Sequelize -- Use options.distinct with a Where Clause? - postgresql

Question

I'm trying to use Sequelize to find all distinct fields in a column, using a query with a where clause. I've searched the Sequelize docs and tried a number of different things, but haven't yet found the correct syntax.
Here's my current draft syntax:
var searchResults = connectors.cars.findAll({
attributes: [
connectors.Sequelize.options.distinct
],
where: {
condition: connectors.Sequelize.where(connectors.Sequelize.fn('LOWER', connectors.Sequelize.col('mfgr')), 'LIKE', '%' + searchString + '%')
}
});
What is the correct way to use options.distinct alongside a where clause?
Note: edited to remove a bunch of extra code that had been requested in the comments, but which in retrospect may have been obfuscating the issue.

Answer 1

Really more of a comment, but what's going on with this code?
}).then((searchResults) => searchResults.map((item) => item.dataValues));
debugger;
return searchResults; <== ERROR IS THROWN HERE
1) What is the destination of the map()? 2) Won't the return occur before the results are returned, as it's not in the .then() block? Should it not be more like this:
}).then((searchResults) => {
var new_results = searchResults.map((item) => item.dataValues));
debugger;
return new_results;
}
Am I missing something?

Answer 2

I'm getting the impression that this is hard to do in raw sql, let alone Sequelize. I found an answer that uses raw sql here:
const myQuery = "WITH cte AS\n" +
"( SELECT id, mfgr, ROW_NUMBER() OVER (PARTITION BY mfgr ORDER BY mfgr DESC) AS rn\n" +
" FROM cars\n" +
" WHERE LOWER(mfgr) like '%" + searchString + "%'\n" +
")\n" +
"SELECT *\n" +
"FROM cte\n" +
"WHERE rn = 1";
let searchResults = connectors.db.query(myQuery);