Syntax problem with html in PHP

Question

im getting a syntax error on this line of code and don't know the correct formatting for it.

This is the part i'm having problems with -

echo " 
   <ul>";  

foreach($photos as $photo) {  

           $farm              = $photo['farm'];  
           $server            = $photo['server'];  
           $photo_id          = $photo['id'];  
           $secret            = $photo['secret'];  
           $photo_title       = $photo['title'];  

<li><img src="http://farm'.$photo['farm'].'.static.flickr.com/'.$photo['server'].'/'.$photo['id'].'_'.$photo['secret'].'_t.jpg" alt="'.$photo['title'].'" ></li>

The problem is with that li tag. How can i format it properly?

and to close the foreach brace. Unless your loop is longer than you have copy/pasted — David Costa
– David Costa, Commented Apr 29, 2011 at 16:12
What's the point of assigning variables if you aren't going to use them? — No Results Found
– No Results Found, Commented Apr 29, 2011 at 16:15

Jason McCreary · Accepted Answer · 2011-04-29 16:11:20Z

2

Per the comments, you probably mean to have the following for your last line:

echo '<li><img src="http://farm'.$photo['farm'].'.static.flickr.com/'.$photo['server'].'/'.$photo['id'].'_'.$photo['secret'].'_t.jpg" alt="'.$photo['title'].'" ></li>';

answered Apr 29, 2011 at 16:11

Jason McCreary

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3 Comments

William Over a year ago

Also, don't forget to close the foreach loop } (this might be in OP's code, but omitted for brevity sake)

Jason McCreary Over a year ago

@William, I didn't want to make any assumptions as the OP only mentioned the line in question.

DIM3NSION Over a year ago

Thanks for that guys. cheers William for pointing out i needed to close my loop :)

Dan Breen · Accepted Answer · 2011-04-29 16:11:41Z

2

The <li> part should be quoted. Try:

echo " 
   <ul>";  

foreach($photos as $photo) {  

           $farm              = $photo['farm'];  
           $server            = $photo['server'];  
           $photo_id          = $photo['id'];  
           $secret            = $photo['secret'];  
           $photo_title       = $photo['title'];  

    echo '<li><img src="http://farm' . $photo['farm'] . 'static.flickr.com/' . $photo['server'] . '/' . $photo['id'] . '_' . $photo['secret'] . '_t.jpg" alt="' . $photo['title'] . '" ></li>';
}
echo '</ul>';

answered Apr 29, 2011 at 16:11

Dan Breen

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