php - MySQL 准备好的语句无法正常工作

Question

我正在尝试使用此查询获取结果

$lat = -14.4428711;
$long = 28.447634;
$userFk = 11;
$stmt = $this->conn->prepare("SELECT p.`id`, p.`storeFk`, p.`name`, p.`price`, p.`quantity`, p.`image`, s.`name` ,(SELECT COUNT(*) FROM tblpromotion WHERE productFk = p.`id`) 

    as promotional , s.`profilepicture`,3956 * 2 * ASIN(SQRT( POWER(SIN((? - s.latitude)*pi()/180/2),2) +COS(? *pi()/180 )*COS(?*pi()/180) *


    POWER(SIN((?-s.longitude)*pi()/180/2),2))) as distance, p.`dateCreated` FROM `tblproduct` p JOIN `tblstore` s ON p.`storeFk` = s.`id` JOIN tblshoppinglist l ON (


    l.name LIKE concat('%',p.name,'%') AND l.userFk = ?) WHERE s.longitude between (?-100/cos(radians(?))*69) and (? +100/cos(radians(?



    ))*69) and s.latitude between (?-(100/69)) and (? +(100 /69)) GROUP BY p.`id` having distance < 100 ORDER BY distance");


$stmt->bind_param("sssssssssss",  $lat, $lat, $lat, $long, $user, $long, $lat, $long, $lat, $lat, $lat);
$stmt->execute();
$response["products"] = array();
$stmt->bind_result($id, $storeFk, $name, $price, $quantity, $itemImage, $shopName, $promotional, $shopProfilePic, $distance, $dateCreated);
 while($row = $stmt->fetch()) 
{
   $product = array();
   $product["id"] = $id;
   $product["storeFk"] = $storeFk;
   $product["itemName"] = $name;
   $product["price"] = $price;
   $product["quantity"] = $quantity;
   $product["itemImage"] = $itemImage;
   $product["shopName"] = $shopName;
   $product["promotional"] = $promotional;
   $product["shopProfilePic"] = $shopProfilePic;
   $product["distance"] = $distance;
   $product["dateCreated"] = $dateCreated;
   $response["products"][] =  $product;

}

但没有得到任何东西，当我用值替换它时，我得到像

这样的结果

SELECT p.`id`, p.`storeFk`, p.`name`, p.`price`, p.`quantity`, p.`image`, s.`name` ,(SELECT COUNT(*) FROM tblpromotion WHERE productFk = p.`id`) as promotional , s.`profilepicture`,3956 * 2 * ASIN(SQRT( POWER(SIN((-14.4428711 - s.latitude)*pi()/180/2),2) +COS(-14.4428711 *pi()/180 )*COS(-14.4428711*pi()/180) *POWER(SIN((28.447634-s.longitude)*pi()/180/2),2))) as distance, p.`dateCreated` FROM `tblproduct` p JOIN `tblstore` s ON p.`storeFk` = s.`id` JOIN tblshoppinglist l ON (l.name LIKE concat('%',p.name,'%') AND l.userFk = 11) WHERE s.longitude between (28.447634-100/cos(radians(-14.4428711))*69) and (28.447634 +100/cos(radians(-14.4428711))*69) and s.latitude between (-14.4428711-(100/69)) and (-14.4428711 +(100 /69)) GROUP BY p.`id` having distance < 100 ORDER BY distance

我是 PHP 准备好的语句的新手，我哪里出错了？

Answer 1

伙计，替换这一行，你的代码有错误:)

$stmt->bind_param("sssssssssss", $lat, $lat, $lat, $long, $userFk, $long, $lat, $long, $lat, $lat, $lat);