We all know this works:
echo p$someVar;
However this does not:
echo p$db-field('name_long');
I know this slight variation will make it work:
echo p . $db-field('name_long');
But it's cumbersome .. Anyway to get the first way to work?
~ Mike
--
Mike Zornek | Project Leader
Apple Student
Try enclosing it in curly braces.
echo p${db-field('name_long')};
Martin Clifford
Homepage: http://www.completesource.net
Developer's Forums: http://www.completesource.net/forums/
Michael Zornek [EMAIL PROTECTED] 07/16/02 01:45PM
We all know this works:
echo p$someVar;
However this does
Twas 7/16/02 1:47 PM, when Martin Clifford [EMAIL PROTECTED] said:
Try enclosing it in curly braces.
echo p${db-field('name_long')};
Parse error :-(
~ Mike
--
Mike Zornek | Project Leader
Apple Student Developers
The Insanely Great Site with the Insanely Long URL
ðÒÉ×ÅÔ!
Michael Zornek wrote:
We all know this works:
echo p$someVar;
However this does not:
echo p$db-field('name_long');
I know this slight variation will make it work:
echo p . $db-field('name_long');
But it's cumbersome .. Anyway to get the first way to work?
~ Mike
In
On Tue, Jul 16, 2002 at 01:45:01PM -0400, Michael Zornek wrote:
I know this slight variation will make it work:
echo p . $db-field('name_long');
That's not an object variable name. That's a function call. You're
asking to echo p and then echoing the value returned by the field()
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