Question

我可以从相同的代码中检索JSON数据，但无法从此URL检索数据，为什么这不会返回任何响应代码或者没有任何数据响应。请帮我解决此问题，或建议我如何从网址中检索非JSON数据。

   URL uniprotFasta = new URL(ur);
  url = "http://www.uniprot.org/uniprot/P69905.fasta";

 URL uniprotFasta = new URL(ur);
 HttpURLConnection conn = (HttpURLConnection) uniprotFasta.openConnection();
    conn.connect();
    int response = conn.getResponseCode();
    Log.e("Response Code * *", String.valueOf(response));
    InputStream is = conn.getInputStream();
    // Convert the InputStream into a string
    String contentAsString = readIt(is, 500);
    Log.e("HTTPURLCON * *",contentAsString);

我也试过这个，但没有得到回应。

String data = "";
URLConnection conn = uniprotFasta.openConnection();
BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String line = "";
while((line = reader.readLine()) != null){
     data += line;
}
Log.e("DATA * *",data);

Answer 1

好的，我已经为你试过了。

在Asyntask中编写代码。

public class TestUrl extends AsyncTask<Void, Void, Void> {

@Override
protected Void doInBackground(Void... params) {
    String url = "http://www.uniprot.org/uniprot/P69905.fasta";
    HttpURLConnection conn = null;
    InputStream is = null;
    try {
        URL uniprotFasta = new URL(url);
        conn = (HttpURLConnection) uniprotFasta.openConnection();
        conn.connect();
        int response = conn.getResponseCode();
        Log.e("Response Code * *", String.valueOf(response));
        is = conn.getInputStream();
    } catch (IOException e) {
        e.printStackTrace();
    }
    // Convert the InputStream into a string
    String contentAsString = convertStreamToString(is);
    Log.e("HTTPURLCON * *", contentAsString);

    return null;
}

将输入流转换为String

的函数

private String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();

String line = null;
try {
    while ((line = reader.readLine()) != null) {
        sb.append(line).append('\n');
    }
} catch (IOException e) {
    e.printStackTrace();
} finally {
    try {
        is.close();
    } catch (IOException e) {
        e.printStackTrace();
    }
}
return sb.toString();

}

在清单文件中授予权限

<uses-permission android:name="android.permission.INTERNET" />

执行将在后台线程中运行代码的Asynctask。
```
new TestUrl().execute();
```

你会得到的回应是

E/Response Code * *: 200
E/HTTPURLCON * *: >sp|P69905|HBA_HUMAN Hemoglobin subunit alpha OS=Homo    sapiens GN=HBA1 PE=1 SV=2
          MVLSPADKTNVKAAWGKVGAHAGEYGAEALERMFLSFPTTKTYFPHFDLSHGSAQVKGHG
          KKVADALTNAVAHVDDMPNALSALSDLHAHKLRVDPVNFKLLSHCLLVTLAAHLPAEFTP
          AVHASLDKFLASVSTVLTSKYR

我希望这会对你有所帮助。

Answer 2

//you Should Declare Header type , Content 

public static void main(String[] args) {

    String urlService="http://www.uniprot.org/uniprot/P69905.fasta";
    DefaultHttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost(urlService);

    try {

        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");
        HttpResponse httpResponse = httpClient.execute(httpPost);
        String result = EntityUtils.toString(httpResponse.getEntity());
        httpClient.getConnectionManager().shutdown();
        System.out.println(result);
    } catch (Exception e) {
        System.out.println("Exception" +e);
    }
}

无法从给定的URL返回任何响应数据

2 个答案: