bash script - unable to set variable with double quotes in value

Question

Need help in fixing this bash script to set a variable with a value including double quotes. Somehow I am defining this incorrectly as my values foo and bar are not enclosed in double quotes as needed.

My script thus far:

#!/usr/local/bin/bash
set -e
set -x

host='127.0.0.1'
db='mydev'
_account="foo"
_profile="bar"
_version=$1
_mongo=$(which mongo);

exp="db.profile_versions_20170420.find({account:${_account}, profile:${_profile}, version:${_version}}).pretty();";
${_mongo} ${host}/${db} --eval "$exp"

set +x

Output shows:

+ host=127.0.0.1
+ db=mydev
+ _account=foo
+ _profile=bar
+ _version=201704112004
++ which mongo
+ _mongo=/usr/local/bin/mongo
+ exp='db.profile_versions_20170420.find({account:foo, profile:bar, version:201704112004}).pretty();'
+ /usr/local/bin/mongo 127.0.0.1/mydev --eval 'db.profile_versions_20170420.find({account:foo, profile:bar, version:201704112004}).pretty();'
MongoDB shell version: 3.2.4
connecting to: 127.0.0.1/mydev
2017-04-22T15:32:55.012-0700 E QUERY    [thread1] ReferenceError: foo is not defined :
@(shell eval):1:36

What i need is account:"foo", profile:"bar" to be enclosed in double quotes.

Answer 1

In bash (and other POSIX shells), the following 2 states are equivalent:

_account=foo
_account="foo"

What you want to do is to preserve the quotations, therefore you can do the following:

_account='"foo"'