use Img source with php variable

Question

I want to use some thumbshots generated images inside a WP loop, and the code looks like this:

if (!$thumbnail && wpbdp_get_option('use-default-picture'))
    $thumbnail = 'http://images.thumbshots.com/image.aspx?cid=xxxxxxxxxx&v=1&w=283&url=URLHERE';

URLHERE is generated by this:

How to replace the URLHERE thing with the output (outputs and URL) of:

<?php echo wpbusdirman_the_listing_meta('single'); ?>

I am very bad at php.

Thanks,

may need to assign the output to a variable, or at least figure out what that is doing/returning. $output=wpbusdirman_the_listing_meta('single'); echo $output; — Dave
– Dave, Commented Sep 28, 2012 at 18:36

esqew · Accepted Answer · 2012-09-28 20:00:31Z

1

$thumbnail = 'http://images.thumbshots.com/image.aspx?cid=HIPFHapOLHw%3d&v=1&w=283&url='.urlencode(wpbusdirman_the_listing_meta('single'));

edited Sep 28, 2012 at 20:00

answered Sep 28, 2012 at 19:22

esqew

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1 Comment

dudewad Over a year ago

It looks like a typo- he put a "$" on the urlencode. Try removing it?

dudewad · Accepted Answer · 2012-09-28 18:44:53Z

0

URL encode the returned value!! You're putting a URL into a URL- if you don't URLEncode it, you're asking for more than a little trouble. I believe it's as simple as:

<?php echo urlencode(wpbusdirman_the_listing_meta('single')); ?>

:)

answered Sep 28, 2012 at 18:44

dudewad

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4 Comments

robertgombos Over a year ago

Thanks a lot but how do I put that in my code: if (!$thumbnail && wpbdp_get_option('use-default-picture')) $thumbnail = 'images.thumbshots.com/…'; - instead oh the yahoo thingy.

dudewad Over a year ago

You could literally insert the code snippet I posted where your "URLHERE" is, or, alternatively (and much cleaner) you could assign it to a variable and insert the variable: <? $imageURL = urlencode(wpbusdirman_the_listing_meta('single')); And then $thumbnail = 'images.thumbshots.com/…';

robertgombos Over a year ago

This is what I did and somehow it doesn't work.. if (!$thumbnail && wpbdp_get_option('use-default-picture')) $imageURL = urlencode(wpbusdirman_the_listing_meta('single')); $thumbnail = 'images.thumbshots.com/…';

dudewad Over a year ago

If you're putting a variable inside a string and expect it to be read as a variable, the string must use double quotes. using single quotes will insert the literal value (in this case $imageURL) . You want to use "images.......$imageURL" instead of 'images....$imageURL'.

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