Question

我在Android studio中编写了一个代码来使用Web服务，如果返回“True”，则跳转到新活动。但是当我点击Login时，应用程序会一直崩溃。我是android studio的新手，我不知道我使用的过程是否正确。

网址 - http://202.54.216.49/logs/test.asmx

Method-loginauth

注意：如果用户名和密码正确，则loginauth返回“True”，如果不正确，则返回“False”。如果返回“True”，我想跳转到新活动（Main2Activity）。

MainActivity.java -

package com.example.abhimanyu.devicecontrol;

import android.content.Context;
import android.content.Intent;
import android.os.AsyncTask;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.PropertyInfo;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapPrimitive;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;

public class MainActivity extends AppCompatActivity {
    EditText textBox, textBox2;
    Button button;
    TextView ans;


    String URL = "http://202.54.216.49/logs/test.asmx";
    String NAMESPACE = "http://tempuri.org/";
    String SOAP_ACTION = "http://tempuri.org/loginauth";
    String METHOD_NAME = "loginauth";
    String PARAMETER_NAME1 = "username";
    String PARAMETER_NAME2 = "password";

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

        textBox = (EditText) findViewById(R.id.txtBox);
        textBox2 = (EditText) findViewById(R.id.txtBox2);
        button = (Button) findViewById(R.id.btn);
        ans = (TextView) findViewById(R.id.answer);
        final Context context;
        button.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                new CallWebService().execute(textBox.getText().toString(), textBox2.getText().toString());
            }
        });
    }

    class CallWebService extends AsyncTask<String, Void, String> {
        @Override
        protected void onPostExecute(String s) {
            if (s.equalsIgnoreCase("True"))
            {
                Intent k = new Intent(MainActivity.this, Main2Activity.class);
                startActivity(k);
            }
            else
            {
                Toast.makeText(getApplicationContext(), "Wrong username or password", Toast.LENGTH_LONG).show();
            }
        }

        @Override
        protected void onPreExecute() {
            Log.i("onPreexecute","running");
            super.onPreExecute();
        }

        @Override
        protected String doInBackground(String... params) {
            String result = null;

            SoapObject soapObject = new SoapObject(NAMESPACE, METHOD_NAME);

            PropertyInfo propertyInfo1 = new PropertyInfo();
            propertyInfo1.setName(PARAMETER_NAME1);
            propertyInfo1.setValue(params[0]);
            propertyInfo1.setType(String.class);
            soapObject.addProperty(propertyInfo1);

            PropertyInfo propertyInfo2 = new PropertyInfo();
            propertyInfo2.setName(PARAMETER_NAME2);
            propertyInfo2.setValue(params[1]);
            propertyInfo2.setType(String.class);
            soapObject.addProperty(propertyInfo2);

            SoapSerializationEnvelope envelope =  new SoapSerializationEnvelope(SoapEnvelope.VER11);
            envelope.dotNet=true;
            envelope.implicitTypes=true;
            envelope.setOutputSoapObject(soapObject);

            HttpTransportSE httpTransportSE =  new HttpTransportSE(URL,30000);

            try {
                httpTransportSE.call(SOAP_ACTION, envelope);
                SoapPrimitive soapPrimitive = (SoapPrimitive)envelope.getResponse();
                result = soapPrimitive.toString();
            } catch (Exception e) {
                e.printStackTrace();
            }

            return result;

        }
    }
}

Main2Activity.java -

package com.example.abhimanyu.devicecontrol;

import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;

public class Main2Activity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main2);
    }
}

的AndroidManifest.xml -

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.example.abhimanyu.devicecontrol">

    <application
        android:allowBackup="true"
        android:icon="@mipmap/ic_launcher"
        android:label="@string/app_name"
        android:roundIcon="@mipmap/ic_launcher_round"
        android:supportsRtl="true"
        android:theme="@style/AppTheme">
        <activity android:name=".MainActivity">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
        <activity android:name=".Main2Activity"></activity>
    </application>

</manifest>

注意：它显示exception- java.net.SocketTimeoutException

Answer 1

<uses-permission android:name="android.permission.WAKE_LOCK" />
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<uses-permission android:name="android.permission.READ_EXTERNAL_STORAGE" />
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.READ_PHONE_STATE" />

将这些行添加到清单文件

后可以正常工作

计划使用Web服务并跳转到下一个活动

1 个答案: