如何解决Android中的此错误？ java.net.MalformedURLException：找不到协议： [英] How to solve this error in Android? java.net.MalformedURLException: Protocol not found:

问题描述

我不知道到底是什么问题，但是我遇到了类似
的错误，即使我也检查了所有路径，它是正确的，但仍然会出错。

I dont know what is exactly problem, But I am getting error like: Even I also checking all paths, and it is correct but still it will get error.

java.net.MalformedURLException：未找到协议：www.w3schools.com/webservices/tempconvert.asmx

java.net.MalformedURLException: Protocol not found: www.w3schools.com/webservices/tempconvert.asmx

我的代码：

package com.example.mytest;

import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapPrimitive;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.AndroidHttpTransport;

import android.app.Activity;
import android.os.Bundle;
import android.util.Log;
import android.view.Menu;
import android.widget.TextView;

@SuppressWarnings("deprecation")
public class MainActivity extends Activity {

    private final String SOAP_ACTION = "http://tempuri.org/CelsiusToFahrenheit";
    private final String METHOD_NAME = "CelsiusToFahrenheit";
    public final String NAMESPACE = "http://tempuri.org";
    public String URL = "www.w3schools.com/webservices/tempconvert.asmx";

    TextView tv;

    public static String TAG = "MyTest";

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

        Log.d(TAG, "MainActivity Started");
        tv = (TextView) findViewById(R.id.txt1);

        SoapObject Request = new SoapObject(NAMESPACE, METHOD_NAME);
        Request.addProperty("Celsius", "32");

        // URL=URL.replaceAll(" ", "%20");

        SoapSerializationEnvelope soapEnvelope = new SoapSerializationEnvelope(
                SoapEnvelope.VER11);
        soapEnvelope.dotNet = true;
        soapEnvelope.setOutputSoapObject(Request);

        Log.d(TAG, "URL:"+URL);

        AndroidHttpTransport aht = new AndroidHttpTransport(URL);

        try {
            // HttpTransportSE aht = new HttpTransportSE(URL);

            Log.d(TAG, "aht:" + aht);

            aht.call(SOAP_ACTION, soapEnvelope);
            SoapPrimitive resultString = (SoapPrimitive) soapEnvelope
                    .getResponse();

            // Object resultString = (Object)soapEnvelope.getResponse();

            Log.d(TAG, "result String=" + resultString);

            tv.setText("Status: " + resultString);
        } catch (Exception e) {
            Log.d(TAG, "Error: ", e);
        }

    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.main, menu);
        return true;
    }

}

我不知道确切的问题是什么但是我遇到了这样的错误：

I dont know what is exact problem but I am getting error like:

02-01 21:07:01.264: D/MyTest(1068): MainActivity Started
02-01 21:07:01.534: D/MyTest(1068): Error: 
02-01 21:07:01.534: D/MyTest(1068): java.net.MalformedURLException: Protocol not found: www.w3schools.com/webservices/tempconvert.asmx
02-01 21:07:01.534: D/MyTest(1068):     at java.net.URL.<init>(URL.java:275)
02-01 21:07:01.534: D/MyTest(1068):     at java.net.URL.<init>(URL.java:159)
02-01 21:07:01.534: D/MyTest(1068):     at org.ksoap2.transport.ServiceConnectionSE.<init>(ServiceConnectionSE.java:65)
02-01 21:07:01.534: D/MyTest(1068):     at org.ksoap2.transport.ServiceConnectionSE.<init>(ServiceConnectionSE.java:61)
02-01 21:07:01.534: D/MyTest(1068):     at org.ksoap2.transport.AndroidServiceConnection.<init>(AndroidServiceConnection.java:27)
02-01 21:07:01.534: D/MyTest(1068):     at org.ksoap2.transport.AndroidHttpTransport.getServiceConnection(AndroidHttpTransport.java:35)
02-01 21:07:01.534: D/MyTest(1068):     at org.ksoap2.transport.HttpTransportSE.call(HttpTransportSE.java:124)
02-01 21:07:01.534: D/MyTest(1068):     at org.ksoap2.transport.HttpTransportSE.call(HttpTransportSE.java:95)

推荐答案

添加 http：// www ... 或 https：// 之前。

Add http:// before the www... or https://.

public String URL = "http://www.w3schools.com/webservices/tempconvert.asmx";

这篇关于如何解决Android中的此错误？ java.net.MalformedURLException：找不到协议：的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！