如何使用PHP / MySQL显示具有相同样式的多个数据库对象？ [关闭]

I have a page with multiple DIVs that all contain different content but all have same styling. I am using PHP on most of my page and would like to create maybe a function where it goes through a table in a database and displays all of them. Here is the DIV I have:

<div class="location-container">
    <a href="http://www.google.com/"><img src="img/img.jpg" /></a>
    <div class="info">
        <a href="http://www.google.com/"><p class="title">Google</p></a>
        <p class="location">Location</p>
        <p class="location">Location</p>
        <p class="phone">Phone</p>
    </div>
</div>

Like I said, I have about 10 of these one after another. I would like to avoid this and maybe have PHP go through my database table and get all of the info from there.

Once you have obtained the $object from the database, you can use PHP to do a simple foreach loop like this:

    <?php

    foreach( $objects as $object ) {

     echo '<div class="location-container">
        <a href="' . $object['link'] . '"><img src="' . $object['imageLink'] .'" /></a>
        <div class="info">
            <a href="http://www.google.com/"><p class="title">' . $object['title'] .'</p></a>
            <p class="location">' . $object['address_1'] .'</p>
            <p class="location">' . $object['address_2'] .'</p>
            <p class="phone">' . $object['phone'] .'</p>
        </div>
      </div>';
    } 

    ?>

You can loop the results that you want, something like:

<?php 
      $servername = "localhost"; 
      $username = "username"; 
      $password = "password"; 
      $dbname = "myDB";

// Create connection 
$conn = mysqli_connect($servername, $username, $password, $dbname); 

//Check connection 
if (!$conn) {
        die("Connection failed: " . mysqli_connect_error()); }



  $sql = "SELECT id, firstname, lastname FROM MyGuests";

$result = mysqli_query($conn, $sql);

    if (mysqli_num_rows($result) > 0) {
    // output data of each row
    echo "div class='location-container'>";
    echo "<a href='http://www.google.com/'><img src='img/img.jpg' /></a>";
        while($row = mysqli_fetch_assoc($result)) {

        echo "<div class='info'>";
        echo "<a href='http://www.google.com/'><p class='title'>" . $row["title"]. "</p></a>";
        echo "<p class='location'>" . $row["location"]. "</p>";
        echo "<p class='location'>" . $row["location"]. "</p>";
        echo "<p class='phone'>" . $row["phone"]. "</p>";
        echo "</div>";
        } 
</div>

    } 
    else {
            echo "0 results"; }

mysqli_close($conn); ?>