How do I validate a PHP integer within a variable? - php

Question

I have integrated Yelp reviews into my directory site with each venue that has a Yelp ID returning the number of reviews and overall score.
Following a successful MySQL query for all venue details, I output the results of the database formatted for the user. The Yelp element is:
while ($searchresults = mysql_fetch_array($sql_result)) {
if ($yelpID = $searchresults['yelpID']) {
require('yelp.php');
if ( $numreviews > 0 ) {
$yelp = '<img src="'.$ratingimg.'" border="0" /> Read '.$numreviews.' reviews on <img src="graphics/yelp_logo_50x25.png" border="0" /><br />';
} else {
$yelp = '';
}
} //END if ($yelpID = $searchresults['yelpID']) {
} //END while ($searchresults = mysql_fetch_array($sql_result)) {
The yelp.php file returns:
$yrating = $result->rating;
$numreviews = $result->review_count;
$ratingimg = $result->rating_img_url;
$url = $result->url;
If a venue has a Yelp ID and one or more reviews then the output displays correctly, but if the venue has no Yelp ID or zero reviews then it displays the Yelp review number of the previous venue.
I've checked the $numreviews variable type and it's an integer.
So far I've tried multiple variations of the "if ( $numreviews > 0 )" statement such as testing it against >=1, !$numreviews etc., also converting the integer to a string and comparing it against other strings.
There are no errors and printing all of the variables returned gives the correct number of reviews for each property with venues having no ID or no reviews returning nothing (as opposed to zero). I've also compared it directly against $result->review_count with the same problem.
Is there a better way to make the comparison or better format of variable to work with to get the correct result?
EDIT:
The statement if ($yelpID = $searchresults['yelpID']) { is not operating as it should. It is identical to other statements in the file, validating row contents which work correctly for their given variable, e.g. $fbID = $searchresults['fbID'] etc.
When I changed require('yelp.php'); to require_once('yelp.php'); all of the venue outputs changed to showing only the first iterated result. Looking through the venues outputted, the error occurs on the first venue after a successful result which makes me think there is a pervasive piece of code in the yelp.php file, causing if ($yelpID = $searchresults['yelpID']) { to be ignored until a positive result is found (a yelpID in the db), i.e. each venue is correctly displayed with a yelp number of reviews until a blank venue is encountered. The preceding venues' number of reviews is then displayed and this continues for each blank venue until a venue is found with a yelpID when it shows the correct number again. The error reoccurs on the next venue output with no yelpID and so on.
Sample erroneous output: (line 1 is var_dump)
string(23) "bayview-hotel-bushmills"
Bayview Hotel
Read 3 reviews on yelp
Benedicts
Read 3 reviews on yelp (note no var_dump output, this link contains the url for the Bayview Hotel entry above)
string(31) "bushmills-inn-hotel-bushmills-2"
Bushmills Inn Hotel
Read 7 reviews on yelp
I suspect this would be a new question rather than clutter/confuse this one further?
END OF EDIT
Note: I'm aware of the need to upgrade to mysqli but I have thousands of lines of legacy code to update. For now I'm working on functionality before reviewing the code for best practice.

Answer 1

Since the yelp.php is sort of a blackbox; the best explanation for this behavior would be that it only set's those variables if it finds a match. Updating your code to this should fix that:
unset($yrating, $numreviews, $ratingimg, $url);
require('yelp.php');
I also noticed this peculiar if-statement, do you realize that's an assignment or is this a copy/paste error? If you want to test (that's what if is for)
if ($yelpID == $searchresults['yelpID']) {