PHP查询问题

PHP code:
<?php
/*
* Created on 2010-3-3
*
* To change the template for this generated file go to
* Window - Preferences - PHPeclipse - PHP - Code Templates
*/
$mysqli=mysqli_connect("localhost","root","x-d","news");
$sql = "SELECT * from news";
$res = mysqli_query($mysqli,$sql);

$a = mysql_fetch_array($res);
$num = mysql_num_rows($a);

echo $num;

?>




报下面的错。。。

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in G:\PHP\news\select.php on line 14

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in G:\PHP\news\select.php on line 15

请大侠指导。。。

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news 既是数据库名又是表名？

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你不要把mysqli和mysql混着用.....

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应该是$res = mysqli_query($mysqli,$sql)这句有问题.楼主看看手册:
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode ] )

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$a = mysql_fetch_array($res);
$num = mysql_num_rows($a);

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把上面的mysql_ 换成mysqli_

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发现函数混用啊,mysql和mysqli.楼主还是查查手册吧

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你得保证mysqli_connect中最后一个参数news是数据库名称
PHP code

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