从目标主机获取JSON对象。错误:目标主机不得为空。

3
我现在正在尝试理解JSON的工作原理,所以我想做一个小的JSON阅读项目来更好地学习。我希望用户可以使用EditText和搜索按钮,并查找用户输入的网站是否有任何JSON数据,然后在文本框中显示它。目前,我正在测试网站:itunes.apple.com/search?term= 正如您在我的代码中所看到的,我只是设置了那些变量,没有使用EditText和Search按钮,直到我把这个东西搞定。我得到了以下错误:
11-03 20:30:58.545: E/AndroidRuntime(8608): FATAL EXCEPTION: AsyncTask #1
11-03 20:30:58.545: E/AndroidRuntime(8608): java.lang.RuntimeException: An error occured while executing doInBackground()
11-03 20:30:58.545: E/AndroidRuntime(8608):     at android.os.AsyncTask$3.done(AsyncTask.java:299)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at java.util.concurrent.FutureTask$Sync.innerSetException(FutureTask.java:273)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at java.util.concurrent.FutureTask.setException(FutureTask.java:124)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:307)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at java.util.concurrent.FutureTask.run(FutureTask.java:137)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1076)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:569)
11-03 20:30:58.545: E/AndroidRuntime(8608):     at java.lang.Thread.run(Thread.java:856)
11-03 20:30:58.545: E/AndroidRuntime(8608): Caused by: java.lang.IllegalStateException:
 Target host must not be null, or set in parameters. scheme=null, host=null, path=itunes.apple.com/search

我相信这是由于以下这行代码引起的:HttpResponse r = client.execute(get);,但我不知道应该如何修复主机为空的错误。我觉得这可能与问号或其他某些东西有关。或者可能我是以错误的方式来解决这个问题。请考虑到这是我第一次使用JSON进行项目开发。也许我需要一个更简单的关于如何使用JSON的教程,以便我能够彻底理解它。谢谢!
public class MainActivity extends Activity implements OnClickListener {

TextView fetchText;
EditText httpEntryBox,jsonEntryBox;
Button search_button_http, search_button_json;
HTTP_Fetcher http_fetch;

HttpClient client;
final static String URL = "itunes.apple.com/search?term=";
JSONObject json;
InputMethodManager imm;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    fetchText = (TextView) findViewById(R.id.fetchText);

    client = new DefaultHttpClient();
    new Read().execute("artistName");


}

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    getMenuInflater().inflate(R.menu.main, menu);
    return true;
}

public JSONObject search( String searchItem ) throws ClientProtocolException, IOException, JSONException, URISyntaxException{
    StringBuilder url = new StringBuilder(URL);
    url.append(searchItem);

    String finished_url = url.toString();

    HttpGet get = new HttpGet(finished_url);
    HttpResponse r = client.execute(get);

    int status = r.getStatusLine().getStatusCode();

    if (status == 200){
        HttpEntity e = r.getEntity();
        String data = EntityUtils.toString(e);
        JSONArray array = new JSONArray(data);
        return array.getJSONObject(0);
    }else{
        Log.e("Search","fetch error");
        return null;
    }


}

public class Read extends AsyncTask<String, Integer, String>{

    @Override
    protected String doInBackground(String... params) {
        try {
            json = search("lilwayne");
            return json.getString(params[0]);
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } catch (JSONException e) {
            e.printStackTrace();
        } catch (URISyntaxException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return null;
    }

    @Override
    protected void onPostExecute(String result) {
        fetchText.setText(result);
    }

}
}
1个回答

3
我认为您遗漏了http协议,所以应该这样写:
final static String URL = "itunes.apple.com/search?term=";

put:

final static String URL = "http://itunes.apple.com/search?term=";

编辑
即使这样做是有效的,但JSON解析将失败,因为这一行:JSONArray array = new JSONArray(data); - 它假设响应是一个数组,但实际上它是一个包含两个条目: resultsCountresults 的JSONObject。你需要将其变成一个JSONObjectJSONObject obj = new JSONObject(data);。互联网上有很多关于android json解析的教程。这是一篇相关文章


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