ElevatedButton(
onPressed:() {
launchWhatsapp(phoneNum,message);
},
child: Text("Message the Buyer",
style: TextStyle(
fontWeight: FontWeight.w400
),
),
),
So this is my button code
launchWhatsapp(String number,String text) async{
String url = "https://wa.me/${number}";
if(await canLaunchUrlString(url)){
await launchUrlString(url);
}
and this is my function method. I used url_laucher to make it work but it will redirect to the browser and show this Image Error but when I try to copy the same link to the browser it works but when trying to open from my app it will show the error as shown in the picture.
CodePudding user response:
Try below code
Your function:
whatsApp() {
return launchUrl(
Uri.parse(
//'https://wa.me/1234567890' //you use this url also
'whatsapp://send?phone=1234567890',//put your number here
),
);
}
Your Widget:
ElevatedButton(
onPressed: () {
whatsApp();
},
child: Text(
"Message the Buyer",
style: TextStyle(fontWeight: FontWeight.w400),
),
),
set internet permission to your AndroidManifest.xml
file path = project_name/android/app/src/main/AndroidManifest.xml
add below line above of application tag
<uses-permission android:name="android.permission.INTERNET"/>