呼唤一个人的问题 [英] Problem in calling a person

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本文介绍了呼唤一个人的问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

先生,我创建了一个连接到特定网站并调用特定人员等的应用程序,我已经成功生成了一个签名的.apk而没有任何错误,我的应用程序也在我的Android设备上运行但是它无法拨打一个特定号码。请帮我解决这个问题。



这就是我所做的:



我创建了一个致电按钮:



 <  按钮  

android:layout_width = wrap_content

android:layout_height = wrap_content

android:text = 给我打电话?

< span class =code-attribute> < span class =code-attribute> android:id = @ + id / button4

android:layout_below = @ + id / editText4
< span class =code-attribute>
android:layout_alignRight = @ + id / textView4

android:layout_alignEnd = @ + id / textView4 / >





添加了一个清单文件:



 <?  xml    版本  =  1.0    encoding   =  utf-8  >  
< 清单 xmlns:android = http://schemas.android.com/apk/res/android

< span class =code-attribute> package = com.vis.vis >
< uses-permission android:name = android.permission.CALL_PHONE / >
< < span class =code-leadattribute> uses-permission android:name = android.permission.RE AD_PHONE_STATE / >





声明并定义要调用的按钮:

  final 按钮调用=(按钮)findViewById(R.id.button4); 
call.setOnClickListener( new View.OnClickListener(){

@ Override
public void onClick(查看v){
意图= new Intent(Intent.ACTION_CALL,Uri.parse( 1234567890));
< span class =code-keyword> try {
startActivity(in);
}

catch (android.content.ActivityNotFoundException ex){
Toast.makeText(getApplicationContext(), 再试一次!,Toast.LENGTH_SHORT)。show();
}
}
});





但它会用于捕获部分弹出再试一次!

我有没有错过了什么?

先生,请帮助我。

解决方案

在你的按钮 onClick 中设置此代码:



 Intent in =  new  Intent(Intent.ACTION_CALL); 
in.setData(Uri.parse( tel:0123456789));
startActivity(in);


更改

意图in = new Intent( Intent.ACTION_CALL,Uri.parse(  1234567890)); 





意图in = new Intent(Intent.ACTION_CALL,Uri.parse(  tel:1234567890)); 





-KR


Sir, I have created an application which connects to a specific website and calls a particular person etc., I have generated a signed ".apk" successfully without any errors and my application also runs on my android device but it fails to call a particular number. Kindly help me to solve this problem.

here is what I have done:

I have created a button for calling:

<Button

        android:layout_width="wrap_content"

        android:layout_height="wrap_content"

        android:text="call me?"

        android:id="@+id/button4"

        android:layout_below="@+id/editText4"

        android:layout_alignRight="@+id/textView4"

        android:layout_alignEnd="@+id/textView4" />



Added a manifest file:

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

    package="com.vis.vis" >
    <uses-permission android:name="android.permission.CALL_PHONE" />
    <uses-permission android:name="android.permission.READ_PHONE_STATE" />



Declared and defined the button to call:

final Button call = (Button) findViewById(R.id.button4);
call.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View v) {
                Intent in=new Intent(Intent.ACTION_CALL,Uri.parse("1234567890"));
                try{
                    startActivity(in);
                }

                catch (android.content.ActivityNotFoundException ex){
                    Toast.makeText(getApplicationContext(),"Try again!",Toast.LENGTH_SHORT).show();
                }
            }
        });



but it goes for catch part pop-upping "Try again!"
Have I missed anything?
Sir, Kindly help me with this.

解决方案

In your button onClick set this code :

Intent in= new Intent(Intent.ACTION_CALL);
in.setData(Uri.parse("tel:0123456789"));
startActivity(in);


Change

Intent in=new Intent(Intent.ACTION_CALL,Uri.parse("1234567890"));


to

Intent in=new Intent(Intent.ACTION_CALL,Uri.parse("tel:1234567890"));



-KR


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