mysql_fetch_array(): supplied argument is not a valid MySQL result resource....
还是结合代码说明一下：
<?php$username = $_POST['username'];$name = $_POST['name'];$pwd = $_POST['pwd'];$email = $_POST['email'];if(!empty($username)){$con = mysql_connect("localhost","root","");if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("test", $con);//连接一个名字叫test的数据库$sql = "select * from t_user";if(!$con){die("Error:".mysql_error());}$result = mysql_query($sql);echo "<table border ='1' bgcolor = 'lightblue'>";echo "<tr><th>用户名</th><th>姓名</th><th>密码</th><th>邮箱</th></tr>";while($row = mysql_fetch_array($result))//记得这里是把$result放入 mysql_fetch_array中，我一开始吧$sql放进去也报的是上面这种错{echo "<tr><td>{$row['f_username']}</td>";//这里字段名写错就可能导致出现上面的错误echo "<td>{$row['f_name']}</td>";echo "<td>{$row['f_password']}</td>";echo "<td>{$row['f_email']}</td></tr>\n";}echo "</table>";mysql_close($con);//关闭连接}?>