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 .=
table
tr
tdimg src=\$myimage1\/td
/tr
tr
tdimg src=\$myimage2\/td
/tr
tr
tdimg src=\$myimage3\/td
/tr
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 .=
table
tr
tdimg src=\image1.jpg\/td
/tr
tr
td/td
/tr
/table
;
ideally I would like to have
$myimage1 = image1.jpg;
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 .=
table
tr
tdimg src=\image1.jpg\/td
/tr
tr
td/td
/tr
/table
;
ideally I would like to have
$myimage1 = image1.jpg;
Hi,
I am making and HTML email. I have 3 images to put in. Currently I have
$body .=
table
tr
tdimg src=\image1.jpg\/td
/tr
tr
td/td
/tr
/table
;
ideally I would like to have
$myimage1 = image1.jpg;
$myimage2 = image2.jpg;
$myimage3 = image3.jpg;
and put them into the HTML
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 .=
table
tr
tdimg src=\image1.jpg\/td
/tr
tr
td/td
/tr
/table
;
ideally I would like to have
$myimage1 =
Hulf [EMAIL PROTECTED] wrote:
Hi,
I am making and HTML email. I have 3 images to put in. Currently I have
$body .=
table
tr
tdimg src=\image1.jpg\/td
/tr
tr
td/td
/tr
/table
;
ideally I would like to have
$myimage1 = image1.jpg;
$myimage2 =
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