CODEWITHSUNDEEP
php
WebbieWorks
WebbieWorks

Reputation: 163

How to not not display html if variable isn't there?

I have a site that has different features but I do not want a feature to show if it does not have a variable assigned to it.

For example, my product has a featured list like this:

<h3>Specifications</h3>
<ul>
    <li class="element element-text first">
        <strong>Manufacturer: </strong><?php echo $manufactuer;?>
    </li>
    <li class="element element-text">
        <strong>Model #: </strong><?php echo $model;?>
    </li>
    <li class="element element-text">
        <strong>Previously Ran: </strong><?php echo $ran;?>
    </li>
    <li class="element element-text last">
</ul>

On the front end, I don't want the Model line displayed if that doesn't have a variable assigned to it.

Upvotes: 0

Views: 244

Answers (5)

Dmitry G
Dmitry G

Reputation: 93

Full code like this:

<h3>Specifications</h3>
        <ul>
            <?php if(isset($manufacturer) && !empty($manufacturer)): ?>
                <li class="element element-text first">
                    <strong>Manufacturer: </strong><?php echo $manufactuer;?>
                </li>
            <?php endif; ?>
            <?php if(isset($model) && !empty($model)): ?>
                <li class="element element-text">
                    <strong>Model #: </strong><?php echo $model;?>
                </li>
            <?php endif; ?>
            <?php if(isset($ran) && !empty($ran)): ?>
                <li class="element element-text">
                    <strong>Previously Ran: </strong><?php echo $ran;?>
                </li>
            <?php endif; ?>
            <li class="element element-text last"></li>
       </ul>

Upvotes: 0

jero
jero

Reputation: 41

<?php if (isset($model) && !empty($model)) : ?>
<li class="element element-text">
  <strong>Model #: </strong>
  <?php echo $model;?>
</li>
<?php endif; ?>

Upvotes: 1

superspartan999
superspartan999

Reputation: 291

You need to use JavaScript to write the HTML when you start the page. You probably want something like this (make sure you have a script called JQuery, you can find it online):

$.ajax({ url: '/my/site.php',
    datatype: "JSON", /* Personal preference */
    type: 'get',
    cache: false,
    success: function(data) {
        alert(output);
        if(data[1] != null){
            document.getElementById('ID').innerHTML = "<li>" + data[1] + "</li>";
        } else {
            console.log("nothing found");
        }
    }
});

You may need to adjust things a bit for your particular code writing, but this would be the approach I think you need. In general, dynamic elements (ones that are not static i.e. ones that respond to stuff outside the HTML) need to dealt with in JavaScript.

Upvotes: 1

Nagendra Rao
Nagendra Rao

Reputation: 7152

You can do,

<?php echo (!empty($model))?$model:'';?>

Read up on: empty, and ternary operator

Upvotes: 0

Shah Rukh
Shah Rukh

Reputation: 3156

you can do this 

  <?php if(isset($model)): ?>
          <li class="element element-text">
          <strong>Model #: </strong><?php echo $model;?>
        </li>
  <?php  endif; ?>

Upvotes: 0

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