Outputting the first image within a query

[Jon E]

Hi,

Could someone give me a helping hand please?

I have the below code:

<?php 
// Show the gallery widget if there are slideshow images attached to this press article
if(count($page->slideshow_images)) {
	echo '<div id="article_slideshow_widget">';
	$n = 0;
	echo "$page->images->first()->url;<img src='http://i.imgur.com/0heduk5.png' class='article_slideshow_widget_zoom'></div>";
	$n++;
	}
	echo "";
?>

I want to output the first image from the images field in the article_slideshow_widget div, but it is only giving me the filename currently (I removed the img tags to see what was being output as I was being given a broken image as the complete URL isn't there!)

Sure I'm close but would really be grateful for a pointer.

Thanks so much

Jon

[titanium]

Does $page->images even exist? The if-condition asks for $page->slideshow_images... 

[Jon E]

Hi,

It does, and I'm getting just the filename.jpg output... not the whole directory.

Thanks

Jon

[diogo]

Try to set the image url to a variable first:

$n = 0;
$image_url = $page->images->first()->url;
echo "{$image_url}<img src...

[Jon E]

Thanks so much

[Soma]

Or if inside a string and multiple chains use {}

echo "{$page->images->first()->url}...";

If only one chain it doesn't need it like

echo "$page->url";

[diogo]

Like Soma said. For more details see here http://php.net/manual/en/language.types.string.php#language.types.string.parsing