Resolved Need help with simple print

[apotd]

Okay so this is what I'm trying to do;

In a loop I want to print different divs that all have a different background image (the background image is stored in a table in a row called "thumbnail").

 print("
<div id=\"thumbnail\" style=\"background: url(\"images/portfolio/{$row['thumbnail']}\"); width: 240px; height: 160px;\">
				
title
				
</div>
");

But only the titles show up, not the background.

I'm totally not familiar with anything of PHP and this seems very simple, but I have no idea on what I do wrong.

Could someone maybe help me out?

[TFlan]

PHP Code:

print('<div id="thumbnail" style="background: url("images/portfolio/'.$row['thumbnail'].'"); width: 240px; height: 160px;">title</div>'); 


Use that (just cleaner), and could you post what the result is?

I have a feeling you may be forgetting the file extension

[jimutt]

I would write it like this:

PHP Code:

echo '<div id="thumbnail" style="background: url("images/portfolio/'.$row['thumbnail'].'"); width: 240px; height: 160px;\">title</div> '; 


It's by the way better practice to use echo instead of print as echo is slightly faster than print and there are therefore no reason to use print. Well if you don't need the return value of print in an expression for example.

EDIT: oops, didn't see you were quicker than me TFlan

[AndrewGSW]

BTW if this is in a loop then you are giving all your divs the same id; id's must be unique on the page.

[apotd]

Mhh I tried it, but it's still not displaying any background image.

My extension is included in the table at this moment, when I print (I'll use echo from now on) them just as images like this:

<img src=\"images/portfolio/{$row['thumbnail']}\" alt=\"{$row['title']}\" title=\"{$row['title']}\" />

they do show up. I'm really confused now.

Originally Posted by AndrewGSW

BTW if this is in a loop then you are giving all your divs the same id; id's must be unique on the page.

Yes I was already thinking that, but how do I do that? For example give it the name of some unique ID in the table or? Or just make a class of it or won't that work?

[AndrewGSW]

If the DIVs don't need IDs then you could just add the same class-name to each for styling. Otherwise you could number them sequentially:

PHP Code:

$i = 0;

echo '<div id="thumbnail"' . $i . 
$i += 1; 


Try clearing your browser's cache; and use the browser's Console - generally, right-click and Inspect Element - to see if the background image is there, but just not displaying for some reason.

[apotd]

Thanks for that code. When inspecting the element I'm getting this:

<div id="thumbnail" style="background: url(" images="" portfolio="" 1.png");="" width:="" 240px;="" height:="" 160px;"="">
				
title
				
</div>

The 1.png of course is the image but I don't understand the rest of it anymore.

[AndrewGSW]

The quotes " are all out-of-whack?! View the page, right-click to View Source, find and copy the relevant HTML; post it here. And re-post your echo statement as it is now.

Try this:

PHP Code:

echo '<div id="thumbnail" style="background: url("images/portfolio/{$row["thumbnail"]}"); width; 240px; height; 160px;">'; 


(it doesn't show errors in my editor )

[TFlan]

style="background: url(" images="" portfolio="" 1.png");=""

What the hell is that?

Please post your code for what assigns the value to $row["thumbnail"]

If you read that, you'll see that it's all wrong. It is trying to find an image at 'images="" portfolio="" 1.png' Which as you might of guessed is not a valid URL

[apotd]

Okay so I just removed the entire print I already had and simple replaced it with the echo posted here, and now it works fine! I'm still confused about why it have this weird result but at least I'm glad it's solved, thanks!