将.json文件转换为JSONArray

Question

我使用cURL以json文件（“twitter-feed.json”）的形式获取一些Twitter提要。我想将这个json文件转换为JSONArray对象。我该怎么做？

我是Java和json的新手。您的建议是最受欢迎的。

FileInputStream infile = new FileInputStream("input/twitter-feed.json"); 

//解析JSON JSONArray jsonArray =新JSONArray（字符串）;

// use 
    for (int i = 0; i < jsonArray.length(); i++) { 
     JSONObject jsonObject = jsonArray.getJSONObject(i); 

     System.out.println(jsonObject.getString("id")); 
     System.out.println(jsonObject.getString("text"));    
     System.out.println(jsonObject.getString("created_at"));  
    } 

谢谢， PD。

Answer 1

您可以尝试Gson：

对于只是阵列可以使用：

Gson gson = new Gson(); 

//(Deserialization) 
int[] ints2 = gson.fromJson("[1,2,3,4,5]", int[].class); 

反序列化对象的数组，你可以这样做：

Container container = new Gson().fromJson(json, Container.class); 

由于shown here

Answer 2

你需要先读取文件，将其转换为String，然后将其送到JSONArray（我假设您使用的是JSON-Java Project。下面的代码演示了如何读取文件并将其设置为JSONArray

// read the source file, source comes from streaming API delimited by newline 
// done by curl https://stream.twitter.com/1/statuses/sample.json?delimited=newline -utwitterUsername:twitterPasswd 
// > /Projects/StackOverflow/src/so7655570/twitter.json 
FileReader f = new FileReader("/Projects/StackOverflow/src/so7655570/twitter.json"); 
BufferedReader br = new BufferedReader(f); 

ArrayList jsonObjectArray = new ArrayList(); 
String currentJSONString = ""; 

// read the file, since I ask for newline separation, it's easier for BufferedReader 
// to separate each String 
while((currentJSONString = br.readLine()) != null) { 
    // create new JSONObject 
    JSONObject currentObject = new JSONObject(currentJSONString); 

    // there are more than one way to do this, right now what I am doing is adding 
    // each JSONObject to an ArrayList 
    jsonObjectArray.add(currentObject); 
} 

for (int i = 0; i < jsonObjectArray.size(); i++) { 
    JSONObject jsonObject = jsonObjectArray.get(i); 

    // check if it has valid ID as delete won't have one 
    // sample of JSON for delete : 
    // {"delete":{"status":{"user_id_str":"50269460","id_str":"121202089660661760","id":121202089660661760,"user_id":50269460}}} 

    if(jsonObject.has("id")) { 
     System.out.println(jsonObject.getInt("id")); 
     System.out.println(jsonObject.getString("text"));    
     System.out.println(jsonObject.getString("created_at") + "\n");  
    } 
} 

步骤说明：

• 流API不提供有效的JSON作为一个整体，而是由delimited field指定一个有效的。这就是为什么，你不能仅仅解析整个结果。
• 为了解析JSON，我用的是delimited使用newline因为BufferedReader有一个方法readLine，我们可以直接用它来获取每个JSONObject的
• 有一次，我从每一行获取每个有效的JSON，我创建JSONObject并添加它到ArrayList
• 然后，我重复每个JSONObject在ArrayList并打印出结果。请注意，如果您想立即使用该结果，并没有需要在以后使用它，你可以做加工本身在while循环没有将它们存储在其中的代码更改为ArrayList：

// read the source file, source comes from streaming API 
// done by curl https://stream.twitter.com/1/statuses/sample.json?delimited=newline -utwitterUsername:twitterPasswd 
// > /Projects/StackOverflow/src/so7655570/twitter.json 
FileReader f = new FileReader("/Projects/StackOverflow/src/so7655570/twitter.json"); 
BufferedReader br = new BufferedReader(f); 

String currentJSONString = ""; 

// read the file, since I ask for newline separation, it's easier for BufferedReader 
// to separate each String 
while((currentJSONString = br.readLine()) != null) { 
    // create new JSONObject 
    JSONObject currentObject = new JSONObject(currentJSONString); 

    // check if it has valid ID as delete status won't have one 
    if(currentObject.has("id")) { 
     System.out.println(currentObject.getInt("id")); 
     System.out.println(currentObject.getString("text"));    
     System.out.println(currentObject.getString("created_at") + "\n");  
    } 
} 

Answer 3

从杰克逊使用ObjectMapper类库是这样的：

//JSON from file to Object 
Staff obj = mapper.readValue(new File("c:\\file.json"), Staff.class); 

//JSON from URL to Object 
Staff obj = mapper.readValue(new URL("http://mkyong.com/api/staff.json"), Staff.class); 

//JSON from String to Object 
Staff obj = mapper.readValue(jsonInString, Staff.class);