PHP警告:mysql\u fetch\u array():提供的参数不是有效的mysql结果资源
将网站上载到web服务器后,我收到以下消息:PHP警告:mysql\u fetch\u array():提供的参数不是有效的mysql结果资源,php,mysql,database,arrays,fetch,Php,Mysql,Database,Arrays,Fetch,将网站上载到web服务器后,我收到以下消息: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home1/m1k3ey/public_html/MikeyDev.com/teamdesire/playersheet.php on line 8 Warning: mysql_num_rows(): supplied argument is not a valid M
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
in /home1/m1k3ey/public_html/MikeyDev.com/teamdesire/playersheet.php on line 8
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource
in /home1/m1k3ey/public_html/MikeyDev.com/teamdesire/playersheet.php on line 9
我的PHP版本是:
PHP版本5.2.17
我看不出我的代码哪里出错了,请任何人帮助:
mysql_select_db('teamdesire', $link);
$query = "SELECT * FROM playershowercase";
$result = mysql_query($query,$link);
$row = array();
Line 8 > while($row[] = mysql_fetch_array($result));
Line 9 > $count = mysql_num_rows($result);
$random = rand(0,$count-1);
试试这个:
mysql_select_db('teamdesire', $link);
$query = "SELECT * FROM playershowercase";
$result = mysql_query($query,$link);
$row = array();
while($result = mysql_fetch_array($result))
{
$row[]=$result;
}
$count = mysql_num_rows($result);
$random = rand(0,$count-1);
也许像下面这样
mysql_select_db('teamdesire', $link);
$query = "SELECT * FROM playershowercase";
$result = mysql_query($query,$link);
$row = mysql_fetch_assoc($result);
$count = mysql_num_rows($result);
$random = rand(0,$count-1);
或
可能的错误源:
- Db未在指定的套接字或端口上运行
- 对表或数据库的权限不足
- 登录凭据不正确
- 表不存在或拼写错误
is_resource($link) or die('Could not connect');
mysql_query(...);
echo mysql_error();
将PHP版本导入服务器???与通常的错误消息一样:您的查询以某种方式失败,您的代码盲目地假设成功,并错误地继续运行。@MikeyT-您是否尝试过
$result=mysql\u query($query,$link)或die(mysql\u error())代码>?或者var\u dump($result
?是的,我试过了,在我的错误日志中,它显示的错误与屏幕上显示的完全相同screen@MikeyT-那么var\u dump($result)
显示了什么?不幸的是,我又遇到了同样的错误
is_resource($link) or die('Could not connect');
mysql_query(...);
echo mysql_error();