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I have database which contains a details of company in a table.I also have the php code in which i can display the searched query data from database.everything works fine, the problem is i want to display all the searched query data in their respective pages,once again am repeating i want to display the query results in a dynamic page(the page is solely responsible for the searched query).

php and html code

    <body>
        <div class="heading">
            <div class="logotext">
            <p style="margin:0px"><a href="../index.php" style="color:black">Clore</a></p>
            </div>
                <div class="searchbox">
                    <div class="SearchBar">

                        <form action="search.php" method="GET">

                            <input style="box-shadow: 0 2px 2px rgba(0,0,0,0.19), 0 2px 5px rgba(0,0,0,0.23);   "  list="myCompanies" id="suggest" type="text" name="q" placeholder="Start typing keywords...." autocomplete="off"><datalist id="myCompanies">
</datalist>
                                <div class="SearchBut">
                                <input style="box-shadow: 0 2px 2px rgba(0,0,0,0.19), 0 2px 5px rgba(0,0,0,0.23);    " type="submit" value="Search">
                                </div>
                        </form>
                    </div>
                </div>
             </div>
        <div>
            <?php
                $conn = mysqli_connect("mysql.hostinger.in","u402378368_clore","clore.cf","u402313368_clore") or die("could not search");
                $set=$_GET['q'];
                if($set){
                $show=("SELECT * FROM table1 where name LIKE '%$set%' OR about LIKE '%$set%' OR founder LIKE '%$set%' OR ceo LIKE '%$set%' OR headquaters LIKE '%$set%' OR established LIKE '%$set%' ") or die("could not search");
                $result = mysqli_query($conn,$show);
                $num_rows = mysqli_num_rows($result);?>
                <div class="nrezdez"><?php
                if($num_rows == 0){
                    echo "No Results Found For * ".$set." * "."<br>";

                }
                else if($num_rows ==1){
                    echo "Found ".$num_rows." Result <br>";
                    echo "Showing Result For * ".$set." *"."<br>";
                }
                else if($num_rows > 1){
                    echo "Found ".$num_rows." Results <br>";
                    echo "Showing Results For * ".$set." *"."<br>";
                }
                ?></div><?php
                while($rows = mysqli_fetch_array($result)){
                ?>
                <div class="redez"><?php
                echo "<strong>Name: </strong><a href=".$rows['link'].">".$rows['name']."</a>"."<br> <br>";
                echo "<strong>About: </strong>".$rows['about']."<br>";
                echo "<strong>Founder: </strong>".$rows['founder']."<br>";
                echo "<strong>CEO: </strong>".$rows['ceo']."<br>";
                echo "<strong>Revenue: </strong>".$rows['revenue']."<br>";
                echo "<strong>Headquaters: </strong>".$rows['headquaters']."<br>";
                echo "<strong>Estd: </strong>".$rows['established']."<br>";

                ?>
                </div>
                <?php
                        }
                     }  
                ?>
            </div>
    </body>

please help me how can i display results in their respective pages. why i need this help because, google will index web pages, but in my case i am displaying data in a single page in which my website will not be indexed by google, but if i display data of the searched query in their respective page google will be indexing my page. thank you

Isaac Bennetch
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