這裏是我想要的,但不能讓它與新的MySQLi工作,只是因爲我的主機沒有全新的PHP等等庫MySQLi的mysqli FECH陣列assoc命令替換NO工作
但是,必須有某種解決方案還是那個所有MYSQLI可以做?
因爲即使是醜陋的名字聽起來像PEDO,我只在mysqli的興趣,併爲這個簡單的事情解決,請不要談PDO。請不要更改我的腳本的結構。問題是唯一的是,如果有東西可以與MySQL工作,或者,如果我也許切換到MySQL的過程,而不是報表
<?php
$sql = new mysqli('localhost','user','pass','db');
$who = $_GET['user'];
$query = $sql->prepare("SELECT * FROM profiles WHERE user=?");
$query->bind_param("s",$who);
$result = $query->execute();
while($row = $result->fetch_array(MYSQLI_ASSOC)) // << HERE IS THE PROBLEM //
// I GET ERROR Fatal error: Call to a member function fetch_array() on a non-object in ... //
「」但必須有某種方式來使其工作像我們做而($行= mysqli_fetch_array($ SQL))//
{
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>title</title>
</head>
<body>
CUSTOM HTML FOR A NICE DESIGN I WANT TO KEEP THE SAME DESIGN LAYOUT ETC...
HELLO <?php echo "$row[USERNAME]"; ?>
YOUR INFO IS <?php echo "$row[JUST_SOME_INFO]"; ?>
</body>
</html>
<?php
}
?>
'不談論PDO,因爲即使是醜陋的名字聽起來像PEDO'你贏的趣聞 – Mark
哈哈哈真正的名稱不好聽 – Firefighter