Php Variable auto increasing number

Question

I am not sure exactly how to word it, but I am trying to input a caption along with photograph. The for() autoincreases and coincides with the field name.

Here is the code I have so far:

for ($i=1;$i<=21;$i++) {
        if (file_exists(ABSPATH.'wp-content/uploads/fsrep/houses/additional/small/'.$ListingDetails->listing_id.'-'.$i.'.jpg')) {
            if (($ListingDetails->listing_caption."_".$i) != '' ) {$ListingCaption = ($ListingDetails->listing_caption.'_'.$i);} else{$ListingCaption = 'View Slideshow';}
            $AdditionalImages .= '<td align="center" valign="center"><a href="'.get_option('home').'/wp-content/uploads/fsrep/houses/additional/large/'.$ListingDetails->listing_id.'-'.$i.'.jpg" title="'.$ListingCaption.'" rel="lightbox[slideshow]" onmouseover="document.getElementById(\'fsrep-main-image-img\').src=\''.get_option('home').'/wp-content/uploads/fsrep/houses/additional/'.$ListingDetails->listing_id.'-'.$i.'.jpg\'; document.getElementById(\'fsrep-main-image-a\').rel=\'lightbox[slideshow]\'" onmouseout="document.getElementById(\'fsrep-main-image-img\').src=\''.get_option('home').'/wp-content/uploads/fsrep/houses/'.$ListingDetails->listing_id.'.jpg\'; document.getElementById(\'fsrep-main-image-a\')"><img src="'.get_option('home').'/wp-content/uploads/fsrep/houses/additional/small/'.$ListingDetails->listing_id.'-'.$i.'.jpg" class="full" /></a></td>';
            if ($i == 7 || $i == 14|| $i == 21) {
                $AdditionalImages .= '</tr><tr>';
            }
        }
    }

I am tying to get it to output the $listing_caption correctly, but it isn't reading the field name as for example $ListingDetails->listing_caption_1 instead I think it is reading it as $ListingDetails->listing_caption, but I am not sure.

The fields in the mysql database are for example:

listing_caption_1, listing_caption_2, listing_caption_3, etc.

what is $ListingDetails? is it row from mysql_fetch_row? — ariefbayu
– ariefbayu, Commented Oct 26, 2011 at 2:02
This in fact has nothing at all to do with photos, or auto-increasing numbers, or databases. It's just a question about how to use a variable with a dynamic name. Please narrow down your question to a minimal issue next time! — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Oct 26, 2011 at 2:30
@TomalakGeret'kal To quote the first thing I wrote "I am not sure exactly how to word it". Thank you for your contribution to my question. — Chris Allington
– Chris Allington, Commented Oct 26, 2011 at 23:38

vzwick · Accepted Answer · 2011-10-26 01:59:47Z

3

Try:

$capt_num = "listing_caption".'_'.$i;
$ListingCaption = $ListingDetails->$capt_num;
...

answered Oct 26, 2011 at 1:59

vzwick

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Josh · Accepted Answer · 2011-10-26 02:02:27Z

2

I'm not sure what your $ListingDetails is, but if it represents a database, and you populate it with say, mysql_fetch_assoc(), you can turn:

$ListingCaption = ($ListingDetails->listing_caption.'_'.$i);

into:

$ListingCaption = $ListingDetails['listing_caption_'.$i];

Assuming that's where you're having trouble.

answered Oct 26, 2011 at 2:02

Josh

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