Why isn't this working?
");
} else {
print("");
}//end if
}//end for loop
?>
_
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Can you not just make it less confusing and have the one variable to
echo to the page?
As you saw from my example, a single variable can have multiple values
if it is produced from a for-loop.
I'm sorry if this doesn't help - often a good night's sleep sorts it
out - exactly what I'm off to d
oops $x should start with "0"
-Original Message-
From: Larry Rivera [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, March 27, 2001 11:04 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] A question - ($file_exists)
or you can put yuou r values into an array the
s
Michell
Sent: Tuesday, March 27, 2001 10:46 AM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] A question - ($file_exists)
Hi there: What about doing it like this:
#---#
"Believe nothing - consider everything&q
Hi there: What about doing it like this:
#---#
"Believe nothing - consider everything"
Russ Michell
Anglia Polytechnic University We
> Whats wrong with this code?
>
> $img1 = "episodeID_a.jpg";
> $img2 = "episodeID_b.jpg";
> $img3 = "episodeID_c.jpg";
> $img$x="blank.jpg";
That doesn't what you expect. I think the syntax is the same as Perl:
$var = "$img$x"; # $var hold the _name_ of the variable
