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  • 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());
        }
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  • 原文地址:https://www.cnblogs.com/chentingk/p/6252072.html
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