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我正在尝试将 mysql 脚本转换为 mysqli 并碰壁。我正在尝试将 mysql_result 转换为 mysqli 但是不确定如何执行此操作,下面是我的代码

$_SESSION['num_user'] = mysql_result(mysqli_query($GLOBALS["___mysqli_ston"], "SELECT COUNT(*) FROM `members` WHERE mem_emailactivated = 1"), 0);
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3 回答 3

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尽管很奇怪,但似乎没有一种方法可以在没有准备/执行的情况下进行获取。

$stmt = mysqli_prepare($GLOBALS['___mysqli_ston'], "SELECT COUNT(*) AS count ...");
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $count);
mysqli_stmt_fetch($stmt);
$_SESSION['num_user'] = $count;
于 2013-04-01T15:34:29.817 回答
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你在这里把事情搞混了。您不会mysql_result使用mysqli. 您将改为使用mysqli遍历返回的数组,然后将此数组存储在$_SESSION.

$result = $mysqli->query("call getUsers()");
if($result){
     // Cycle through results
    while ($row = $result->fetch_object()){
        $user_arr[] = $row;
    }

    $result->close();
}

$_SESSION['num_user'] = $user_arr;
于 2013-04-01T15:27:47.583 回答
0

使用SafeMysql

$sql = "SELECT COUNT(*) FROM `members` WHERE mem_emailactivated = 1";
$_SESSION['num_user'] = $db->getOne($sql);
于 2013-04-01T15:31:29.157 回答