<?php if ($this->checkPosition('image')) : ?>
<?php
echo "<table class=\"remove-margin-t\" cellpadding=\"0\" cellspacing=\"0\" width=\"97%\" align=\"center\" border=\"0\" style=\"max-width:625px; border:1px solid;\" background=\"..\images";
?>
<?php
echo $this->renderPosition('image')
<?php
echo ".png\">";
?>
<?php endif; ?>
I am trying to figure out how to call the image properly. echo for image is called and has a specific name like 'pink','blue','green' etc. However, it depends on the position part...
This is what it is supposed to look like in html.
<table cellpadding="0" cellspacing="0" width="97%" align="center" border="0" style="max-width:625px; border:1px solid #CCC" background="http://localhost/images/[insert color name here].png" >
Here is the original php
<?php if ($this->checkPosition('color')) : ?>
<?php echo $this->renderPosition('color'); ?>
<?php endif; ?>
Any help would be appreciated. I am sure it must be a '\' or '"' issue.
Best,
Steven
To Jared:
Do you mean like this?
<?php if ($this->checkPosition('image')) : ?>
<?php
echo "<table class=\"remove-margin-t\" cellpadding=\"0\" cellspacing=\"0\" width=\"97%\" align=\"center\" border=\"0\" style=\"max-width:625px; border:1px solid;\" background=\"../images/";
echo $this->renderPosition('image')
echo ".png\">";
?>
<?php endif; ?>
background=\"../images/"echoor not toecho, this is the question).HEREDOCand a$colorvariable: codepad.org/akAfjKKl (If I'm not using a template technique.)