return id;
    }
    public void setId(long id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public Map<Long, CategoryMap> getChildren() {
        return children;
    }
    public void setChildren(Map<Long, CategoryMap> children) {
        this.children = children;
    }
}

　　　　在对乱序数组生成三级n叉树这个过程中，最快的方式是用map。在生成过程中，只能一级级生成，因为没有父节点不可能有子节点跟随这个逻辑的存在。

    //集合存储
    public void test1(){
        tool.connection2MYSQL();
        Connection conn = tool.getCon();
        String sql = "select * from category";
        Statement stm = null;
        List<category> clist = new ArrayList<>();
        try {
            stm = conn.createStatement();
            ResultSet rs = stm.executeQuery(sql); 
            while(rs.next()){
                category c = new category();
                c.setId(rs.getLong("id"));
                c.setName(rs.getString("name"));
                c.setParentId(rs.getLong("parent_id"));
                c.setIsRoot(rs.getInt("is_root"));
                clist.add(c);
            }
            tool.closeConn(conn);
        } catch (SQLException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }
        /*
         * 检查集合
         */
//        for(category ca:clist){
//            System.out.println("id: "+ca.getId()+" name: "+ca.getName()+" parentId: "+ca.getParentId()+" isRoot: "+ca.getIsRoot());
//        }
        /**
         * 逻辑尝试=====
         */
        List<CategoryTree> roots = new ArrayList<>();
        List<CategoryTree> second = new ArrayList<>();
        List<CategoryTree> third = new ArrayList<>();
        //一次遍历 添加根节点
        int i = 0;
        while(i != clist.size()-1){
            if(clist.get(i).getParentId() == 0){
                CategoryTree ct = new CategoryTree();
                ct.setId(clist.get(i).getId());
                ct.setName(clist.get(i).getName());
                roots.add(ct);
                clist.remove(i);
            }else
                i++;
        }
        //二次遍历 添加二级节点
        for(int j=0;j<roots.size();j++){
            i = 0;
            while(i < clist.size()){
                if(clist.get(i).getParentId() == roots.get(j).getId()){
                    CategoryTree ct = new CategoryTree();
                    ct.setId(clist.get(i).getId());
                    ct.setName(clist.get(i).getName());
                    roots.get(j).getChildren().add(ct);
                    second.add(ct);//用空间换
                    clist.remove(i);
                }else
                    i++;
            }
        }
        //三次遍历 添加三级节点
        for(int j=0;j<second.size();j++){
            i = 0;
            while(i < clist.size()){
                if(clist.get(i).getParentId() == second.get(j).getId()){
                    CategoryTree ct = new CategoryTree();
                    ct.setId(clist.get(i).getId());
                    ct.setName(clist.get(i).getName());
                    second.get(j).getChildren().add(ct);
                    third.add(ct);//用空间换
                    clist.remove(i);
                }else
                    i++;
            }
        }
        for(category ca:clist){
            System.out.println("id: "+ca.getId()+" name: "+ca.getName()+" parentId: "+ca.getParentId()+" isRoot: "+ca.getIsRoot());
        }
        for(CategoryTree ct:roots){
            System.out.println("id: "+ct.getId()+" name: "+ct.getName());
            {
            for(CategoryTree ct1:ct.getChildren())
            {
                System.out.println("二级 id: "+ct1.getId()+" name: "+ct1.getName());
                for(CategoryTree ct2:ct1.getChildren())
                    System.out.println("三级 id: "+ct2.getId()+" name: "+ct2.getName());
            }
                
            }
        }/**
         * 逻辑尝试=====
         */
        
    }

　　　　我对每一级的节点做了一个额外的存储，在第三级生成的时候简化一个循环，也就是n平方的复杂度。

　　　　使用map生成的话，仅在生成这个过程，就可以把问题化简成n的复杂度。因为Map“知道”自己存储的对象的id，而List要通过遍历才知道它的自己存的元素的id。

public void test2(){
        tool.connection2MYSQL();
        Connection conn = tool.getCon();
        String sql = "select * from category";
        Statement stm = null;
        List<category> clist = new ArrayList<>();
        try {
            stm = conn.createStatement();
            ResultSet rs = stm.executeQuery(sql); 
            while(rs.next()){
                category c = new category();
                c.setId(rs.getLong("id"));
                c.setName(rs.getString("name"));
                c.setParentId(rs.getLong("parent_id"));
                c.setIsRoot(rs.getInt("is_root"));
                clist.add(c);
            }
            tool.closeConn(conn);
        } catch (SQLException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }
        /**
         * 逻辑尝试=====
         */
        Map<Long,CategoryMap> rootMap = new HashMap<>();
        Map<Long,CategoryMap> secondMap = new HashMap<>();
        //遍历一级
        int i = 0;
        while(i < clist.size()){
            if(clist.get(i).getParentId() == 0){
                CategoryMap cm = new CategoryMap();
                cm.setId(clist.get(i).getId());
                cm.setName(clist.get(i).getName());
                rootMap.put(cm.getId(),cm);
                clist.remove(i);
            }else
                i++;
        }
        //遍历二级
        i = 0;
        while (i < clist.size()) {
            if (rootMap.get(clist.get(i).getParentId()) != null) {
                CategoryMap cm = new CategoryMap();
                cm.setId(clist.get(i).getId());
                cm.setName(clist.get(i).getName());
                rootMap.get(clist.get(i).getParentId()).getChildren().put(cm.getId(), cm);
                secondMap.put(cm.getId(), cm);
                clist.remove(i);
            } else
                i++;
        }
        //遍历三级
        i = 0;
        while (i < clist.size()) {
            if (secondMap.get(clist.get(i).getParentId()) != null) {
                CategoryMap cm = new CategoryMap();
                cm.setId(clist.get(i).getId());
                cm.setName(clist.get(i).getName());
                secondMap.get(clist.get(i).getParentId()).getChildren().put(cm.getId(), cm);
                clist.remove(i);
            } else
                i++;
        }
//        for (Map.Entry<Long, CategoryMap> entry : rootMap.entrySet()) {  
//            System.out.println("Key = " + entry.getKey() + ", id : " + entry.getValue().getId()+" name : "+entry.getValue().getName());
//            for (Map.Entry<Long, CategoryMap> entry1 : entry.getValue().getChildren().entrySet()){
//                System.out.println("二级 Key = " + entry1.getKey() + ", id : " + entry1.getValue().getId()+" name : "+entry1.getValue().getName());
//                  for (Map.Entry<Long, CategoryMap> entry2 : entry1.getValue().getChildren().entrySet()){
//                      System.out.println("三级 Key = " + entry2.getKey() + ", id : " + entry2.getValue().getId()+" name : "+entry2.getValue().getName());
//                  }
//            }
//            
//        }
        JSONArray json = new JSONArray();
        for (CategoryMap entry : rootMap.values()) { 
            JSONObject job1 = new JSONObject();
            job1.put("id", entry.getId());
            job1.put("name", entry.getName());
            JSONArray joa1 = new JSONArray();
//            System.out.println("id : " + entry.getId()+" name : "+entry.getName());
            for (CategoryMap entry1 : entry.getChildren().values()){
                JSONObject job2 = new JSONObject();
                job2.put("id", entry1.getId());
                job2.put("name", entry1.getName());
                JSONArray joa2 = new JSONArray();
//                System.out.println("二级 id : " + entry1.getId()+" name : "+entry1.getName());
                  for (CategoryMap entry2 : entry1.getChildren().values()){
                    JSONObject job3 = new JSONObject();
                    job3.put("id", entry2.getId());
                    job3.put("name", entry2.getName());
                    joa2.add(job3);
//                    System.out.println("三级 id : " + entry2.getId() + " name : "+entry2.getName());
                  }
                  job2.put("chird", joa2);
                  joa1.add(job2);
            }
            job1.put("chird", joa1);
            json.add(job1);
        }
        
        for(int k=0;k<json.size();k++){
            JSONObject jo = json.getJSONObject(k);
            System.out.println(jo.toString());
        }
        /**
         * 逻辑尝试=====
         */
    }

　　　　最后的生成json的时候，仍然需要三次方的复杂度，我在考虑如何能在整理过程中顺带生成json，现在还没做出来，不过优化成n应该还是有机会的。

　　　　另外，遍历源数据的时候，把抽掉的节点remove掉也是一种减少重复遍历的方式。

　　　　最后生成的结果如同预期。

连接成功
{"chird":[{"chird":[{"name":"衬衣","id":16}],"name":"七匹狼","id":6},{"chird":[{"name":"运动服","id":17}],"name":"阿迪达斯","id":7}],"name":"男装","id":1}
{"chird":[{"chird":[{"name":"毛衣","id":18}],"name":"zara","id":8},{"chird":[{"name":"包包","id":19}],"name":"普拉达","id":9}],"name":"女装","id":2}
{"chird":[{"chird":[{"name":"笔记本电脑","id":20}],"name":"dell","id":10},{"chird":[{"name":"台式电脑","id":21}],"name":"lenovo","id":11}],"name":"电脑","id":3}
{"chird":[{"chird":[{"name":"note","id":22}],"name":"三星","id":12},{"chird":[{"name":"iPhone","id":23}],"name":"苹果","id":13}],"name":"手机","id":4}
{"chird":[{"chird":[{"name":"第一版","id":24}],"name":"Java程序设计","id":14},{"chird":[{"name":"第三版","id":25}],"name":"C++程序设计","id":15}],"name":"图书","id":5}

• 总结

　　　　对于表设计和这个问题的解决不知道有没有更加规范一点的方式，还需要继续探讨。对于我们项目来说，这个方法已经可以适应了，商品分类我打算用个分类表来做映射，类目表叶子节点的id和商品的id存入，检索的时候根据叶子节点id检索所有商品。本次问题，在于对乱序数组转化为多叉树的模型建立，解决了这个问题，对转化思维又有了进一步的提高。（在这里其实问题的阶并不重要，因为这是有穷集合，类目表的条目基本不会超过几百，考虑的最多的就是访问量的问题，因为小电商不会做cache，每一次访问都生成三级树，所以本次问题重在如何用更好的运算模型去解决同一个问题）

• 追加

　　　　关于三级关系树的运用，我做了一个小的地域管理页面，与类目树是一样的结构。

<%@ page language="java" contentType="text/html; charset=UTF-8"
    pageEncoding="UTF-8"%>
    <%@include file="/common.jsp" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
        <script type="text/javascript" src="js/jquery.js" ></script>
        <script type="text/javascript" src="js/jquery.form.js"></script>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>地域信息</title>
        <style>
            .chosen{
                background:blue;
            }
            .select{
                width:30px;
            }
        </style>
</head>
<body>
<div>
    <div style="margin-left:20px;float:left;">
        <h2>一级类别</h2>
        <h3>当前：${current1.name}</h3>
    <form action="letao_DevelopmentModelArea_addArea" id="addForm1">
        编码:<input size="8" name="code"> name:<input size="8" name="name"> <input size="8" name="parent" type="hidden" value="0">
        <br/>
    </form>
    <br/>
    <button class="add1">添加</button>
        <table style="border: 2px solid BLACK;">
            <thead>
                <tr>
                    <th>选择</th>
                    <th>id</th>
                    <th>name</th>
                    <th>编码</th>
                    <th>操作</th>
                </tr>
            </thead>
            <s:iterator id="r1" value="#rank1">
            <s:if test="#current1.id == #r1.id">
            <tr class="chosen" type="${ r1.id}">
                <td><button class="select" id="${ r1.id}" type="1">选择</button></td>
                <td>${ r1.id}</td>
                <td><input value ="${ r1.name}" id="name${ r1.id}" size="8" ></td>
                <td><input value ="${ r1.code}" id="code${ r1.id}" size="8"></td>
                <td><button type="${ r1.id}" class="update">更新</button> | <button type="${ r1.id}" class="del">删除</button></td>
            </tr>
            </s:if>
            <s:else>
            <tr type="${ r1.id}">
                <td><button class="select" id="${ r1.id}" type="1">选择</button></td>
                <td>${ r1.id}</td>
                <td><input value ="${ r1.name}" id="name${ r1.id}" size="8" ></td>
                <td><input value ="${ r1.code}" id="code${ r1.id}" size="8"></td>
                <td><button type="${ r1.id}" class="update">更新</button> | <button type="${ r1.id}" class="del">删除</button></td>
            </tr>
            </s:else>
            </s:iterator>
        </table>
    </div>
    
    <div style="margin-left:20px;float:left;">
        <h2>二级类别</h2>
        <h3>上级：${current1.name}</h3>
        <form action="letao_DevelopmentModelArea_addArea" id="addForm2">
        编码:<input size="8" name="code"> name:<input size="8" name="name"><input size="8" name="parent" type="hidden" value="${current1.id}">
        <br/>
        </form>
        <br/>
        <button class="add2">添加</button>
        <table style="border: 2px solid BLACK;">
            <thead>
                <tr>
                    <th>选择</th>
                    <th>id</th>
                    <th>name</th>
                    <th>编码</th>
                    <th>操作</th>
                </tr>
            </thead>
            <s:iterator id="r2" value="#rank2">
            
                <s:if test="#current2.id == #r2.id">
                <tr class="chosen" type="${ r2.id}">
                    <td><button class="select" id="${ r2.id}" type="2">选择</button></td>
                    <td>${ r2.id}</td>
                    <td><input value ="${ r2.name}" id="name${ r2.id}" size="8"></td>
                    <td><input value ="${ r2.code}" id="code${ r2.id}" size="8"></td>
                    <td><button type="${ r2.id}"  class="update">更新</button> | <button type="${ r2.id}" class="del">删除</button></td>
                </tr>
                </s:if>
                <s:else>
                <tr  type="${ r2.id}">
                    <td><button class="select" id="${ r2.id}" type="2">选择</button></td>
                    <td>${ r2.id}</td>
                    <td><input value ="${ r2.name}" id="name${ r2.id}" size="8"></td>
                    <td><input value ="${ r2.code}" id="code${ r2.id}" size="8"></td>
                    <td><button type="${ r2.id}"  class="update">更新</button> | <button type="${ r2.id}" class="del">删除</button></td>
                </tr>
                </s:else>

            </s:iterator>
        </table>
    </div>
        <div style="margin-left:20px;float:left;">
        <h2>三级类别</h2>
        <h3>上级：${current1.name}->${current2.name}</h3>
        <form action="letao_DevelopmentModelArea_addArea" id="addForm3">
        编码:<input size="8" name="code"> name:<input size="8" name="name"> <input size="8" name="parent" type="hidden" value="${current2.id }">
        <br/>
        </form>
        <br/>
        <button class="add3">添加</button>
        <table style="border: 2px solid BLACK;">
            <thead>
                <tr>
                    <th>id</th>
                    <th>name</th>
                    <th>编码</th>
                    <th>操作</th>
                </tr>
            </thead>
            <s:iterator id="r3" value="#rank3">
            <tr>
                <td>${ r3.id}</td>
                <td><input value ="${ r3.name}" id="name${ r3.id}" size="8"></td>
                <td><input value ="${ r3.code}" id="code${ r3.id}" size="8"></td>
                <td><button type="${ r3.id}" class="update">更新</button> | <button type="${ r3.id}" class="del">删除</button></td>
            </tr>
            </s:iterator>
        </table>
    </div>
</div>
<form action="letao_DevelopmentModelArea_updateInfo" method="post" id="updateForm" style="display:none;">
    <input id="hideId" type="hidden" name="id" type="hidden">
    <input id="updateCode" type="hidden" name="code">
    <input id="updateName" type="hidden" name="name">
</form>
<form action = "letao_DevelopmentModelArea_listArea" method="post" id="listForm">
    <input id="firstRankId" type="hidden" name="firstRankId" value="${current1.id }">
    <input id="secondRankId" type="hidden" name="secondRankId" value="${current2.id }">
</form>
</body>
<script>
$('.update').click(function(e){
    var dataId = $(e.target).attr("type");
    $("#hideId").val(dataId);
    $("#updateCode").val($("#code"+dataId).val());
    $("#updateName").val($("#name"+dataId).val());
    $('#updateForm').ajaxSubmit({
            type:'post',
            dataType: 'json',
            success: function (data) {
                if(data.status==1){
                    alert(data.info);
                    location.reload();
                }else{
                    alert(data.info);
                }
            },
            error: function (XMLResponse) {
                //alert(XMLResponse.responseText);
                var ss =JSON.stringify(XMLResponse);
                alert('操作失败！'+ss);
            }
        });
});
$('.del').click(function(e){
    var dataId = $(e.target).attr("type");
                $.getJSON("letao_DevelopmentModelArea_deleteInfo?id=" + dataId,
                        function(data) {
                            if (data.status == 1) {
                                alert(data.info);
                                location.reload();
                            } else {
                                alert(data.info);
                            }
                        });
            });
$('.add1').click(function(){
    $('#addForm1').ajaxSubmit({
        type:'post',
        dataType: 'json',
        success: function (data) {
            if(data.status==1){
                alert(data.info);
                location.reload();
            }else{
                alert(data.info);
            }
        },
        error: function (XMLResponse) {
            //alert(XMLResponse.responseText);
            var ss =JSON.stringify(XMLResponse);
            alert('操作失败！'+ss);
        }
    });
});
$('.add2').click(function(){
    $('#addForm2').ajaxSubmit({
        type:'post',
        dataType: 'json',
        success: function (data) {
            if(data.status==1){
                alert(data.info);
                location.reload();
            }else{
                alert(data.info);
            }
        },
        error: function (XMLResponse) {
            //alert(XMLResponse.responseText);
            var ss =JSON.stringify(XMLResponse);
            alert('操作失败！'+ss);
        }
    });
});
$('.add3').click(function(){
    $('#addForm3').ajaxSubmit({
        type:'post',
        dataType: 'json',
        success: function (data) {
            if(data.status==1){
                alert(data.info);
                location.reload();
            }else{
                alert(data.info);
            }
        },
        error: function (XMLResponse) {
            //alert(XMLResponse.responseText);
            var ss =JSON.stringify(XMLResponse);
            alert('操作失败！'+ss);
        }
    });
});
$('.select').click(function(e){
    if($(e.target).attr("type") === "1"){
        $('#firstRankId').val($(e.target).attr("id"));
    }else if($(e.target).attr("type") === "2"){
        $('#secondRankId').val($(e.target).attr("id"));
    }
    $('#listForm').submit();
});
</script>
</html>

 

　　　　查询用的sql语句使用in来构造。页面控制使用一级当前id和二级当前id作为请求参数刷新页面。

　　　　action的设计也是一般操作数据库的形式，不过列表作为页面展示，而其他作为REST接口。

package com.dijing.letao.action;

import java.util.List;

import com.dijing.letao.DJActionSupport;
import com.dijing.letao.NoNeedLogin;
import com.dijing.letao.dao.AreaDao;
import com.dijing.letao.model.Headquarters.Area;
import com.dijing.server.web.DJAction;
import com.dijing.server.web.action.JsonResult;
import com.opensymphony.xwork2.ActionContext;

/**
 * 地区开发模式校准控制器
 */
public class DevelopmentModelAreaAction extends DJActionSupport{
    private long id;
    private String name;
    private long parent;
    private long firstRankId;
    private long secondRankId;
    private long code;
    /**
     * letao_DevelopmentModelArea_listArea
     * @return
     * @throws Exception
     */
    @NoNeedLogin
    public String listArea() throws Exception {
        return executePage(() -> {
            AreaDao aDao = new AreaDao();
            if(firstRankId == 0 && secondRankId == 0){
                List<Area> rank1 = aDao.findByParent(0);
                List<Area> rank2 = null;
                List<Area> rank3 = null;
                if(rank1.size() > 0){
                    rank2 = aDao.findByParent(rank1.get(0).getId());
                    ActionContext.getContext().put("current1", rank1.get(0));
                }
                if(rank2.size() > 0){
                    rank3 = aDao.findByParent(rank2.get(0).getId());
                    ActionContext.getContext().put("current2", rank2.get(0));
                }
                ActionContext.getContext().put("rank1", rank1);
                ActionContext.getContext().put("rank2", rank2);
                ActionContext.getContext().put("rank3", rank3);
                return DJAction.SHOW;
            }else if(firstRankId != 0 && secondRankId == 0){
                List<Area> rank1 = aDao.findByParent(0);
                List<Area> rank2 = null;
                List<Area> rank3 = null;
                if(rank1.size() > 0){
                    rank2 = aDao.findByParent(rank1.get(0).getId());
                    ActionContext.getContext().put("current1", aDao.findByById(firstRankId));
                }
                if(rank2.size() > 0){
                    rank3 = aDao.findByParent(rank2.get(0).getId());
                    ActionContext.getContext().put("current2", rank2.get(0));
                }
                ActionContext.getContext().put("rank1", rank1);
                ActionContext.getContext().put("rank2", rank2);
                ActionContext.getContext().put("rank3", rank3);
                return DJAction.SHOW;
            }else if(firstRankId != 0 && secondRankId != 0){
                System.out.println("===================");
                List<Area> rank1 = aDao.findByParent(0);
                List<Area> rank2 = null;
                List<Area> rank3 = null;
                if(rank1.size() > 0){
                    rank2 = aDao.findByParent(rank1.get(0).getId());
                    ActionContext.getContext().put("current1", aDao.findByById(firstRankId));
                }
                if(rank2.size() > 0){
                    rank3 = aDao.findByParent(aDao.findByById(secondRankId).getId());
                    ActionContext.getContext().put("current2", aDao.findByById(secondRankId));
                }
                ActionContext.getContext().put("rank1", rank1);
                ActionContext.getContext().put("rank2", rank2);
                ActionContext.getContext().put("rank3", rank3);
                return DJAction.SHOW;
            }else
                return DJAction.FAILURE;
        });
    }
    /**
     * letao_DevelopmentModelArea_updateInfo
     * @return
     * @throws Exception
     */
    @NoNeedLogin
    public String updateInfo() throws Exception {
        return execute(() -> {
            if(name == null || code == 0 || id == 0)
                return ok(JsonResult.jsonResultError("请求为空"));
            AreaDao aDao = new AreaDao();
            Area a = aDao.findByById(id);
            a.setName(name);
            a.setCode(code);
            if(a.update())
                return ok(JsonResult.jsonResultSuccess("更新成功:"));
            else
                return ok(JsonResult.jsonResultError("更新失败"));
        });
    }
    /**
     * letao_DevelopmentModelArea_deleteInfo
     * @return
     * @throws Exception
     */
    @NoNeedLogin
    public String deleteInfo() throws Exception {
        if(id == 0)
            return ok(JsonResult.jsonResultError("请求为空"));
        AreaDao aDao = new AreaDao();
        Area a = aDao.findByById(id);
        if(a == null)
            return ok(JsonResult.jsonResultError("条目不存在"));
        List<Area> son1 = aDao.findByParent(a.getId());
        
        if(son1.size() >0){
            List<Area> son2 = aDao.findByParentList(son1);
            for(Area s:son2)
                s.delete();
        }
        for(Area s:son1)
            s.delete();
        a.delete();
        return ok(JsonResult.jsonResultSuccess("删除成功"));
    }
    /**
     * letao_DevelopmentModelArea_addArea
     * @return
     * @throws Exception
     */
    @NoNeedLogin
    public String addArea() throws Exception {
        return execute(() -> {
            if(name == null)
                return ok(JsonResult.jsonResultError("请求为空"));
            if(parent >0){
                AreaDao aDao = new AreaDao();
                if(aDao.findByParent(parent) == null)
                    ok(JsonResult.jsonResultError("父级不存在 添加不合法"));
            }
            Area a = new Area();
            a.setName(name);
            a.setCode(code);
            a.setParent(parent);
            if(a.save())
                return ok(JsonResult.jsonResultSuccess("添加成功:"));
            else
                return ok(JsonResult.jsonResultError("添加失败，已存在"));
        });
    }
    public long getId() {
        return id;
    }
    public void setId(long id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public long getParent() {
        return parent;
    }
    public void setParent(long parent) {
        this.parent = parent;
    }
    public long getFirstRankId() {
        return firstRankId;
    }
    public void setFirstRankId(long firstRankId) {
        this.firstRankId = firstRankId;
    }
    public long getSecondRankId() {
        return secondRankId;
    }
    public void setSecondRankId(long secondRankId) {
        this.secondRankId = secondRankId;
    }
    public long getCode() {
        return code;
    }
    public void setCode(long code) {
        this.code = code;
    }
}

　　　　dao层查询的方法，也是很简单的单表查询。

    public List<Area> findByParent(long parent) {
        return mysql.queryListSql(model, " where parent="+parent);
    }
    public List<Area> findByParentList(List<Area> list){
        StringBuilder sb = new StringBuilder();
        sb.append("where parent in (");
        sb.append(list.get(0).getId());
        for(int i=1;i<list.size();i++){
            sb.append(",");
            sb.append(list.get(i).getId());
        }
        sb.append(")");
        return mysql.queryListSql(model, sb.toString());
    }