| 发表于:2007-04-09 16:34:21 楼主 |
小弟是新手,刚刚开始学习php,在调试程序中需要对数据库进行两次查询,建立了三个页面conn.php、getsettingphp、index.php,需要在index.php中调用conn.php和getsetting.php,代码如下: conn.php <?php //这里设置连接数据库 $myconn=mysql_connect( 'localhost ', 'root ', '123456 '); //连接数据库服务器 mysql_select_db( "tdababase "); //连接指定的数据库 mysql_query( "set names 'gbk ' "); //设置编码类型 ?> getsetting.php <?php //这里获得数据库中的系统信息 $str_sql= "select * from setting "; $result=mysql_query($str_sql,$myconn); while ($row=mysql_fetch_object($result)){ global $sysnane,$syscompany,$sysinfo,$sysediter,$sysemail; $sysnane=$row-> l_sysname; $syscompany=$row-> l_company; $sysinfo=$row-> l_sysinfo; $sysediter=$row-> l_sysediter; $sysemail=$row-> l_sysemail; } mysql_free_result($result); //这里获取数据库中的样式信息 $str_sql= "select * from style where l_stylemain=1 "; $result=mysql_query($str_sql,$myconn); while ($row=mysql_fetch_object($result)){ global $stylenane,$stylepatch,$stylefile; $stylenane=$row-> l_stylename; $stylepatch=$row-> l_stylepatch; $stylefile=$row-> l_stylefile; } mysql_free_result($result); ?> index.php <?php require_once "conn.php "; require_once "getsetting.php "; echo $sysnane; echo $syscompany; echo $sysinfo; echo $sysediter; echo $sysemail; echo $stylenane; echo $stylepatch; echo $stylefile; ?> 但是运行后提示: warning: mysql_fetch_object(): supplied argument is not a valid mysql result resource in /3wdoc/getsetting.php on line 17 warning: mysql_free_result(): supplied argument is not a valid mysql result resource in /3wdoc/getsetting.php on line 23 代码上是不是那里还需要做什么?请老鸟指点迷津!!!! |
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