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6 posts by 2 authors in: Forums > CMS Builder
Last Post: October 9, 2017

Is it possible to show the label for a category queried in the URL?

"Burgers" has a category num of "16", and is selected from a field called "meal_category".

Example of a category search for meals:

http://example.com/meals.php?meal_category=16

When the meal_category=16 is queried in the URL, I'd like to show the label for the category on the on page - as seen below:

Burgers

Thanks, Zicky

Hi Zick,

What happens when you use the :label pseudo on the detail page?

Something like:

<?php echo $your_sectionRecord['meal_category:label'] ?>

Jerry Kornbluth

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Hey Jerry,

The problem with something like <?php echo $your_sectionRecord['meal_category:label'] ?> is that this is a multi-section list page of menu items. When someone clicks on a link from the menu categories page, they arrive to a meals.php file which returns a listing of all meal items listed under the category as defined by the URL meals.php?menu_category=XX .  This works fine, but listed the category selected on the meals.php page based on the URL query produces results similar to this Example: Salads, Salads, Salads, Salads... and so on for every salad listed under the category "Salads". And if the user ever removed "?menu_category=XX"  from the URL query of meals.php, then they would see all the menu_category listed for all meals... Example: Salads, Salads, Salads, Salads, Burgers, Burgers, Burgers, Burgers, Steaks, Steaks, Steaks, Desserts, Desserts, Desserts... and so on for every meal items listed.

Hope this all makes since.

However, thanks for the suggestion... it got me thinking about a different approach to this problem, and I wrote the solution below which is working and accomplishes what I needed.

First thing I'm doing below is checking to see if the URL is lacking "?menu_category=XX", and if it is, then I provide a default solution as defined by $menu_pageRecord["title"]

<?php $url = $_SERVER['REQUEST_URI']; ?>
      <?php if ($url == '/meals.php'): ?>
         <?php echo htmlspecialchars($menu_pageRecord["title"]); ?>

ELSE if "?menu_category=XX" is defined, then load the "meal_category:label" as you suggested.

I also added a counter++ to break after the first occurrence of the meal_category to eliminate the repeated meal_category instance such as Salad, Salad, Salad, so it only produces one instance of Salad.

<?php else: ?>

             <?php $counter = 0; ?>
                <?php foreach ($menuRecords as $record): ?>
                   <?php $counter++;
                       if($counter > 1) {break;} 
                   ?>
                        <?php echo $record['meal_category:label'] ?>
                 <?php endforeach ?>

      <?php endif; ?>

Put this all together and here's the final solution.

<h2>    
    <?php $url = $_SERVER['REQUEST_URI']; ?>
      <?php if ($url == '/meals.php'): ?>
         <?php echo htmlspecialchars($menu_pageRecord["title"]); ?>
      <?php else: ?>

             <?php $counter = 0; ?>
                <?php foreach ($menuRecords as $record): ?>
                   <?php $counter++;
                       if($counter > 1) {break;} 
                   ?>
                        <?php echo $record['meal_category:label'] ?>
                 <?php endforeach ?>

      <?php endif; ?>
 <h2>

Thanks, Zicky

I'll check out your ajax post and see if there's anything I can suggest once I return from dinner.

Zicky

Thanks Zick

The first CMS Builder reference book is now available on-line!
Take advantage of a free 3 month trial subscription, only for CMSB users, at: http://www.thecmsbcookbook.com/trial.php