Php Image variable

Question

I am trying to assign an image src to a variable, that doesn't seem to work.
I am pretty sure I am missing a " or a ' but I have just about tried every combination now.
The second issue I have is that I want to use two variables and concatenate doesn't seem to work.

see below: $item.$image

Any help would be appreciated.

<li class="first">                                    
<?php
$image =  "<img src="<?php echo base_url() . "assets/" . "menuicon.png" />

if (!empty($pcategory)) 
{
   foreach ($pcategory as $key => $item) 
   {
      echo "<li><a href='" . site_url() . "cat/$key'>$item.$image</a></li>"; 
   } 
}
?>
</li>

so does it work as $image = "<img src=" . <?php echo base_url() . "assets/" . "menuicon.png" /> — Angst
– Angst, Commented Feb 15, 2014 at 13:37
If you have a problem with quotation marks, then why don't you simply look at the finished html markup you output? — arkascha
– arkascha, Commented Feb 15, 2014 at 13:37
arkascha, the page just goes blank and doesn't load otherwise I would. — Johan Fourie
– Johan Fourie, Commented Feb 15, 2014 at 13:39

Fabio · Accepted Answer · 2014-02-15 14:00:10Z

2

You are opening php tag twice, one inside the other, you should instead concatenate them as follow

<?php
$image = '<img src="'. base_url() . 'assets/menuicon.png" />';
                   //^ here you need to concatenate
if (!empty($pcategory)) {
    foreach ($pcategory as $key => $item) {
       echo "<li><a href='" . site_url() . "cat/$key'>$item.$image</a></li>"; 
    } 
}

?>

edited Feb 15, 2014 at 14:00

answered Feb 15, 2014 at 13:37

Fabio

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1 Comment

James Over a year ago

$image = "<img src=". base_url() . "assets/" . "menuicon.png />" = PHP Parse error: syntax error, unexpected end of file

Hett · Accepted Answer · 2014-02-15 13:37:33Z

1

Try this

$image =  '<img src="' . base_url() . '"assets/enuicon.png" />';
if (!empty($pcategory)) 
{
    $site_url = site_url();
    foreach ($pcategory as $key => $item) {
        echo "<li><a href='{$site_url}cat/{$key}'>{$item}{$image}</a></li>"; 
    } 
}

answered Feb 15, 2014 at 13:37

Hett

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Comments

Johan Fourie · Accepted Answer · 2014-02-15 13:47:41Z

0

Thank you Hett and Fabio. Both of you pointed me in the right direction but it didn't work.

Here is the right solution:

<li class="first">                                    
       <?php
$image =  '<img src="' . base_url() . 'assets/menuicon.png" />';

if (!empty($pcategory)) {
 foreach ($pcategory as $key => $item) {
  echo "<li><a href='" . site_url() . "cat/$key'>$item.$image</a></li>"; 
                                    } 
                                }

                                ?>
                          </li>

I appreciate your help, wouldn't have figured it out without you guys. Thanks!

answered Feb 15, 2014 at 13:47

Johan Fourie

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James · Accepted Answer · 2014-02-15 13:53:30Z

0

You open the PHP tag twice in a row:

<?php // HERE
$image =  "<img src="<?php echo base_url() . "assets/" . "menuicon.png" /> // HERE

Looks like you've been trying to do this in HTML then switched to it all being in PHP (at a guess).

You should always check your PHP error log file, as it's invaluable as a coder - However, it wont produce any error in this case as PHP will simply think you're stating the image is at <?php echo base_url() . "assets/" . "menuicon.png, which of course it wont be.

You should also be checking (among other things, like your browser source code/output)) your browser page info. This shows you media and images, and you would have seen in the media tab that the link location to this image was <?php echo base_url() . "assets/" . "menuicon.png, and would have lead you to the issue.

This is the code you need:

<li class="first">                                    
  <?php

    $image =  "<img src='".base_url() . "assets/menuicon.png' />";

    if (!empty($pcategory))
      {
        foreach ($pcategory as $key => $item)
          {
            echo "<li><a href='" . site_url() . "cat/$key'>$item.$image</a></li>"; 
          }
       }
  ?>
</li>

You don't need to break in and out of the $image code as much as you did.
i.e. assets and /menuicon.png are both HTML and not PHP, so no need to break in/out to PHP.

edited Feb 15, 2014 at 13:53

answered Feb 15, 2014 at 13:47

James

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1 Comment

Johan Fourie Over a year ago

Thank you man. I really appreciate your help. Will keep it in mind in the future.

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