Question

有人可以解释为什么以及如何修复我在Eclipse编程或Android时遇到的错误消息？

无法将HTTP解析为类型

代码示例如下。

try {
                // Call class "HTTP" passing an URL using Async "execute"
                // Save returned string in "result"
                String result = new HTTP().execute("http://domain.com/file.php?ID=1&R=T").get();

                Log.i("EmonLog", "Result: "+result);

                result = result.replaceAll("\"","");
                TextView powerval = (TextView) findViewById(R.id.information_message);
                powerval.setText("Result: " + result);
            } catch (Exception e) {
                Log.i("EmonLog", "Error: "+e);
            }

该程序编译和工作完美，但我得到此错误和HTTP下的红线，我想了解原因，以及如何解决它。

编辑...添加了更多信息。这是执行调用的线程。

    class HTTP extends AsyncTask<String, Void, String>
{
    @Override
    protected String doInBackground(String... params) {
        String result = "";
        try {
            // Get the first parameter of the "params" which has been related to "String"
            String urlstring = params[0];
            // Log the string that contains the url
            Log.i("EmonLog", "HTTP Connecting: "+urlstring);
            // Convert the String into an actual URL
            URL url = new URL(urlstring);
            // Make the HTTP connection to the Internet
            HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
            // Once the connection is done, try to store the information received into "reader"
            try {
                InputStream reader = new BufferedInputStream(urlConnection.getInputStream());
                // Create a new String which will extract and contain the information received into the Buffered Input Stream
                String text = "";
                int i = 0;
                while((i=reader.read())!=-1)
                {
                    text += (char)i;
                }
                Log.i("EmonLog", "HTTP Response: "+text);
                result = text;

            } catch (Exception e) {
                Log.i("EmonLog", "HTTP Exception: "+e);
            }
            finally {
                Log.i("EmonLog", "HTTP Disconnecting");
                urlConnection.disconnect();
            }

        } catch (Exception e) {
            e.printStackTrace();
            Log.i("EmonLog", "HTTP Exception: "+e);
        }

        return result;
    }
}

我的进口商品：

import android.support.v7.app.ActionBarActivity;
import android.os.Bundle;
import android.os.Handler;
import android.util.Log;
import android.view.Menu;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;

非常感谢你的帮助。

Answer 1

您不能将HTTP用作AsyncTask的名称。

你已经有了

org.apache.http.protocol.HTTP

作为HTTP。

请将您的AsyncTask重命名为其他内容。

可能是，

class HTTPTask extends AsyncTask<String, Void, String>

会很好。

Android Eclipse：无法将HTTP解析为某种类型

1 个答案: