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我有一個MySQL的句子就像如果我在phpMyAdmin的執行它的魅力：問題與MySQL的句子包括兩個臨時表創建
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
我想執行通過Ajax請求的PHP函數，即構建dinamically SQL語句以上，但我總是得到這個錯誤在我的瀏覽器的控制檯：
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
就這樣，下面的錯誤太：
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in /www/htdocs/example/inc/functions.php on line 538
對應於該行的我的PHP代碼：
while ($row = mysqli_fetch_assoc($result))
也許克隆HASH2表從HASH1表
CREATE TEMPORARY TABLE hash2
SELECT * FROM hash1;
聽起來不可思議，但如果我不這樣做的這樣，在我的phpMyAdmin我得到這個錯誤：
#1137 - Can't reopen table: 'b'
我不明白爲什麼我的sql語句在我的phpMyadmin中工作正常，但是當我在我的PHP文件中構建它時，它不起作用。有人可以幫我嗎？
如需進一步信息，這是我的PHP代碼：
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
}
};