如何将MySQL语句分配给PHP变量

Question

我已经浏览了整个网络，似乎无法找到将MySQL查询的结果分配给PHP变量的正确方法。我目前拥有的代码返回“资源ID＃3”错误。这是它的样子。

//Select the 'aQID' of the question that has it's BOOL set to "true"

$currentQ = mysql_query("SELECT aQID FROM approvedQuestions WHERE status='1'");

$cuQ = mysql_result($currentQ,1,"status");

echo $cuQ;

我知道查询只返回一条记录（活动问题）。但我似乎无法弄清楚使用什么功能。

Answer 1

$currentQ = mysql_query("SELECT aQID FROM approvedQuestions WHERE status='1'");

    $rs = mysql_fetch_array($currentQ);

    var_dump($rs);