从json文本文件加载JSONObject的最佳方法是什么？

Question

将包含JSON的文件加载到JSONObject中的最简单方法是什么.

目前我正在使用json-lib.

这就是我所拥有的,但它抛出异常:

XMLSerializer xml = new XMLSerializer();
JSON json = xml.readFromFile("samples/sample7.json”);     //line 507
System.out.println(json.toString(2));

输出是:

Exception in thread "main" java.lang.NullPointerException
    at java.io.Reader.<init>(Reader.java:61)
    at java.io.InputStreamReader.<init>(InputStreamReader.java:55)
    at net.sf.json.xml.XMLSerializer.readFromStream(XMLSerializer.java:386)
    at net.sf.json.xml.XMLSerializer.readFromFile(XMLSerializer.java:370)
    at corebus.test.deprecated.TestMain.main(TestMain.java:507)

Answer 1

谢谢@Kit Ho的回答.我使用了你的代码,发现我一直遇到错误,我的InputStream总是为null,而在创建JSONObject时则是ClassNotFound异常.这是我的代码版本,它为我提供了一些技巧:

import java.io.File;
import java.io.FileInputStream;
import java.io.InputStream;
import org.apache.commons.io.IOUtils;

import org.json.JSONObject;
public class JSONParsing {
    public static void main(String[] args) throws Exception {
        File f = new File("file.json");
        if (f.exists()){
            InputStream is = new FileInputStream("file.json");
            String jsonTxt = IOUtils.toString(is, "UTF-8");
            System.out.println(jsonTxt);
            JSONObject json = new JSONObject(jsonTxt);       
            String a = json.getString("1000");
            System.out.println(a);   
        }
    }
}

我发现这个答案对于FileInputStream和getResourceAsStream之间的区别很有启发性.希望这也有助于其他人.

Answer 2

试试这个:

import net.sf.json.JSONObject;
import net.sf.json.JSONSerializer;
import org.apache.commons.io.IOUtils; 

    public class JsonParsing {

        public static void main(String[] args) throws Exception {
            InputStream is = 
                    JsonParsing.class.getResourceAsStream( "sample-json.txt");
            String jsonTxt = IOUtils.toString( is );

            JSONObject json = (JSONObject) JSONSerializer.toJSON( jsonTxt );        
            double coolness = json.getDouble( "coolness" );
            int altitude = json.getInt( "altitude" );
            JSONObject pilot = json.getJSONObject("pilot");
            String firstName = pilot.getString("firstName");
            String lastName = pilot.getString("lastName");

            System.out.println( "Coolness: " + coolness );
            System.out.println( "Altitude: " + altitude );
            System.out.println( "Pilot: " + lastName );
        }
    }

这是你的sample-json.txt,应该是json格式

{
 'foo':'bar',
 'coolness':2.0,
 'altitude':39000,
 'pilot':
     {
         'firstName':'Buzz',
         'lastName':'Aldrin'
     },
 'mission':'apollo 11'
}

Answer 3

使用java 8,您可以尝试这样:

import org.json.JSONException;
import org.json.JSONObject;

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;

public class JSONUtil {

    public static JSONObject parseJSONFile(String filename) throws JSONException, IOException {
        String content = new String(Files.readAllBytes(Paths.get(filename)));
        return new JSONObject(content);
    }

    public static void main(String[] args) throws IOException, JSONException {
        String filename = "path/to/file/abc.json";
        JSONObject jsonObject = parseJSONFile(filename);

        //do anything you want with jsonObject
    }
}