assigning jq result to makefile variable - bash

Question

I have the following command as a makefile command:
update-env:
echo "{ \"Variables\":${ENV_VALUE}}" > ./new_env.json && \
UPDATE_ENVVARS=$$(jq -s '.[0] as $$a |.[1] as $$b | $$a * $$b' old_env.json new_env.json | jq -c . ) && \
echo "${UPDATE_ENVVARS}"
the ${ENV_VALUE} is taken when
make update-env ENV_VALUE="{\\\"HOST_URL\\\": \\\"https:\\\/\\\/test.com\\\"}"
the file new_env.json is generated properly
when executing jq -s '.[0] as $a |.[1] as $b | $a * $b' old_env.json new_env.json | jq -c . it generates the appropriate compact json result desired.
When running everything in sequence (the assignment and echo for validation), I get an empty result.
My goal for the command is to merge two json output and assign it to the UPDATE_ENVVARS for it to be reused as an input for another command that will accept the json. Per testing, it came out empty when echo, when I execute the jq solo, the merge output is functional.

Answer 1

Only a minor bit of editing was needed:
update-env:
echo '{ "Variables":${ENV_VALUE}}' > ./new_env.json && \
UPDATE_ENVVARS=$$(jq -s '.[0] as $$a |.[1] as $$b | $$a * $$b' old_env.json new_env.json | jq -c . ) && \
echo "$${UPDATE_ENVVARS}"
Note:
We're using single quotes in the first-line echo -- the substitution is performed by make, not by the shell, so single quotes don't suppress it.
We're doubling up the $$ in the last line, so we're expanding the shell variable set in the second line of the recipe, not a make variable that nothing ever set at all.
See this working at https://replit.com/#CharlesDuffy2/RoyalIdolizedProfile
By contrast, if you want to assign to a make variable instead of a shell variable, this question is duplicative of Makefile command substitution problem, and the answer by Ignacio Vazquez-Abrams is appropriate.