Display Specific Image Based On SQL Query Result

Question

I have a page that displays products based on a previous user query. The search results contain sets of metadata, most importantly a "product rating" on a 1-10 scale.

Here's the code that pulls the product rating metadata

<?php echo get_post_meta( get_the_ID(), 'product_rating', true ); ?>

This pulls the 'rating' metadata from our dB and simply displays the rating as a numeral (1, 1.5, 2, 2.5 ect).

I'm looking to take the value of the 'product_rating' and set an image to display for each value. A rating of 1 will display one image, while a rating of 2 display a different image ect ect.

The current code I have in place substitutes an image for the value, but the image is displaying multiple times (up to 10 times for a single value). My code is below, using a rating value of "4" as an example.

<?php $rating = get_post_meta( get_the_ID(), 'product_rating', true ); ?>

<?php For($i=4; $i <= $rating; $i++){
echo "<img src='http://www.domain.com/lpimage/lock50.png' />";
}
?>

Is my code close to where I need to be, or am I completely off base for the function I want?

Are your ratings real numbers, i.e. you could have a 1.234, or are they guaranteed to be in steps of 0.5? — Bob Brown
– Bob Brown, Commented Jul 26, 2014 at 1:54
And also, how many images do you expect to use, and what is the maximum rating? Is the minimum zero, or can it become negative? — Bob Brown
– Bob Brown, Commented Jul 26, 2014 at 1:59
Ratings are in steps of .5, with 0 being the lowest. I'm going to use around 20 images, which will simply be stars that are correspondingly filled to match the product rating. A rating of 4 will have 4 gold stars, with 6 empty stars. — mn8809
– mn8809, Commented Jul 26, 2014 at 2:14

Brian DeMilia · Accepted Answer · 2014-07-26 01:59:45Z

1

Assuming your rating variable is correctly defined I think you can handle the conditional display of various images as follows:

<?php if ($rating == 1) echo '<img src="urlofimage1.jpg"></img>';
  elseif ($rating == 2) echo '<img src="urlofimage2.jpg"></img>';
  elseif ($rating == 3) echo '<img src="urlofimage3.jpg"></img>';
  elseif ($rating == 4) echo '<img src="urlofimage4.jpg"></img>';
?>

You use the double equal sign when comparing a value, you only use the single equal sign (=) when setting a variable's value.

Use >= and <= for ranges, ie. 1.3 or 2.7

edited Jul 26, 2014 at 1:59

answered Jul 26, 2014 at 1:53

Brian DeMilia

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1 Comment

mn8809 Over a year ago

Absolutely perfect, I really appreciate the time and extra info!

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