PHP又出错.

19821119

又出错,汗一下,这是最后一个错误....

源码如下:

<?php
if ($login_utype=="0")
$sqlquery1 ="select ExpertInfo.*,Member.Conn,ExpertCate.CateName as CateName2 from ExpertInfo,ExpertCate,Member where ExpertInfo.CateId=ExpertCate.CateId and ExpertInfo.Uid=Member.Uid and ExpertInfo.Uid=".$login_uid ;
if ($login_utype=="1")
$sqlquery1 = "select * from CompanyInfo where Uid=".$login_uid ;
if ($login_utype=="2")

$sqlquery1 = "select * from CompanyInfo where Uid=".$login_uid ;
$getmyself = mysql_db_query($dbname1,$sqlquery1, $mysql_local_link);
$myselfrow = mysql_fetch_array($getmyself);

if ($login_utype=="0")
$extrastr = "&MyType=0&myinfo=1&MyType1=0" ;
else
$extrastr = "&MyType=0&myinfo=1";

$i = mysql_num_fields($getmyself)-1 ;
while ($i>=0)
{
   $fieldname = mysql_field_name($getmyself,$i) ;       
   $value = ltrim(trim($myselfrow[$fieldname])) ;
   $extrastr = $extrastr."&".$fieldname."1=".$value ;
   $i-- ;
}


$sqlquery2 = "select * from MoneySetting where MSId=10" ;
$getmyself2 = mysql_db_query($dbname1,$sqlquery2, $mysql_local_link);
$myselfrow2=mysql_fetch_array($getmyself2);
$MSUnit = $myselfrow2[MSUnit] ;
?>
浏览器出错:

Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in d:\website\userinfo\selfinfo.php on line 9

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in d:\website\userinfo\selfinfo.php on line 10

Warning: mysql_num_fields(): supplied argument is not a valid MySQL result resource in d:\website\userinfo\selfinfo.php on line 17

Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in d:\website\userinfo\selfinfo.php on line 28

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in d:\website\userinfo\selfinfo.php on line 29


谢谢各位.:em02:

carset

推荐使用 switch 结构，或者加上 {} 来区分代码块.

这样看起来太累.还是不看了..

james.liu

代码命名最好简单点

看的头晕

chf99cn

晕,你最好整理一下.