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I have a html code:

  <div class="col-lg-4 col-sm-6" id="filterDiv">
                            <a href="<?php echo $result['company_link']; ?>" class="feature feature-4 feature-center animated" data-animate="fadeInUp" data-delay=".3">
                                <div class="feature-img">
                                    <img src="images/products/a.jpg" alt="product">
                                </div>
                                <h5 class="title title-sm title-dark"><?php echo $result['company_name']; ?></h5>
                                <span id="myCategory">Category: </span> <?php echo $result['company_category']; ?> ||
                                <span id="myLocation">Location: </span> <?php echo $result['company_location']; ?>
                            </a>
                        </div>

I want to display this code in PHP. Can anybody Help me out in this? Thanks in advance.

I have tried this code but it didn't work:

echo "<div class='nk-block nk-block-features-4 mgb-m30'>";
 echo  " <div class='row'>";
 echo "<div class=\"col-lg-4 col-sm-6\" id=\"filterDiv\">";

Parse error:

syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:\xampp\htdocs\asaworkholding\ajax\category-search.php on line 31
Nick
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1 Answers1

-3

Syntax Error, Please use this code

  echo "<div class='nk-block nk-block-features-4 mgb-m30'>";
     echo  " <div class='row'>";
     echo "<div class='col-lg-4 col-sm-6' id='filterDiv'>";

Hope this helps you

Balvinder Singh
  • 310
  • 2
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