Question

我的机器人应用程序在主页面上有一个按钮，为您提供一个对话弹出页面，并在对话框页面中询问您是否要拨打电话

我已经尝试了一遍又一遍，并且还没有弄明白如何让按钮拨打电话

我打开的java页面打包your.package2.phonecalls;

import android.app.Activity;
import android.app.Dialog;
import android.content.Intent;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;

public class ButtonphonecallsActivity extends Activity {
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    Button TheCallButton = (Button) findViewById(R.id.phonecallbutton);  
    TheCallButton.setOnClickListener(new View.OnClickListener() {  

        public void onClick(View v) {
            final Dialog dialog = new Dialog(ButtonphonecallsActivity.this);
            dialog.setContentView(R.layout.alertphonecall);
            dialog.setTitle("Zohan contact number");

            Button Nosir = (Button) dialog.findViewById(R.id.nobutton);  
            Nosir.setOnClickListener(new View.OnClickListener() {  
                @Override               
                public void onClick(View v) {
                    dialog.cancel();
                    } 
            });   

            Button Yessir = (Button) dialog.findViewById(R.id.yesbutton);
            Yessir.setOnClickListener(new OnClickListener() {
                @Override               
                public void onClick(View v) {
                    dialog.cancel();
                    startActivity(new Intent(ButtonphonecallsActivity.this, makingaphonecall.class));        
                }
            });

            dialog.show();
        }
    });
}
}

电话活动

package your.package2.phonecalls;

import android.app.Activity;
import android.content.ActivityNotFoundException;
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.util.Log;

public class makingaphonecall extends Activity {
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    call();
}

private void call() {
try {
    Intent callIntent = new Intent(Intent.ACTION_CALL);
    callIntent.setData(Uri.parse("tel:7578263678"));
    startActivity(callIntent);
} catch (ActivityNotFoundException activityException) {
     Log.e("helloandroid dialing example", "Call failed");
}
}

}

android manifest

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="your.package2.phonecalls"
android:versionCode="1"
android:versionName="1.0" >

<uses-sdk android:minSdkVersion="8" />
<uses-permission android:name="android.permission.CALL_PHONE"/>

<application
    android:icon="@drawable/ic_launcher"
    android:label="@string/app_name" 
    android:debuggable="true" >

    <activity
        android:name=".ButtonphonecallsActivity"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>

    <activity
        android:name=".makeaphonecall"
        android:label="@string/app_name" >
    </activity>

    </application>

</manifest>

Answer 1

为什么要进行单独的活动来拨打电话？只需将以下两行添加到您的onclick中。

Intent dialer = new Intent(Intent.ACTION_CALL,Uri.parse("tel:7578263678"));
startActivity(dialer);

Answer 2

Button b = (Button) this.findViewById(R.id.callButton);
b.setOnClickListener(new OnClickListener() {

    @Override
    public void onClick(View v) {
        Intent i = new Intent(Intent.ACTION_CALL);
        i.setData(Uri.parse("tel:123456"));
        CallTestAndroidActivity.this.startActivity(i);              
    }
});

这应该可以正常工作。清单就像你拥有它一样好

希望这有帮助。

我正在尝试制作一个简单的应用程序，让您可以选择拨打电话

2 个答案: