How to make a phone call programmatically?

Question

I'm passing to an activity the number to call by a bundle

and then, in such activity, I have a button to call to that number, this is the code:

callButton.setOnClickListener(new OnClickListener() {
            public void onClick(View v) {
                Intent intent = new Intent(Intent.ACTION_CALL, Uri.parse(bundle.getString("mobilePhone")));
            }
        });

Something is wrong, because when I press the button nothing happens...

What am I doing wrong?

PD: I'm using Android 1.5 compatible project... maybe phone call is incompatible to 1.5?

In the selected answer, there is not check for marshmallow permission. It will not work directly in marshmallow 6.0 or above device. **I know I am too late but this question has large vote so I thought it will help to others in future.** In marshmallow devices we need to take run time permission for call... Here is example to make call in marshmallow or above. [How to make call in android marshmallow 6.0 or above](http://androiderstack.com/index.php/2017/08/02/how-to-make-call-in-android-marshmallow-6-0-or-above/) — Vishal Chhodwani, Aug 14 '17 at 10:07

score 271 · Accepted Answer · edited Oct 09 '13 at 15:17

271

You forgot to call startActivity. It should look like this:

Intent intent = new Intent(Intent.ACTION_CALL);

intent.setData(Uri.parse("tel:" + bundle.getString("mobilePhone")));
context.startActivity(intent);

An intent by itself is simply an object that describes something. It doesn't do anything.

Don't forget to add the relevant permission to your manifest:

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

edited Oct 09 '13 at 15:17

The Alpha

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answered Jan 27 '11 at 13:16

Lior

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Hi @Lior how could i do in-case that device has dual-sim. Is it possible to make call through a particular sim? – Dinash Feb 20 '13 at 11:24
3

@Dinash: Take a look at this already answered question: http://stackoverflow.com/questions/13231962/call-from-second-sim – Lior Mar 01 '13 at 13:21
I have used the same code but not working in the Galaxy S7 edge. This is my code Intent intentCall = new Intent(Intent.ACTION_CALL); String uri = "tel:" + number.trim(); intentCall.setData(Uri.parse(uri)); if (ActivityCompat.checkSelfPermission(getApplicationContext(), Manifest.permission.CALL_PHONE) == PackageManager.PERMISSION_GRANTED) { startActivity(intentCall); } – AngelJanniee Dec 08 '16 at 04:14
Can this simulator be used to make calls without physical device – user3754136 Jun 21 '17 at 07:13
Using this intent requires flag FLAG_ACTIVITY_NEW_TASK because it is outside the app. – donmj Sep 24 '20 at 10:55

Anirudh · Answer 2 · 2013-04-13T07:56:52.547

25

Tried this on my phone and it works perfectly.

Intent intent = new Intent(Intent.ACTION_CALL);
intent.setData(Uri.parse("tel:900..." ));
startActivity(intent);

Add this permission in manifest file.

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

edited Apr 13 '13 at 07:56

answered Aug 24 '12 at 04:58

Anirudh

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number is like 1689,,3,2,1,1 . I am getting issue it will be only 1689..Please help me if you have any solution for the same. – Teraiya Mayur Apr 21 '15 at 11:16
@TeraiyaMayur you can remove any non-numeric characters first. So if you have phone number as `String number = "1689,,3,2,1,1` do `nubmer = "tel:" + number.replaceAll("[^0-9]", "");` – Walk Apr 16 '17 at 13:20
@Walk, a problem is that we want to call with special symbols like `,*`. They must be among digits. On an emulator it works right, but on a device it drops additional symbols and digits. – CoolMind Sep 26 '19 at 15:15

Ashish Dwivedi · Answer 3 · 2012-04-16T09:44:29.900

13

 Intent callIntent = new Intent(Intent.ACTION_CALL, Uri.parse("tel:"+198+","+1+","+1)); 
             startActivity(callIntent);

for multiple ordered call

This is used to DTMF calling systems. If call is drop then, you should pass more " , " between numbers.

edited Apr 16 '12 at 09:44

answered Aug 26 '11 at 11:53

Ashish Dwivedi

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Dwivedi Ji : number is like 1689,,3,2,1,1 . I am getting issue it will be only 1689..Please help me if you have any solution for the same. – Teraiya Mayur Apr 21 '15 at 11:17
@TeraiyaMayur, It's depends upon IVRS call system. Its work for me because of my calling extinction support take early response. – Ashish Dwivedi Apr 23 '15 at 05:32
Yes you are right i also get proper valid IVR number and it works fine. thanks – Teraiya Mayur Apr 27 '15 at 10:12

score 3 · Answer 4 · answered Jan 27 '11 at 13:16

3

Take a look there : http://developer.android.com/guide/topics/intents/intents-filters.html

DO you have update your manifest file in order to give call rights ?

answered Jan 27 '11 at 13:16

ykatchou

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CoolMind · Answer 5 · 2019-09-27T14:35:56.753

This doesn't require a permission.

val intent = Intent(Intent.ACTION_DIAL, Uri.parse("tel:+123456"))
startActivity(intent)

Or

val intent = Intent(Intent.ACTION_DIAL, Uri.fromParts("tel", "+123456", null))
startActivity(intent)

But it shows one more dialog (asking whether you want to call phone just once or always). So it would be better to use ACTION_CALL with a permission (see Revoked permission android.permission.CALL_PHONE).

score 3 · Answer 6 · edited Apr 20 '23 at 06:33

3

It is working very well. In this way, you do not need to have permission from user. You can open directly phone calling part.

Trick point, use ACTION_DIAL instead of ACTION_CALL.

private void callPhoneNumber() {
    String phone = "***********";
    Intent callIntent = new Intent(Intent.ACTION_DIAL);
    callIntent.setData(Uri.parse("tel:" + phone));
    startActivity(callIntent);
}

edited Apr 20 '23 at 06:33

Hafeez Ul Haq

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answered Sep 05 '20 at 19:26

canerkaseler

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score 2 · Answer 7 · edited Jul 18 '13 at 14:05

2

Here I will show you that how you can make a phone call from your activity. To make a call you have to put down this code in your app.

try {
    Intent my_callIntent = new Intent(Intent.ACTION_CALL);
    my_callIntent.setData(Uri.parse("tel:"+phn_no));
    //here the word 'tel' is important for making a call...
    startActivity(my_callIntent);
} catch (ActivityNotFoundException e) {
    Toast.makeText(getApplicationContext(), "Error in your phone call"+e.getMessage(), Toast.LENGTH_LONG).show();
}

edited Jul 18 '13 at 14:05

B770

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answered Oct 20 '12 at 23:01

Pir Fahim Shah

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@Astor i have defined it very clearly, an i have already gotten points through this, which means that my answer is some what better than the other for some people. – Pir Fahim Shah Jan 13 '13 at 19:08

score 1 · Answer 8 · edited Sep 26 '19 at 15:15

1

If anyone is looking for in Kotlin

    val  uri = "tel:+800******"
    val call_customer_service = Intent(Intent.ACTION_CALL)
    call_customer_service.setData(Uri.parse(uri))
    startActivity(call_customer_service)

Like some other solutions it requires android.permission.CALL_PHONE permission.

edited Sep 26 '19 at 15:15

CoolMind

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answered Apr 15 '19 at 05:38

Mughil

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score 0 · Answer 9 · edited Mar 06 '18 at 22:35

protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main); 
   final Button button = (Button) findViewById(R.id.btn_call);
    button.setOnClickListener(new View.OnClickListener() {
        public void onClick(View v) {
            String mobileNo = "123456789";
            String uri = "tel:" + mobileNo.trim();
            Intent intent = new Intent(Intent.ACTION_CALL);
            intent.setData(Uri.parse(uri));
            startActivity(intent);
        }
    });*
 }

score 0 · Answer 10 · answered Oct 03 '20 at 15:20

If you end up with a SecurityException (and the call does not work), You should consider requesting the user permission to make a call as this is considered a dangerous permission:

ActivityCompat.requestPermissions(
    activity,
    new String[] {Manifest.permission.CALL_PHONE},
    1
);

Note this has nothing to do with the manifest permission (that you must have as well)

How to make a phone call programmatically?

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