getData(): Observable<string[]> {
return this._userService.getUser()
.pipe(
switchMap(userInfo=> {
return this.getPrivateGroup();
}),
catchError(this.someErrorHandler)
);
}
如何使这条线正常工作?
答案 0 :(得分:0)
此代码将在您单击按钮时发送短信,而无需转到消息框
String sendMessage = "Your Message";
SmsManager sms = SmsManager.getDefault();
PendingIntent sentPI;
String SENT = "SMS_SENT";
sentPI = PendingIntent.getBroadcast(MainActivity.this, 0, new Intent(SENT), 0);
sms.sendTextMessage("MOBILE NUMBER", null, sendMessage, sentPI, null);
而且Yaa不会忘记在清单中添加权限
<uses-permission android:name="android.permission.SEND_SMS"/>