use*_*729 7 android android-intent android-implicit-intent
call按钮怎么做?我把我的号码作为一个字符串EditText.这是我的示例代码:
String phone = editPhone.getText().toString();
btnPhone.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View arg0) {
call();
}
});
public void call() {
try {
Intent callIntent = new Intent(Intent.ACTION_CALL);
callIntent.setData(Uri.parse(phone));
startActivity(callIntent);
} catch (ActivityNotFoundException activityException) {
Log.e("myphone dialer", "Call failed", e);
}
}
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我添加了所有permissions清单文件.
但我得到了 NullPointerexception
MKJ*_*ekh 27
这种简单的方法应该适合您.
防爆.
public class CallActivity extends Activity{
String phone = "";
onCreate()
{
btnPhone.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View arg0) {
phone = editPhone.getText().toString();
call();
}
});
}
public void call() {
Intent callIntent = new Intent(Intent.ACTION_CALL);
callIntent.setData(Uri.parse("tel:"+phone));
startActivity(callIntent);
}
}
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您可能正在使用String变量phone超出范围.
Sha*_*dne 10
我想你错过了URI中的"tel:"部分.
更换以下..
Intent callIntent = new Intent(Intent.ACTION_CALL);
callIntent.setData(Uri.parse(phone));
startActivity(callIntent);
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同
Intent callIntent = new Intent(Intent.ACTION_CALL,Uri.parse("tel:"+phone));
startActivity(callIntent);
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要么
Intent callIntent = new Intent(Intent.ACTION_CALL);
callIntent.setData(Uri.parse("tel:"+phone));
startActivity(callIntent);
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看下面的代码可能对你有帮助.
打电话
EditText num = (EditText)findViewById(R.id.number_edit);
String uri = "tel:" + num.trim();
Intent intent = new Intent(Intent.ACTION_CALL);
intent.setData(Uri.parse(uri));
startActivity(intent);
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用于打开拨号器
Intent dial = new Intent();
dial.setAction("android.intent.action.DIAL");
dial.setData(Uri.parse("tel:"));
startActivity(dial);
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