Filter output and set variable

I would like to run a bash script which invokes a command and then sets a variable in the script with a part of the output from the first command. In my case I would like to set the variable URL to https://loady-7jmiymbtlg-uc.a.run.app.

my-script.sh:

gcloud run deploy loady

# echo $URL <--- how to set this with the output from the above command

When I run my script (example output):

karl@Karls-MacBook-Pro ~ $ ./my-script.sh
Deploying container to Cloud Run service [loady] in project [loady] region [us-central1]
✓ Deploying new service... Done.                                                               
  ✓ Creating Revision...                                                                       
  ✓ Routing traffic...                                                                         
  ✓ Setting IAM Policy...                                                                      
Done.                                                                                          
Service [loady] revision [loady-00001-nod] has been deployed and is serving 100 percent of traffic.
Service URL: https://loady-7jmiymbtlg-uc.a.run.app

So as you can see, the last line there.

This should work:

URL=$(gcloud run deploy loady 2>&1 |grep -o -m1 "https://\S*")

The 2>&1 merges the stdout and stderr outputs of the gcloud command into stdout, needed for | to filter both through grep should that be necessary.

grep will output only (-o) the first (-m1) url matching the regular expression https://\S* (meaning https://everything.until.before/next/space). The final stdout of the contents of $() is invisibly stored in $URL.

Since stderr isn't captured by $() or |, you can choose to still also display the output of the gcloud command by duplicating it all back to stderr with tee.

URL=$(gcloud run deploy loady 2>&1 |tee /dev/stderr |grep -o -m1 "https://\S*")