将json字符串从外层到最内层依次连接，平铺成一个List＜String＞

Posted 2023-03-10 天涯泪小武

就是将json平铺的功能，如

String jsonString = "\\"a\\": \\"b\\": \\"c\\": 1, \\"d\\": [2, 3]";

变成

[a.b.c=1, d[0]=2, d[1]=3]

这样能得到整个json的所有key，如果需要排重，可以用Set

import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.JSONArray;
import com.alibaba.fastjson.JSONObject;

import java.util.ArrayList;
import java.util.List;

public class JSONUtils 

    public static List<String> flattenJson(String jsonString) 
        List<String> flattened = new ArrayList<>();
        flattenJson(JSON.parseObject(jsonString), "", flattened);
        return flattened;
    

    private static void flattenJson(JSONObject obj, String prefix, List<String> flattened) 
        for (String key : obj.keySet()) 
            Object value = obj.get(key);
            String newPrefix = prefix + key;
            if (value instanceof JSONObject) 
                flattenJson((JSONObject) value, newPrefix + ".", flattened);
             else if (value instanceof JSONArray) 
                JSONArray arr = (JSONArray) value;
                for (int i = 0; i < arr.size(); i++) 
                    Object arrValue = arr.get(i);
                    if (arrValue instanceof JSONObject) 
                        flattenJson((JSONObject) arrValue, newPrefix + "[" + i + "].", flattened);
                     else 
                        flattened.add(newPrefix + "[" + i + "]=" + arrValue.toString() + "(" + arrValue.getClass().getSimpleName() + ")");
                    
                
             else 
                flattened.add(newPrefix + "=" + value.toString() + "(" + value.getClass().getSimpleName() + ")");