为什么我这句
$char_sql="select * from $table3 where select=20030005";
和
$char_sql="select * from $table3 where select='20030005' ";
都会出下面的错误？

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in e:\inetpub\wwwroot\getlans\viewsingleup.php on line 29

我试过了，用$char_sql="select * from $table3 ";读取全部数据是没问题的