将查询结果转换为PHP变量

Question

下面的查询在页面上，其中$submissionid只有一个值。 因此，场points仅具有一个值。

如何将下面查询中的points转换为常规PHP变量？

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

编辑：这是从MySQL数据库中提取的。

编辑II：我要points等于整个页面上的变量，而不仅仅是在while循环内。

Answer 1

这样，您可以读取查询返回的所有点。

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

Answer 2

我将使用ORM库（我使用Kohana附带的库）来消除SQL注入的可能性。 但是，假设已经建立了数据库连接，则此代码将执行您要查找的操作。

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

如果您尚未建立MySQL数据库连接，请签出mysql_connect 。

Answer 3

好吧，您需要从该查询中获取资源，然后将其提供给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，该查询也不安全（如果$ submissionid来自用户输入，则容易受到SQL注入的攻击），您应该使用iloveitaly提到的ORM库。 （我使用Zend DB ，尽管从技术上讲它并不是ORM本身）

编辑：

正如评论所指出的，这取决于您是否实际使用Mysql。 如果不是，则可以使用PDO库进行查询。

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