What is $rsVendorJobs, where is it populated?  I assumed, from the way
you were using it - that it was a class object with a 'fields' function
that returns the value of a column when passed the column name.  Which,
looking back, seems pretty excessively  abstracted :)

Is $rsVendorJobs just the result of a mysql_query()? If so, you probably
just want to be using:

$to = $rsVendorJobs['Conmail'];

Also note - the case sensitivity may or may not be a problem (you may
need to use 'conmail')
Otherwise, let's see some code to flesh this out!

Cheers,
- Martin Norland, Database / Web Developer, International Outreach x3257
The opinion(s) contained within this email do not necessarily represent
those of St. Jude Children's Research Hospital.


-----Original Message-----
From: Stuart Felenstein [mailto:[EMAIL PROTECTED] 
Sent: Monday, December 06, 2004 1:37 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] stumped-mail and database



--- "Norland, Martin" <[EMAIL PROTECTED]>
wrote:


> 1) change
> $to = "$rsVendorJobs->Fields('Conmail')";
>       To
> $to = "{$rsVendorJobs->Fields('Conmail')}";
>       Or just
> $to = $rsVendorJobs->Fields('Conmail');
> 
> 2) change
> $body = '$cl';
>       To
> $body = $cl;
> 

Well I'm sure this is not the syntax, but since I've
written it both ways I get a :

Fatal error: Call to a member function on a non-object
in /home/xxxxxxx/public_html/appjobb.php on line 39

Which I'm guessing means it is written correctly but
now the variable doesn't exist ?

Stuart

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