android - 通过 HTTP 访问 Web 服务时未获得所需的输出

Question

我正在尝试使用 Web 服务将摄氏度转换为华氏度，但未获得所需的输出。谁能帮帮我……这是我的代码。

        HttpClient httpClient = new DefaultHttpClient(); 
        HttpContext localContext = new BasicHttpContext(); 
        httpClient.getParams().setParameter("Celsius", "32");

        HttpPost httpGet = new HttpPost("http://www.w3schools.com/webservices/tempconvert.asmx"); 
        HttpResponse response = httpClient.execute(httpGet, localContext); 

       tv.setText(response.toString());
    }
    catch (Exception e) {
        tv.setText(e.toString());
    }
}

Answer 1

首先，HttpParams 不是查询/GET 参数的集合。正如文档所说，它用于“HTTP 协议(protocol)和框架参数”。因此，在 GET 请求的情况下，您可以通过将“?Celsius=32”附加到该 URL 或使用 Uri.Builder 来添加查询参数。对于 POST 请求，您必须使用 setEntity 方法。就像这个例子( source ):

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);

// add POST params
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("foo", "123"));
nameValuePairs.add(new BasicNameValuePair("bar", "456"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

HttpResponse response = httpclient.execute(httppost);

其次，为了将响应读取为 String，您需要这样的东西:

HttpResponse response = httpClient.execute(request);
InputStream content = response.getEntity().getContent();
String json = convertStreamToString(content);

// ...

private String convertStreamToString(InputStream input) throws IOException {
    Reader inputReader = new InputStreamReader(input);
    BufferedReader reader = new BufferedReader(inputReader, 8192);
    StringBuilder string = new StringBuilder();
    String line = null;

    try {
      while ((line = reader.readLine()) != null) {
        string.append(line + "\n");
      }
    } finally {
      input.close();
    }

    return string.toString();
}