Optimize result page to query number of result first

Here is my code. But I have some problem about loading time for result more than 1000 rows.

How can I change code to check number of query result for create paginate first. Then, get result only that paginate page?

function showResult(req, res){

    var n = req.query.query;
    mysql_conn.query('SELECT query_text FROM catalogsearch_query WHERE query_text LIKE "%' + n + '%" ORDER BY popularity DESC LIMIT 0 , 10', function (error, rows) {
        mysql_crawl.query('SELECT prod_name, full_price, discount_price, quantity, fulltext_id,prod_link, images, prod_desc, status, web_name,web_logo FROM `catalogsearch_fulltext` WHERE MATCH(data_index) AGAINST("'+n+'") ', function(error, product_data) {

            var totalItems = product_data.length, itemts=product_data;

    //set default variables
    var totalResult = totalItems,
        pageSize = 10,
        pageCount = Math.floor(totalResult / pageSize)
        currentPage = 1


    //set current page if specifed as get variable (eg: /?page=2)
    if (typeof req.query.page !== 'undefined') {
        currentPage = +req.query.page;
    }

    //render index.ejs view file
            res.render('result.html', {result: n, 
            related: rows.map(row => row.query_text), 
            page_num: p,
            product_data: product_data,
            totalItems: totalItems,
            pageSize: pageSize,
            pageCount: pageCount,
            currentPage: currentPage})
        });
    });
}

via Moomoo Soso