Ahh..that seems to have fixed it. Its now pulling NO data, but is not giving
error msg....
thanks, 1/2 way there now :)

brad


> From: Bogdan Stancescu <[EMAIL PROTECTED]>
> Date: Mon, 11 Feb 2002 05:46:46 +0200
> To: Brad Wright <[EMAIL PROTECTED]>
> Cc: PHP General List <[EMAIL PROTECTED]>
> Subject: Re: [PHP] 2 conditions why wont they work??
> 
> No, with your syntax it won't. You have to quote it in the query, as in
> $query2 = "select * from Table where  userNo = $userNo and clientID =
> '$clientID'";
> 
> Bogdan
> 
> Brad Wright wrote:
> 
>> Thanks,
>> 
>> $clientID is a string but is not empty...already tried echo($clientID) and
>> it is not empty.
>> 
>> Did you mean that if the value of $cientID is a string it wont work????
>> 
>> Thanks
>> Brad
>> 
>>> From: Bogdan Stancescu <[EMAIL PROTECTED]>
>>> Date: Mon, 11 Feb 2002 05:38:31 +0200
>>> To: Brad Wright <[EMAIL PROTECTED]>
>>> Cc: PHP General List <[EMAIL PROTECTED]>
>>> Subject: Re: [PHP] 2 conditions why wont they work??
>>> 
>>> Do an echo($clientID) before - it most probably is either empty or it's a
>>> string and MySQL actually issues errors there.
>>> 
>>> Bogdan
>>> 
>>> Brad Wright wrote:
>>> 
>>>> Hi all,
>>>> Im was sure you could select a row from a mySQL database based on 2
>>>> conditions. My code:
>>>> 
>>>> $query2 = "select * from Table where  userNo = $userNo and clientID =
>>>> $clientID";
>>>> $result2 = mysql_query($query2,$db);
>>>> 
>>>> This returns :
>>>> Warning: Supplied argument is not a valid MySQL result resource
>>>> 
>>>> b ut if i change the line to:
>>>> $query2 = "select * from Table where  userNo = $userNo;
>>>> $result2 = mysql_query($query2,$db);
>>>> 
>>>> it works (but get all the rows with userNo = $userNo.
>>>> 
>>>> HELP!!!! Im sure this should work...what am i doing wrong???
>>>> 
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>>> 
>>> 
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