On Tuesday 10 August 2004 22:54, Henri Marc wrote:
> > Variables in single-quoted strings are not
> > evaluated. Either user double
> > quotes or concatination:
>
> Thank you very much all for your help, specially Kevin
> Waterson for his complete program.
> It was simple, I always make some mista
On 10 August 2004 15:55, Henri Marc wrote:
> Hello,
>
> > Variables in single-quoted strings are not
> > evaluated. Either user double
> > quotes or concatination:
> Thank you very much all for your help, specially Kevin
> Waterson for his complete program.
> It was simple, I always make some mis
Hello,
> Variables in single-quoted strings are not
> evaluated. Either user double
> quotes or concatination:
Thank you very much all for your help, specially Kevin
Waterson for his complete program.
It was simple, I always make some mistakes with those
quotes :-(
Another problem still related t
$myimage = 'hi.gif';
echo "";
Henri marc wrote:
Hello,
I would like to use a variable instead of an image
file name in a html page with this instruction:
';
?>
I tried but the image doesn't show up. Is it
impossible or do I do something wrong?
My goal is to have a random image print in the page,
th
"Henri marc" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Hello,
>
> I would like to use a variable instead of an image
> file name in a html page with this instruction:
>
> echo '';
> ?>
>
> I tried but the image doesn't show up. Is it
> impossible or do I do something wrong?
> M
Try this:
http://pear.php.net/gifs/pearsmall.gif";;
echo '';
?>
Probably you are not initializing the $myimage var.
To see if the html generated is ok, you can use your browser "view
source" capability.
Henri marc wrote:
Hello,
I would like to use a variable instead of an image
file name in a
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