mySQL句子的问题包括两个临时表创建

I have a mySQL sentence that works like a charm if I execute it in my phpMyAdmin:

CREATE TEMPORARY TABLE hash1
      SELECT * FROM
      (
        (
            SELECT DISTINCT feature_id AS fl, feature_value AS fv FROM gf_product_features WHERE feature_id = '1' AND feature_value = 'No frost total'
        ) UNION 
        (
            SELECT DISTINCT feature_id AS fl, feature_value AS fv FROM gf_product_features WHERE feature_id = '3' AND feature_value = '43'
        )) AS q;


      CREATE TEMPORARY TABLE hash2
        SELECT * FROM hash1;

        SELECT 
          p.id AS id, 
          p.main_image AS main_image, 
          p.type AS taxonomy, 
          p.name AS model, 
          p.sku AS sku, 
          p.price AS price, 
          b.brand_name AS brand_name, 
          b.brand_image AS brand_logo,
          pf.feature_value AS feature_value, 
          f.feature AS feature_label,
          f.id AS feature_id
        FROM
        (
          SELECT  a.*
          FROM    gf_product AS a
          INNER JOIN
          (
            SELECT product_id
            FROM
            (
              SELECT a.product_id , count(*) AS commons
              FROM   gf_product_features AS a
              INNER JOIN hash1 AS b 
                ON    a.feature_id = b.fl 
                AND   a.feature_value = b.fv 
              GROUP BY a.product_id 
              ) AS features
              WHERE commons = (SELECT count(*) AS count FROM hash2)  
            ) b1 ON a.id = b1.product_id 
          ) AS p
        INNER JOIN  gf_brands AS b 
            ON p.brand_id = b.id
        INNER JOIN  gf_product_features AS pf 
            ON pf.product_id = p.id   
        INNER JOIN  gf_features AS f 
            ON pf.feature_id = f.id
        ORDER BY    price ASC, 
                    feature_id ASC

I want to execute a php function through Ajax request, that constructs dinamically the sql sentence above, but I'm always getting this error in my browser's console:

    You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'CREATE TEMPORARY TABLE hash2
        SELECT * FROM hash1;

        SELECT 
       ' at line 12

And thus, the following error too:

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in /www/htdocs/example/inc/functions.php on line 538

Which corresponds to this line of my php code:

while ($row = mysqli_fetch_assoc($result))

Maybe clone hash2 table from hash1 table

CREATE TEMPORARY TABLE hash2
        SELECT * FROM hash1;

sounds weird, but if I don't do this in that way, in my phpMyAdmin I get this error:

 #1137 - Can't reopen table: 'b'

I can't realize why my sql sentence works fine in my phpMyadmin but, when I construct it on my php file it doesn't works. Can anybody help me, please?

For further information, this is my PHP code:

    function getProductsFromFilteredQuery($connection, $filters, &$html)
{
    $sql = '';
    $m = count($filters); // $filters are an array of values like this: ['value1A, value2A', 'value1B, value2B', ...]

    $sql = 'CREATE TEMPORARY TABLE hash1
      SELECT * FROM
      (';

    for ($n = 0; $n < $m; $n++)
    {
        $string                 = explode(', ', $filters[$n]);
        $feature_id         = $string[0];
        $feature_value  = $string[1];

        $sql .= "
        (
            SELECT DISTINCT feature_id AS fl, feature_value AS fv FROM gf_product_features WHERE feature_id = '" . $feature_id . "' AND feature_value = '" . $feature_value . "'
        )";

        if ($n < ($m - 1))
        {
            $sql .= ' UNION ';
        }
    }


    $sql .=  ') AS q;


      CREATE TEMPORARY TABLE hash2 -- In this line I get an error
        SELECT * FROM hash1;

        SELECT 
          p.id AS id, 
          p.main_image AS main_image, 
          p.type AS taxonomy, 
          p.name AS model, 
          p.sku AS sku, 
          p.price AS price, 
          b.brand_name AS brand_name, 
          b.brand_image AS brand_logo,
          pf.feature_value AS feature_value, 
          f.feature AS feature_label,
          f.id AS feature_id
        FROM
        (
          SELECT  a.*
          FROM    gf_product AS a
          INNER JOIN
          (
            SELECT product_id
            FROM
            (
              SELECT a.product_id , count(*) AS commons
              FROM   gf_product_features AS a
              INNER JOIN hash1 AS b 
                ON    a.feature_id = b.fl 
                AND   a.feature_value = b.fv 
              GROUP BY a.product_id 
              ) AS features
              WHERE commons = (SELECT count(*) AS count FROM hash2)  
            ) b1 ON a.id = b1.product_id 
          ) AS p
        INNER JOIN  gf_brands AS b 
            ON p.brand_id = b.id
        INNER JOIN  gf_product_features AS pf 
            ON pf.product_id = p.id   
        INNER JOIN  gf_features AS f 
            ON pf.feature_id = f.id
        ORDER BY    price ASC, 
                    feature_id ASC';

    $result = mysqli_query($connection, $sql);

    while ($row = mysqli_fetch_assoc($result)) // In this line I get an error too
    {
        // Do some stuff... and at last, return the resulting $html
    }
};