In your code below, mysql will throw an error if $clientID is empty or
non-numeric because it is not surrounded with quotation marks.. mysql will
treat it as a field name (or just throw a syntax error if it's empty).

-----Original Message-----
From: Brad Wright [mailto:[EMAIL PROTECTED]]
Sent: 11 February 2002 03:45
To: Bogdan Stancescu
Cc: PHP General List
Subject: Re: [PHP] 2 conditions why wont they work??


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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