Query Problem

[daksport00@gmail.com]

Im not sure if Im posting this in the proper Group, if not, please
direct me to the correct place.

I am working on a website for a gaming league/ladder. I have one
particular line of code that is giving me a hell of an issue.


$mbrguid = (mysql_query(SE LECT guid FROM users WHERE id='$inviteid'" )
if (is_null($mrbgu id)){go into error page}


The table Users has a text field GUID which defaults to a Null. Id is
a numberic player ID which is typed in by the end-user.


All of my SQL knowledge tells me that this code should pull the GUID
field form the Users table based on the ID.


The problem I am having is that when I echo the $mrbguid variable, it
return a value of "Resource ID #19". This value doesn't appear in the
Users table anywhere, let alone in ANY table.


Any suggestions as to what is causing this error, or perhaps a way to
work around this code.


What I am looking to do, is pull the GUID field form the table. If it
is a null or blank value, throw the program into error.


Yes, I know that Null and Blank(empty) are not the same thing, but I
can change the table setup to suit the easier variant.


Thanks


-Dave

[Andy Hassall]

Re: Query Problem

On 26 Jun 2006 12:31:59 -0700, daksport00@gmai l.com wrote:
[color=blue]
>$mbrguid = (mysql_query(SE LECT guid FROM users WHERE id='$inviteid'" )
>if (is_null($mrbgu id)){go into error page}
>
>The problem I am having is that when I echo the $mrbguid variable, it
>return a value of "Resource ID #19". This value doesn't appear in the
>Users table anywhere, let alone in ANY table.[/color]

This is how mysql_query works, it returns a result set resource.

http://uk.php.net/mysql_query

--
Andy Hassall :: andy@andyh.co.u k :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

[Geoff Berrow]

Re: Query Problem

Message-ID: <1151350319.546 164.145600@i40g 2000cwc.googleg roups.com> from
daksport00@gmai l.com contained the following:
[color=blue]
>$mbrguid = (mysql_query(SE LECT guid FROM users WHERE id='$inviteid'" )
>if (is_null($mrbgu id)){go into error page}
>
>
>The table Users has a text field GUID which defaults to a Null. Id is
>a numberic player ID which is typed in by the end-user.
>
>
>All of my SQL knowledge tells me that this code should pull the GUID
>field form the Users table based on the ID.[/color]


$result = mysql_query("SE LECT guid FROM users WHERE id='$inviteid'" );
$mbrguid=mysql_ fetch_array($re sult);
if (is_null($mrbgu id['guid'])){go into error page}

--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/

[Alvaro G. Vicario]

Re: Query Problem

*** daksport00@gmai l.com escribió/wrote (26 Jun 2006 12:31:59 -0700):[color=blue]
> $mbrguid = (mysql_query(SE LECT guid FROM users WHERE id='$inviteid'" )
> if (is_null($mrbgu id)){go into error page}[/color]

Next time please post real code. That's where errors are 99% of the times.

[color=blue]
> All of my SQL knowledge tells me that this code should pull the GUID
> field form the Users table based on the ID.[/color]

It's normally better to just try the actual SQL code: open your favourite
MySQL client and paste it there.



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