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
}
};
