use Img source with php variable? - php

Question

i have the following code in my detailansicht.php:
<div class="file_description_box">
<b>Beschreibung:</b><br /><br />
<?php
if(!empty($beschreibung))
echo '<div align="center">';
echo '<img src="$coverlink" alt="Cover">';
echo '</div><br />';
echo format_content($beschreibung);
**else**
echo '<i>Keine Beschreibung vorhanden</i>';
?>
</div>
but i think there has to be a mistake in the img tag. everytime i trie to open the page, he shows me an error: "Parse error: syntax error, unexpected T_ELSE in bla/bla/include/detailansicht.php on line 123" (Line 123 is the else marked bold).
I tried several methods but i always get this error.
It would be nice, if someone could help me with this.
-sTarbuZz

Answer 1

You are missing some curly brackets and your PHP variable wasn't embedded property. Your code should look like this:
<div class="file_description_box">
<b>Beschreibung:</b><br /><br />
<?php
if(!empty($beschreibung)){
echo '<div align="center">';
echo '<img src="'.$coverlink.'" alt="Cover">';
echo '</div><br />';
echo format_content($beschreibung);
}else{
echo '<i>Keine Beschreibung vorhanden</i>';
}
?>
</div>
Just on a side note it really doesn't matter what you use PHP for, if there was a fault with the image tag and it wasn't being displayed properly it would generally be a HTML problem assuming you have outputted it correctly.
In this case it was a PHP problem because there were a few errors in the code, it hasn't really got anything to do with the image tag.

Answer 2

<?php
if(!empty($beschreibung)) {
echo "<div align=\"center\">";
echo "<img src=\"$coverlink\" alt=\"Cover\">";
echo "</div><br />";
echo format_content($beschreibung);
} else {
echo "<i>Keine Beschreibung vorhanden</i>";
}
?>

Answer 3

You are missing curly braces:
if () {
...
} else {
...
}
so your PHP is not syntactically correct and will not be parsed by PHP hypertext pre-processor.

Answer 4

Try using curly braces:
<?php
if(!empty($beschreibung)) {
echo '<div align="center">';
echo '<img src="$coverlink" alt="Cover">';
echo '</div><br />';
echo format_content($beschreibung);
} else {
echo '<i>Keine Beschreibung vorhanden</i>';
}
?>

Answer 5

use
echo '<img src="', $coverlink', " alt="Cover">';
PHP variables inside in single quote won't be evaluated

Answer 6

Yep, you're missing the curly braces. Simply formatting the code nicely with tabs won't make it a block.
Also you missed the ending of the img tag (/>), but that's unrelated to your question.

Answer 7

thanks for all your answers...
I'm trying Kris' method now...
And the thing with the brackets... i don't know how this could happen... i didn't write the original script (i just added the part with the cover) and i didn't recognize that there were missing brackets, because $beschreibung is never empty and i think php ignores the if and else if the brackets aare missing...

Answer 8

I would go for the alternative syntax which i feel is easier on the eyes intermixed with html:
<? if(!empty($beschreibung)) : ?>
<div align="center">
<img src="<?= $coverlink; ?>" alt="Cover">
</div><br />
<?= format_content($beschreibung); ?>
<? else : ?>
<i>Keine Beschreibung vorhanden</i>
<? endif ; ?>
PS: i am not advocating putting logic inside the markup, just noting it can be done.
edit: fixed syntax error (; after endif instead of :)