MySQL的預處理語句無法正常工作

我想這個查詢MySQL的預處理語句無法正常工作

$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; 

}

獲取結果，但沒有得到任何東西，當我和值替換它，我得到的結果一樣

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

是新來的Php準備了報表，我哪裏出錯了？

來源

2017-06-01 Sandy McCathy

H你檢查你的SQL查詢 - 它是否返回MySQL中的東西？ –

絕對值的第二個是返回值，但是當我將它重寫爲準備語句時，它不返回任何內容。 –

$ stmt-> bind_param（「sssssssssss」，$ lat，$ lat，$ lat，$ long，$ userFk，$ long，$ lat，$ long，$ lat，$ lat，$ lat）;更換這條線 –

好傢伙，替換這一行，你必須在代碼:)

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

來源

2017-06-01 00:43:30

你改變了什麼？我無法發現任何區別。「替換這一行」，爲什麼？一個好的答案總是會包含一個解釋，說明改變了什麼 - 以及爲什麼。 – Qirel

有一個錯誤在那一行中，他使用了一個未定義的變量 –

個錯誤 - 絃樂

d - 雙

我 - 爲整數

b - 用於blob

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

來源

2017-06-01 04:26:06

MySQL的預處理語句無法正常工作

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