How to write php inside return html?

Question

I'm a real newbie to php and I just cant figure this out! I want different images to appear depending on what rating value is returned from the database.How do I do that?! This is the code:

    return '
        <div class="comment">
            <div class="name">'.$link_open.$d['name'].$link_close.'</div>
            <div class="date" title="Added at '.date('H:i \o\n d M Y',$d['dt']).'">'.date('d M Y',$d['dt']).'</div>
            <div class="rating">'.$d['rating'].'</div>
            <p>'.$d['body'].'</p>
        </div>
    ';

So if the value is 5 I want the image 5stars.png to show up. I thought maybe you could write something like:

<img src="img/<?php echo $row['rating']; ?>stars.png">

But that won't work inside the return statement. What do I have to write to be able to write php inside the ? Is it even possible?

can you show me how you mean? lol

Jenetic
– Jenetic

2015-10-16 14:54:45 +00:00
Commented Oct 16, 2015 at 14:54 — Jenetic
– Jenetic, Commented Oct 16, 2015 at 14:54

elixenide · Accepted Answer · 2015-10-16 15:04:07Z

3

You would do it like this:

return '
    <div class="comment">
        <div class="name">'.$link_open.$d['name'].$link_close.'</div>
        <div class="date" title="Added at '.date('H:i \o\n d M Y',$d['dt']).'">'.date('d M Y',$d['dt']).'</div>
        <div class="rating">'.$d['rating'].'</div>
        <p><img src="img/' . $d['rating'] . 'stars.png"></p>
    </div>
';

You can't use echo inside a string, as you have noticed. Instead, you use the . operator to concatenate strings.

Please note: this is a pretty basic question and not really suited for this site. I would recommend that you find a good tutorial on PHP basics and start there.

edited Oct 16, 2015 at 15:04

answered Oct 16, 2015 at 14:53

elixenide

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2 Comments

Jenetic Over a year ago

I'm sorry, I've searched for the answer but haven't find it. This seemed so complicated so I thought my best shot at getting this right was to ask here. When I try that code, it wont read the rating value at all and just try to show "stars.png".

elixenide Over a year ago

@Jenetic Well, you refer to $row in your question, but your return value originally had $d. You haven't posted enough info to make clear what you need, but I suspect you need to change $row['rating'] to $d['rating'].

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