I do not have the die statment -- the DB is the same except the name and I
have double checked that I just did a back up of the previous db and moved
it to the new server. I have done some more digging, my query appears to
work, but I almost appears that I have a problem with selecting the
database --
if I comment out the mysql_select -- statment I get the same error, perhaps
I just can't get to that DB.
intresting.

I guess I can assume at this point there is something flaky with my DB


"Jeffrey N Dyke" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
> Do you have an 'or die(mysql_error())'  statement following your
> mysql_query($result) line.  99% of the time, this error means your query
> failed.  if it works on another server....are the fields the same, the
> dbname, the tablename?
>
> hth
> jeff
>
>
>
>                       "Greg Hetrick"
>                       <[EMAIL PROTECTED]        To:
[EMAIL PROTECTED]
>                       net>                     cc:
>                                                Subject:  [PHP-DB]
mysql_fetch_array issues.
>                       07/09/2003 12:04
>                       PM
>
>
>
>
>
>
> I am getting the following error when attempting to pull data out of a
> mysql
> DB
>
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result
> resource in /home/pffl/public_html/pffl/webpage/html/dataentry.php on line
> 125
>
> I can take this same code to a different web server and it is pulling
> correctly!
>
> I am currently running Apache 2.0.46 with PHP 4.3.2 I was running Apache
> 1.3.x with PHP 4.3.1 and getting the same thing, any ideas? here is the
> chunk of code where I use the function.
>
> while ($myrow=mysql_fetch_array($result))
>         {
>                 ?>
>                 <option value="<?print $myrow['name'];?>, <?print
> $myrow['team'];?>"> <? echo $myrow["name"]; ?></option>
>                 <?
>         }
>
> Any Ideas.
>
>
>
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>
>
>
>



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