如果有人能告诉我如何为 mysqli 替换/修复它，我将不胜感激。这里有两个例子，所以我可以从中学习并解决所有需要的问题。
示例 1：
include dirname(__FILE__) . '/database.php';
mysql_select_db(databasename, mysql_connect(server, user, pass));
$q_settings = mysql_query("SELECT * FROM settings");
$settings = array();
while ($setting_one = mysql_fetch_array($q_settings)) {
$settings[$setting_one["name"]] = $setting_one["value"];
}
示例 2：
$one_blog = mysql_fetch_array(mysql_query("SELECT * FROM blog WHERE id = '" . mysql_real_escape_string($_GET["id"]) . "' LIMIT 1"));
感谢您的任何提示和示例。我已经尝试了 php.net 用户评论中的一些示例，但在某些情况下我失败了。这就是我所做的。
$db_select = mysqli_connect(server, user, pass, databasename) or die("Error " . mysqli_error($db_select));
$db_query = "SELECT * FROM settings";
$db_result = $db_select->query($db_query) or die("Error in the consult..." . mysqli_error($db_select));
$settings = array();
while ($setting_one = mysqli_fetch_array($db_result)) {
$settings[$setting_one["name"]] = $setting_one["value"];
}
对于您的第一个示例：
$mysqli = mysqli_connect("server", "user", "password", "database");
$q_settings = $mysqli->query("SELECT * FROM settings");
$settings = array();
while($row = $q_settings->fetch_assoc()){
$settings[$row["name"]] = $row["value"];
}
我一遍又一遍地尝试，我相信这是我的第二个示例的正确方法。
$one_blog = mysqli_fetch_array($mysqli->query("SELECT * FROM blog WHERE id = '" . $mysqli->real_escape_string($_GET["id"]) . "' LIMIT 1"));
感谢韦恩提供您的第一个样品。这真的很有帮助。