以Java获取数据 [英] Retrieving Data in Java

问题描述

我是一个java新手。是否可以从网站获取数据，然后将其存储在某种数据结构中？例如，该计划在给定时间从雅虎金融机构获得股票的价值并存储。就像我说的，我不是那么熟练的Java，我想知道这是否可以完成。如果可以，是否很难做到？

解决方案

  public class GetYahooData 
 {
 public ArrayList< JSONObject> getOutputFromUrl（String url）
 {
 ArrayList< JSONObject> output = new ArrayList< JSONObject>（）; 
 HttpClient httpClient = new DefaultHttpClient（）; 
 HttpPost httpPost = new HttpPost（url）; 
 HttpResponse响应; 
 StringBuilder builder = new StringBuilder（）; 
 JSONObject myjson; 
 JSONArray the_json_array; 
 try 
 {
 response = httpClient.execute（httpPost）; 
 BufferedReader in = new BufferedReader（new InputStreamReader（response.getEntity（）。getContent（），UTF-8））; 
 char [] buf = new char [8000]; 
 int l = 0; 
 while（l> = 0）
 {
 builder.append（buf，0，l）; 
 l = in.read（buf）; 
} 
 myjson = new JSONObject（{child：+ builder.toString（）+}）; 
 JSONObject mmm = new JSONObject（builder.toString（））; 
 JSONArray mmmArr = mmm.getJSONArray（status）; 
 the_json_array = myjson.getJSONArray（child）; （int i = 0; i< the_json_array.length（）; i ++）
 {
 JSONObject another_json_object = the_json_array.getJSONObject（i）; // the_json_array.getJSONObject（i）; 
 output.add（another_json_object）; 
} 
} catch（ClientProtocolException e）{
 System.out.println（ClientProtocolException：+ e）; 
 e.printStackTrace（）; 
} catch（IOException e）{
 System.out.println（IOException：+ e）; 
 e.printStackTrace（）; 
} catch（JSONException e）{
 System.out.println（JSONException hussain：+ e）; 
 e.printStackTrace（）; 
} 
返回输出; 
} 
} 
 
 public class useYahoo 
 {
 public static void main（String args []）
 {
 String url =你想要的响应的url; 
 getYahooData object = new GetYahooData（）; 
 ArrayList< JSONObject> output = object.getOutputFromUrl（url）; 
} 
} 
  

I'm a java novice. Is it possible to get data from a website and then store it in some sort of data structure? For example, the program gets the value of a stock from yahoo finance at a given time and stores it. Like I said, I'm not that proficient with Java and I'd like to know if this could be done. If it can be, is it very hard to do it?

解决方案

    public class GetYahooData
    {
        public ArrayList<JSONObject> getOutputFromUrl(String url) 
        {
            ArrayList<JSONObject> output = new ArrayList<JSONObject>();
            HttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            HttpResponse response;
            StringBuilder builder= new StringBuilder();
            JSONObject myjson ;
            JSONArray the_json_array;
            try 
            {
                response = httpClient.execute(httpPost);
                BufferedReader in = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));
                char[] buf = new char[8000];
                int l = 0;
                    while (l >= 0) 
                    {
                        builder.append(buf, 0, l);
                        l = in.read(buf);
                    }
                myjson = new JSONObject("{child:"+builder.toString()+"}");
                JSONObject mmm = new JSONObject(builder.toString());
                JSONArray mmmArr = mmm.getJSONArray("status");
                the_json_array = myjson.getJSONArray("child");
                for (int i = 0; i < the_json_array.length(); i++) 
                {
                    JSONObject another_json_object =  the_json_array.getJSONObject(i);//the_json_array.getJSONObject(i);
                    output.add(another_json_object);
                }
            } catch (ClientProtocolException e) {
                System.out.println("ClientProtocolException :"+e);
                e.printStackTrace();
            } catch (IOException e) {
                System.out.println("IOException :"+e);
                e.printStackTrace();
            } catch (JSONException e) {
                System.out.println("JSONException hussain :"+e);
                e.printStackTrace();
            }
            return output;
        }
    }

public class useYahoo
{
    public static void main(String args[])
    {
        String url = "the url you want the response from";
        getYahooData object = new GetYahooData();
        ArrayList<JSONObject> output = object.getOutputFromUrl(url);
    }
}

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