Tuesday, 15 July 2014

PHP If Not syntax -



PHP If Not syntax -

trying utilize php if not syntax getting parse error. help in finding error in code?

<?php $file = get_field('fl_brochure' ); if(trim($file) != "") { echo '<a target=_blank href="' . $file['url' ] . '">' . '<img src="http://www.domain.com/images/brochure.png" alt="brochure" />' . brochure . '</a>' } ?>

your missing ; (semi-colon) @ end of echo statement.

$file = get_field('fl_brochure' ); if(trim($file) != ""){ echo '<a target=_blank href="' . $file['url' ] . '">' . '<img src="http://www.domain.com/images/brochure.png" alt="brochure" />' . brochure . '</a>'; }

this assuming brochure working constant.

php

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