mysqli_connect不报错，但此后操作都显示not a valid MySQL-Link resource

<code>$connect = mysqli_connect("host","user","password","db") or die("Error " . mysqli_error($connect));
$result = mysql_query('select * from admin',$connnet);
print(mysql_num_rows($result));
mysql_close();
</code>

报错：

<code>Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource
Warning: mysql_close(): no MySQL-Link resource supplied
</code>

而使用面向对象方法则没有问题
请问这是什么问题？

回复内容：

<code>$connect = mysqli_connect("host","user","password","db") or die("Error " . mysqli_error($connect));
$result = mysql_query('select * from admin',$connnet);
print(mysql_num_rows($result));
mysql_close();
</code>

报错：

<code>Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource
Warning: mysql_close(): no MySQL-Link resource supplied
</code>

而使用面向对象方法则没有问题
请问这是什么问题？

因为你写错字母了...
第一个变量是 $connect
第二个变量是 $connnect
多了一个n，好不？
当然，也要用mysqli_query而不是mysql_query

mysql_query -> mysqli_query
mysql_num_rows -> mysqli_num_rows
mysql_close -> mysqli_close

哥们你用mysqli_connect创建的连接得用mysqli_xxx来用
mysqli_xxx和mysql_xxx是不能混用的