How Do You Pass a Variable to Curl in Bash Script

Question

How Do You Pass a Variable to Curl in Bash Script

I'm not able to get this curl to work in a bash script using an environment variable. The API key gets misinterpreted somehow when passed in via a variable and I'm getting authentication errors on submit. If I plug in the API key with no surrounding quotes as plaintext, this works just fine. I've tried various forms of escaping quotes and other combinations. Any help would be great.

Code

#!/bin/bash

source .env

curl -X POST https://api.easypost.com/v2/shipments \
  -u "$EASYPOST_TEST_API_KEY": \
  -d 'shipment[to_address][name]=Dr. Steve Brule' \
  -d 'shipment[to_address][street1]=179 N Harbor Dr' \
  -d 'shipment[to_address][city]=Redondo Beach' \
  -d 'shipment[to_address][state]=CA' \
  -d 'shipment[to_address][zip]=90277' \
  -d 'shipment[to_address][country]=US' \
  -d 'shipment[to_address][phone]=8573875756' \
  -d 'shipment[to_address][email]=dr_steve_brule@gmail.com' \
  -d 'shipment[from_address][name]=EasyPost' \
  -d 'shipment[from_address][street1]=417 Montgomery Street' \
  -d 'shipment[from_address][street2]=5th Floor' \
  -d 'shipment[from_address][city]=San Francisco' \
  -d 'shipment[from_address][state]=CA' \
  -d 'shipment[from_address][zip]=94104' \
  -d 'shipment[from_address][country]=US' \
  -d 'shipment[from_address][phone]=4153334445' \
  -d 'shipment[from_address][email]=support@easypost.com' \
  -d 'shipment[parcel][length]=20.2' \
  -d 'shipment[parcel][width]=10.9' \
  -d 'shipment[parcel][height]=5' \
  -d 'shipment[parcel][weight]=65.9' \

Answer 1

Per Gordon Davisson's suggestion to break it down into a minimal request, it started sending right. Then I rebuilt it just as in my question above and for whatever reason it worked like a charm. The -x flag was also incredibly helpful in troubleshooting this but come to find out there wasn't anything inherently wrong with it in the first place. Thanks for everyone's responses!