Return variable from website Bash command

I used to have a script which basically, returned the current temperature from a website. This was done with wget then awk to return the "Temperature" as a variable then I used sox to create a file saying the Temperature is etc.

The website has changed and I am having trouble re-writing it.

This is what I have:


URL='https://wttr.in/rhyl'

temp="wget -q  -O- "$URL" | awk -F\' 'data-value/{print $1 }'| head -1)"

sox -V1 -c 1  silence.wav  base.wav $temp.wav temp-dry.wav

sox -V1 -m temp-bed2.wav temp-dry.wav tempfx.wav
sox -V1 tempfx.wav tempfx+15db.wav vol 9 db
sox -V1 temp-dry.wav temp-dry+10db.wav vol 10 db

I'm happy with the sox bit so far, I just cant seem to return the variable i.e. "12" from the temperature section of wttr.in

The desired output of $temp is just number no special characters i.e $temp = 12

Solution

This is a version bash file that updates and audio file for the current temperature, for the radio station jingle.

#/////Get weather for where you want thanks to  User3439894 and Bengamin W. /////

    temp="$(curl -s https://wttr.in/rhyl?format=%t | grep -Eo [0-9]+)"

#   ////use sox to make the dry file outof pre-defind wav files/////

    sox -V1 -c 1  silence.wav  base.wav $temp.wav temp-dry.wav

#  //// Here I use sox again to create a wav file with start and end jingle /////

    sox -V1 -m temp-bed2.wav temp-dry.wav tempfx.wav

#     //// Using sox again I boost the audio by what is required 15db and 10 db///

     sox -V1 tempfx.wav tempfx+15db.wav vol 9 db
     sox -V1 temp-dry.wav temp-dry+10db.wav vol 10 db

I then use crontab to run this bash script as a cronjob.

thanks everyone

Alex