在每个图像中回声不同，并在数据库中的最后2行中进行一个行

Question

我的代码在单个图像中显示数据库中的所有值，我希望每个图像从最后2个值开始在数据库中仅显示一个值。是否有帮助？以下是我的代码

    <div class="col-xs-3 col-sm-3 col-md-3">
                <img src="grey.jpg">            
       <div class="top-left"> <p><?php $servername = "localhost";
           $username = "root";
           $password = "";
           $dbname = "software";
            // Create connection
           $conn = new mysqli($servername, $username, $password, $dbname);
           // Check connection
           if ($conn->connect_error) {
           die("Connection failed: " . $conn->connect_error);
            }
            $sql = "SELECT ID, description FROM test";
            $result = $conn->query($sql);
             //if ($result->num_rows == 1) {
           // output data of each row
           while($row = $result->fetch_assoc()) {
            echo " ". $row["description"]. "<br>";
           }
         //} else {
          // echo "0 results";
         //} ?> </p></div>
   </div>
    <div class="col-xs-3 col-sm-3 col-md-3">
     <img src="imagesb.jpg">
    </div>

Answer 1

您可以使用

之类的东西从数据库中获得最后两行

SELECT ID, description FROM test ORDER BY ID DESC LIMIT 2

(或您要订购的任何列(。

至于您想要的输出方式，尚不清楚您在问什么，但听起来您想在DIV标签外移动时循环。类似...

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "software";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT ID, description FROM test ORDER BY ID DESC LIMIT 2"; // Last 2 rows
$result = $conn->query($sql);
while($row = $result->fetch_assoc()) : ?>
    <div class="col-xs-3 col-sm-3 col-md-3">
        <img src="grey.jpg">
        <div class="top-left">
            <p><?php echo htmlspecialchars($row["description"], ENT_QUOTES); ?></p>
        </div>
    </div>
    <div class="col-xs-3 col-sm-3 col-md-3">
        <img src="imagesb.jpg">
    </div>
<?php endwhile;

请记住，每当您输出任何变量/DB内容时，您需要逃脱它(在这种情况下为HTMLSpecialChars(，否则您的页面可能会被黑客入侵。这是我发现的一个随机链接，涵盖了基础知识https://www.inanimatt.com/php-actput-escaping.html