將查詢結果轉換為PHP變量

Question

下面的查詢在頁面上，其中$submissionid只有一個值。 因此，場points僅具有一個值。

如何將下面查詢中的points轉換為常規PHP變量？

  $sqlStra = "SELECT points, submissionid
               FROM submission 
              WHERE submissionid = '$submissionid'";

編輯：這是從MySQL數據庫中提取的。

編輯II：我要points等於整個頁面上的變量，而不僅僅是在while循環內。

Answer 1

這樣，您可以讀取查詢返回的所有點。

while ($row = mysql_fetch_array($sqlStra, MYSQL_ASSOC)) {
    $points = $row["points"]);
}

Answer 2

我將使用ORM庫（我使用Kohana附帶的庫）來消除SQL注入的可能性。 但是，假設已經建立了數據庫連接，則此代碼將執行您要查找的操作。

$resource = mysql_query("SELECT points, submissionid FROM submission WHERE submissionid = '$submissionid'");
$result = mysql_fetch_assoc($resource);
echo $result["points"];

如果您尚未建立MySQL數據庫連接，請簽出mysql_connect 。

Answer 3

好吧，您需要從該查詢中獲取資源，然后將其提供給mysql_fetch_assoc，如下所示：

$res = mysql_query($sqlStra);

// If you **Know for sure** you'll only have one row:
$row = mysql_fetch_assoc($res);
$points = $row['points'];

//Otherwise, you're going to need to loop.
$array_of_all_points = array();

while ($row = mysql_fetch_assoc($res)) {
    // $row['points'] now has what you want. You can assign it normally, or can use
    // extract($row) to turn it into $points, which I would advise against personally.
    $points = $row['points'];
    $array_of_all_points[$row['submissionid']] = $row['points'];
}

echo $points; // This will be the last $row['points'] that was executed.
echo $array_of_all_points['thatsubmissionid']; // Will output the points for the given session id. 

此外，該查詢也不安全（如果$ submissionid來自用戶輸入，則容易受到SQL注入的攻擊），您應該使用iloveitaly提到的ORM庫。 （我使用Zend DB ，盡管從技術上講它並不是ORM本身）

編輯：

正如評論所指出的，這取決於您是否實際使用Mysql。 如果不是，則可以使用PDO庫進行查詢。

Answer 4

$sqlStra = "SELECT points FROM submission WHERE submissionid = ?"; 

$statement = $connection->prepare($sqlStra);
$statement->bind_param('i', $submissionid);
$statement->bind_result($points);
$statement->execute();
$statement->fetch(); // this will create / populate $points