Ross,
If I understand correctly what you want to do, you're almost there...
You need:
$myimage1 = "image1.jpg";
$myimage2 = "image2.jpg";
$myimage3 = "image3.jpg";
$body .="
";
Cheers,
Mark
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Le lundi 28 mars 2011 à 12:06 +0100, Hulf a écrit :
> Hi,
>
> I am making and HTML email. I have 3 images to put in. Currently I have
>
> $body .="
>
>
>
>
>
>
>
>
>
> ";
>
>
> ideally I would like to have
>
> $myimage1 = "image1.jpg";
> $myimage2 = "image2.jpg";
> $my
On 3/28/2011 4:06 AM, "Hulf" <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am making and HTML email. I have 3 images to put in. Currently I have
>
> $body .="
>
>
>
>
>
>
>
>
>
> ";
>
>
> ideally I would like to have
>
> $myimage1 = "image1.jpg";
> $myimage2 = "image2.jpg";
Hulf <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am making and HTML email. I have 3 images to put in. Currently I have
>
> $body .="
>
>
>
>
>
>
>
>
>
> ";
>
>
> ideally I would like to have
>
> $myimage1 = "image1.jpg";
> $myimage2 = "image2.jpg";
> $myimage3 = "ima
On Mon, Mar 28, 2011 at 7:06 AM, Hulf <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am making and HTML email. I have 3 images to put in. Currently I have
>
> $body .="
>
>
>
>
>
>
>
>
>
> ";
>
>
> ideally I would like to have
>
> $myimage1 = "image1.jpg";
> $myimage2 = "ima
Hi,
I am making and HTML email. I have 3 images to put in. Currently I have
$body .="
";
ideally I would like to have
$myimage1 = "image1.jpg";
$myimage2 = "image2.jpg";
$myimage3 = "image3.jpg";
and put them into the HTML body variable. I have tried escaping them in
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