【问题标题】:Letting AsyncTask have a List<NameValuePair> as Parameter让 AsyncTask 有一个 List<NameValuePair> 作为参数
【发布时间】:2013-08-17 19:11:03
【问题描述】:

我想以JSON 的形式将数据发送到我的服务器。首先,它运行良好,但现在我不得不更改类以使用AsyncTask 进行一些网络操作,因为我遇到了android.os.NetworkOnMainThreadException 的问题,解决方案是使用AsyncTask。但是,我遇到了构造函数UrlEncodedFormEntity(List&lt;NameValuePair&gt;[]) 未定义的问题。那么,我应该在这门课中改变什么?

我的代码:

public class JSONParser extends AsyncTask<List<NameValuePair>, Void, String> {
private static String registerURL = "http://sit-edu4.sit.kmutt.ac.th/csc498/53270327/Boss/sftrip/index.php/register";
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

protected String doInBackground(List<NameValuePair>... params) {
    DefaultHttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost(registerURL);
    // Making HTTP request
    try {
        // defaultHttpClient

        httpPost.setEntity(new UrlEncodedFormEntity(params));

        HttpResponse httpResponse = httpClient.execute(httpPost);
        StatusLine statusLine = httpResponse.getStatusLine();
        int statusCode = statusLine.getStatusCode();
        if (statusCode == 200) {
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();
        } else {
            Log.e("Log", "Failed to download result..");
        }

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        if (is != null) {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        json = sb.toString();
        Log.e("JSON", json);
        } else {
            Log.e("Log", "Something wrong with IS");
        }
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }
    return json;
}

}

【问题讨论】:

    标签: android parameters arraylist android-asynctask


    【解决方案1】:

    这样做:

    public class JSONParser extends AsyncTask<NameValuePair, Void, String> {
    
         protected String doInBackground(NameValuePair... params) {
    
         }
    }
    

    (NameValuePair...params) 实际上意味着方法 doInBackground 可以有一个未指定数量的参数,因此您可以将例如一个数组传递给它。

    new JSONParser().execute(new NameValuePair[] { .... your namevaluepairs ... });
    

    此外,您可以考虑更改此行:

    httpPost.setEntity(new UrlEncodedFormEntity(params));
    

    到这里:

    httpPost.setEntity(new UrlEncodedFormEntity(params[0]));
    

    这将获得 NameValuePair ArrayLists 的“params”数组的第一项。

    【讨论】:

      【解决方案2】:

      您始终可以将 ArrayList 作为参数传递给构造函数,并将其作为字段存储在 JSONParser 中:

      public JSONParser(List <NameValuePair> list) { this.list = list; }
      

      同时添加私有字段:

      private ArrayList <NameValuePair> list = null;
      

      现在您可以在 AsyncTask 中随意操作列表。

      【讨论】:

        【解决方案3】:
        public static String getHttpResponse(String url, List<NameValuePair> nameValuePairs) {
        
                HttpClient httpClient = new DefaultHttpClient();
                HttpProtocolParams.setUseExpectContinue(httpClient.getParams(),true);
        
                HttpPost httpPost = new HttpPost(url);
                UrlEncodedFormEntity entity;
                try {
                    entity = new UrlEncodedFormEntity(nameValuePairs);
                    httpPost.setEntity(entity);
                    HttpResponse response = httpClient.execute(httpPost);
        
                    HttpEntity resEntity = response.getEntity();
                    String res =  EntityUtils.toString(resEntity);
                    return res;
                } catch (UnsupportedEncodingException e) {
                    e.printStackTrace();
                } catch (ClientProtocolException e) {
                    e.printStackTrace();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                return null;
            }
        

        【讨论】:

        • 感谢您的回答。您还应该在代码中添加简短描述。
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