如何从Java中的URL XML响应，并将其存储在MongoDB中

问题描述：

我想请求一个URL将返回类似以下内容的XML响应：如何从Java中的URL XML响应，并将其存储在MongoDB中

<SERVERPERFORMANCEMETRICS> 
<SERVERNAME>abc</SERVERNAME> 
<SERVERSTARTUPTIME>11/25/2016 11:00 AM</SERVERSTARTUPTIME> 
<TOTALMASTERREQUESTS>4</TOTALMASTERREQUESTS> 
<TOTALRENDERERREQUESTS>13</TOTALRENDERERREQUESTS> 
<FAILEDREQUESTS>10</FAILEDREQUESTS> 
</SERVERPERFORMANCEMETRICS>

，现在我想分析和存储该XML的数据在MONGODB。

 URL url; 
     HttpURLConnection urlConnection = null; 


     MongoClientURI mongoClientURI = new MongoClientURI(dbURL); 
     MongoClient mongoClient = new MongoClient(mongoClientURI); 

     /* To connect database, you need to specify the database name, if the database 
      doesn't exist then MongoDB creates it automatically. */ 
     MongoDatabase mongoDatabase = mongoClient.getDatabase("myDb"); 
     System.out.println("Connected to database successfully"); 

     MongoCollection<Document> mongoCollection = mongoDatabase.getCollection("myCollection"); 
     System.out.println("Collection created successfully"); 

     url = new URL(targetURL); 
     urlConnection = (HttpURLConnection) url.openConnection(); 
     InputStream inputStream = urlConnection.getInputStream(); 
     InputStreamReader inputStreamReader = new InputStreamReader(inputStream);

请，张贴XML作为代码，而不是作为图片找到了解决办法。 –

' ABC 2016年11月25日上午11时 SERVERSTARTUPTIME> 4 TOTALMASTERREQUESTS> 13 TOTALRENDERERREQUESTS> 10 SERVERPERFORMANCEMETRICS>' –

答

我用自己JSON-java的罐子

 if (inputStream != null) { 
      BufferedReader bufferedReader = new BufferedReader(inputStreamReader); 
      //BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8")); 
      StringBuilder responseStrBuilder = new StringBuilder(); 

      String inputStr; 
      while ((inputStr = bufferedReader.readLine()) != null) { 
       responseStrBuilder.append(inputStr); 
      } 

      JSONObject jsonObject = XML.toJSONObject(responseStrBuilder.toString()); 
      System.out.println(jsonObject); 

      org.bson.Document jsonDocument = org.bson.Document.parse(jsonObject.toString()); 
      mongoCollection.insertOne(jsonDocument); 
      System.out.println("Collection inserted successfully"); 
     }

答

使用W3C的DocumentBuilder：

DocumentBuilder builder = DocumentBuilderFactory.newInstance().newDocumentBuilder(); 
Document doc = builder.parse(inputStream);

如何从Java中的URL XML响应，并将其存储在MongoDB中

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