mysqli什么都不返回

Question

我有一个像这样的mysql表

name    adults  children
single  1        0  
double  2        0  
suite   3        0  

但我无法使用此功能检索“成人”和“儿童”的值

function getInfo($mysqli, $c, $query){

    // WARNING! THIS LINE IS NOT THE SQL QUERY BUT JUST AN ERROR LOG!
    error_log("query is: SELECT " . $query . " FROM categories WHERE `name` = " . $c);

    $mysqli->select_db("hotel");

    $sql = $mysqli->prepare("SELECT ? FROM categories WHERE `name` = ?
    LIMIT 0, 999999");

    $sql->bind_param('ss', $query, $c);



    $sql->bind_result($result);

    while ($sql->fetch())
        return($result);

}

错误日志显示变量已正确传递给函数：

[15-Apr-2013 19:36:10] SELECT adults FROM categories WHERE `name` = single

为什么这个函数返回NULL？

Answer 1

你错过了$sql->execute();

$sql->bind_param('ss', $query, $c);
$sql->execute();
$sql->bind_result($result);