php程序为啥警告啊,程序没有错啊
这是程序代码
$sql = "SELECT id, cname FROM usercountry ORDER BY cname ASC";
$result=mysql_query($sql,$conn);
$rows=mysql_num_rows($result);
if ($rows<=0) { ----这是市82行,没有错误啊,可是警告啊
echo "<tr><td>Country:</td><td width='65%' height='35' align='left'>no country</td></tr>";
}else {
?> 错误如下:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\Debug\php\phphuanjing\htdocs\agoodic\admin\adduser.php on line 82
请高手指教一下,如何处理啊
这是程序代码
$sql = "SELECT id, cname FROM usercountry ORDER BY cname ASC";
$result=mysql_query($sql,$conn);
$rows=mysql_num_rows($result);
if ($rows<=0) { ----这是市82行,没有错误啊,可是警告啊
echo "<tr><td>Country:</td><td width='65%' height='35' align='left'>no country</td></tr>";
}else {
?> 错误如下:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\Debug\php\phphuanjing\htdocs\agoodic\admin\adduser.php on line 82
请高手指教一下,如何处理啊
SELECT id, cname FROM usercountry ORDER BY cname ASC 这句去数据库运行下看看!
贴出来大家看看!
$rows=mysql_num_rows($result); 下加上
echo $rows;//看看是否有值
$result=mysql_query($sql,$conn);
$rows=mysql_num_rows($result);
之间加一个判断,如下:
$result=mysql_query($sql,$conn);
if(count($result) <= 0)
{
echo " <tr> <td>Country: </td> <td width='65%' height='35' align='left'>no country </td> </tr>"; }else
{
$rows=mysql_num_rows($result);
}
这样做就不会有警告了!