package com.lyr.easymybatis.util;
import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.JSONArray;
import com.alibaba.fastjson.JSONObject;
import java.util.Collection;
import java.util.List;
import java.util.Map;
/**
* @Author lyr
* @create 2020/12/16 23:13
*/
public class MarkdownUtil {
private static void json2MarkdownList(StringBuilder output,Object obj,int stack,String key) {
if (obj instanceof Collection) {
//数组类型
output.append(pre(stack));
//嵌套json类型
output.append(formatName(key,"list",stack));
output.append(endl());
// for(Object nested: (Collection)obj) {
if (((Collection) obj).size()>0) {
json2MarkdownList(output, ((List)obj).get(0),stack+1,"[+]");
}
// }
}else if (obj instanceof Map) {
output.append(pre(stack));
//嵌套json类型
output.append(formatName(key,"Object",stack));
output.append(endl());
((Map) obj).forEach((k,v) -> {
json2MarkdownList(output, v, stack+1,k.toString());
});
}else{
//单值类型
output.append(pre(stack));
String type = "number";
if (obj instanceof String) type = "String";
output.append(formatName(key,type,stack));
output.append(endl());
}
}
private static String formatName(String key,String type,int stack) {
return key+" 【 type= "+type+" ,required = true , value = "+key +" 】";
}
private static String endl() {
return "\r\n";
}
private static String pre(int stack) {
if (stack<=0) return "- ";
int len = (stack)*4;
StringBuilder s = new StringBuilder();
for (int i=0;i<len;++i) s.append(" ");
s.append("- ");
return s.toString();
}
public static String json2MarkdownList(String json) {
JSONObject jsn = JSON.parseObject(json);
StringBuilder sb = new StringBuilder();
json2MarkdownList(sb,jsn,-1,"");
return sb.toString();
}
}
@Test
void outPUTMarkdown() {
String json = "{\n" +
" \"data\": {\n" +
" \"total\": 15,\n" +
" \"pages\": 3,\n" +
" \"recommend\": [\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 29,\n" +
" \"title\": \"二叉树最大路径和\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"题目描述\\r\\n给定一个二叉树，请计算节点值之和最大的路径的节点值之和是多少。\\r\\n这个路径的开始节点和结束节点可以是二叉树中的任意节点\\r\\n例如：\\r\\n给出以下的二叉树，\\r\\n \\r\\n返回的结果为6\\r\\n示例1\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" },\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 28,\n" +
" \"title\": \"分割链表\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前\\r\\n\\r\\n给定一个链表的头指针 ListNode* pHead，请返回重新排列后的链表的头指针。注意：分割以后保持原来的数据顺序不变。\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" },\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 27,\n" +
" \"title\": \"dp之 找零\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"题目描述\\r\\nZ国的货币系统包含面值1元、4元、16元、64元共计4种硬币，以及面值1024元的纸币。现在小Y使用1024元的纸币购买了一件价值为N (0 < N \\\\le 1024)N(0<N≤1024)的商品，请问最少他会收到多少硬币？\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" },\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 26,\n" +
" \"title\": \"之字形层序遍历\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"给定一个二叉树，返回该二叉树的之字形层序遍历，（第一层从左向右，下一层从右向左，一直这样交替）\\r\\n例如：\\r\\n给定的二叉树是{3,9,20,#,#,15,7},\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" },\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 25,\n" +
" \"title\": \"合并K个已排序的链表\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"题目描述\\r\\n合并\\\\ k k 个已排序的链表并将其作为一个已排序的链表返回。分析并描述其复杂度。\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" },\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 24,\n" +
" \"title\": \"返回一颗二叉搜索树的第k小节点\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"给定一棵二叉搜索树，请找出其中的第k小的结点。例如， （5，3，7，2，4，6，8） 中，按结点数值大小顺序第三小结点的值为4。\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" },\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 23,\n" +
" \"title\": \"#栈#括号匹配问题\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"判断括号序列是否合法\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" },\n" +
" {\n" +
" \"blog\": {\n" +
" \"commentCount\": null,\n" +
" \"id\": 22,\n" +
" \"title\": \"ps将图片设计稿调节为 横向750px\",\n" +
" \"firstPicture\": null,\n" +
" \"flag\": null,\n" +
" \"views\": null,\n" +
" \"appreciation\": null,\n" +
" \"shareStatement\": null,\n" +
" \"commentabled\": null,\n" +
" \"published\": null,\n" +
" \"recommend\": null,\n" +
" \"createTime\": null,\n" +
" \"updateTime\": null,\n" +
" \"typeId\": null,\n" +
" \"userId\": null,\n" +
" \"tagIds\": null,\n" +
" \"content\": null,\n" +
" \"description\": \"ps 调节 设计稿 横向 750px\"\n" +
" },\n" +
" \"type\": null,\n" +
" \"tags\": [],\n" +
" \"user\": null,\n" +
" \"commentCount\": null,\n" +
" \"tagIds\": \"\"\n" +
" }\n" +
" ],\n" +
" \"contentList\": [\n" +
" {\n" +
" \"commentCount\": 0,\n" +
" \"id\": 29,\n" +
" \"title\": \"二叉树最大路径和\",\n" +
" \"firstPicture\": \"http://img.itshare.club/20200915222524.png\",\n" +
" \"flag\": \"原创\",\n" +
" \"views\": 4,\n" +
" \"appreciation\": 1,\n" +
" \"shareStatement\": 1,\n" +
" \"commentabled\": 1,\n" +
" \"published\": 1,\n" +
" \"recommend\": 1,\n" +
" \"createTime\": \"2020-09-15T14:25:49.000+00:00\",\n" +
" \"updateTime\": \"2020-09-15T14:25:49.000+00:00\",\n" +
" \"typeId\": 13,\n" +
" \"userId\": null,\n" +
" \"tagIds\": \"12\",\n" +
" \"content\": null,\n" +
" \"description\": \"题目描述\\r\\n给定一个二叉树，请计算节点值之和最大的路径的节点值之和是多少。\\r\\n这个路径的开始节点和结束节点可以是二叉树中的任意节点\\r\\n例如：\\r\\n给出以下的二叉树，\\r\\n \\r\\n返回的结果为6\\r\\n示例1\"\n" +
" },\n" +
" {\n" +
" \"commentCount\": 0,\n" +
" \"id\": 22,\n" +
" \"title\": \"ps将图片设计稿调节为 横向750px\",\n" +
" \"firstPicture\": \"http://img.itshare.club/20200902000615.webp\",\n" +
" \"flag\": \"原创\",\n" +
" \"views\": 1,\n" +
" \"appreciation\": 1,\n" +
" \"shareStatement\": 1,\n" +
" \"commentabled\": 1,\n" +
" \"published\": 1,\n" +
" \"recommend\": 1,\n" +
" \"createTime\": \"2020-09-01T16:07:01.000+00:00\",\n" +
" \"updateTime\": \"2020-09-08T06:06:16.000+00:00\",\n" +
" \"typeId\": 15,\n" +
" \"userId\": null,\n" +
" \"tagIds\": \"13\",\n" +
" \"content\": null,\n" +
" \"description\": \"ps 调节 设计稿 横向 750px\"\n" +
" },\n" +
" {\n" +
" \"commentCount\": 0,\n" +
" \"id\": 25,\n" +
" \"title\": \"合并K个已排序的链表\",\n" +
" \"firstPicture\": \"http://img.itshare.club/20200903190846.png\",\n" +
" \"flag\": \"原创\",\n" +
" \"views\": 1,\n" +
" \"appreciation\": 1,\n" +
" \"shareStatement\": 1,\n" +
" \"commentabled\": 1,\n" +
" \"published\": 1,\n" +
" \"recommend\": 1,\n" +
" \"createTime\": \"2020-09-03T11:09:44.000+00:00\",\n" +
" \"updateTime\": \"2020-09-08T06:06:03.000+00:00\",\n" +
" \"typeId\": 14,\n" +
" \"userId\": null,\n" +
" \"tagIds\": \"12\",\n" +
" \"content\": null,\n" +
" \"description\": \"题目描述\\r\\n合并\\\\ k k 个已排序的链表并将其作为一个已排序的链表返回。分析并描述其复杂度。\"\n" +
" },\n" +
" {\n" +
" \"commentCount\": 0,\n" +
" \"id\": 23,\n" +
" \"title\": \"#栈#括号匹配问题\",\n" +
" \"firstPicture\": \"http://img.itshare.club/20200903011814.png\",\n" +
" \"flag\": \"原创\",\n" +
" \"views\": 4,\n" +
" \"appreciation\": 1,\n" +
" \"shareStatement\": 1,\n" +
" \"commentabled\": 1,\n" +
" \"published\": 1,\n" +
" \"recommend\": 1,\n" +
" \"createTime\": \"2020-09-02T17:18:55.000+00:00\",\n" +
" \"updateTime\": \"2020-09-08T06:05:54.000+00:00\",\n" +
" \"typeId\": 14,\n" +
" \"userId\": null,\n" +
" \"tagIds\": \"12\",\n" +
" \"content\": null,\n" +
" \"description\": \"判断括号序列是否合法\"\n" +
" },\n" +
" {\n" +
" \"commentCount\": 0,\n" +
" \"id\": 19,\n" +
" \"title\": \"给定链表，返回环的入口节点\",\n" +
" \"firstPicture\": \"http://img.itshare.club/20200829013310.webp\",\n" +
" \"flag\": \"原创\",\n" +
" \"views\": 0,\n" +
" \"appreciation\": 1,\n" +
" \"shareStatement\": 1,\n" +
" \"commentabled\": 1,\n" +
" \"published\": 1,\n" +
" \"recommend\": 1,\n" +
" \"createTime\": \"2020-08-28T17:33:26.000+00:00\",\n" +
" \"updateTime\": \"2020-09-08T06:05:42.000+00:00\",\n" +
" \"typeId\": 14,\n" +
" \"userId\": null,\n" +
" \"tagIds\": \"12\",\n" +
" \"content\": null,\n" +
" \"description\": \"对于一个给定的链表，返回环的入口节点，如果没有环，返回null\"\n" +
" }\n" +
" ]\n" +
" },\n" +
" \"code\": 200,\n" +
" \"message\": \"ok\"\n" +
"} ";
System.out.println(MarkdownUtil.json2MarkdownList(json));
}
效果预览：
