Question

我是编程的新手，我在修复这部分代码时遇到了一些麻烦，因为我的慈善网站显示了不同家庭的案例，如果有人能帮助我，我将非常感激。这是我的代码..

    <div id="Families"><div id="arrow"></div>
    <?php
$server = "localhost";
$user = "root";
$pass = "";
$dbname = "help_someone";

//Creating connection for mysqli

$conn = new mysqli($server, $user, $pass, $dbname);

//Checking connection

if($conn->connect_error){
 die("Connection failed:" . $conn->connect_error);
}
$sql = "SELECT * FROM familjet";
$result = mysqli_query($conn, $sql)or die(mysqli_error());
$count = mysqli_num_rows($result);
while($row = mysqli_fetch_array($result)) {  ?>

<div class="cases"><?php echo '<img src="data:image/jpeg;base64,' . base64_encode($row['image']) . '" width="350" height="200">'?>               
<button type="button" class="show-modal1">SHIKO VIDEON</button>
<script src="http://code.jquery.com/jquery-1.11.3.min.js"></script>
<script src="js/jquery.showYtVideo.js"></script>
<script>
 jQuery(document).ready(function ($) {
            $('.show-modal1').on('click', function () {
                $.showYtVideo({
                    videoId: '<?php echo $row["videoID"];?>'
                });
            });
        });  
</script>
   <p><?php echo $row["pershkrimi"];?></p>
       </div><!--end cases-->
   <?php }
?>  
  </div><!--end Families-->

问题是，当我点击按钮时，它只显示最后一个videoID，因此我只获得了我在数据库中的最后一条记录的视频。其余的代码工作得很好。有人可以建议我该怎么办？

Answer 1

在您的代码中，您正在循环脚本元素。可能你的循环中的最后一个脚本回显是正在执行的脚本。

将脚本放在<body>标记的底部，并将其设为静态内容。

<body>
 <?php
  //php loop here ...
 ?>
 <script src="http://code.jquery.com/jquery-1.11.3.min.js"></script>
 <script src="js/jquery.showYtVideo.js"></script>
</body>

对于下一个修复，我们将使用数据属性，每次单击带有.show-modal类的按钮时，我们都会获得data-video-id属性并使用它来显示视频元素;

<body>
 <div id="Families">
  <div id="arrow"></div>
  <?php
   $server = "localhost";
   $user = "root";
   $pass = "";
   $dbname = "help_someone";

   //Creating connection for mysqli
   $conn = new mysqli($server, $user, $pass, $dbname);

   //Checking connection
   if($conn->connect_error){
    die("Connection failed:" . $conn->connect_error);
   }
   $sql = "SELECT * FROM familjet";
   $result = mysqli_query($conn, $sql)or die(mysqli_error());
   $count = mysqli_num_rows($result);

   while($row = mysqli_fetch_array($result)) {
    echo"<div class='cases'>";
    echo"<img src='data:image/jpeg;base64,".base64_encode($row['image'])."' width='350' height='200'>";
    echo"<button type='button' class='show-modal' data-video-id='".$row['video_id']."'>SHIKO VIDEON</button>"; //use data-attribute, data-video-id
    echo"<p>".$row["pershkrimi"]."</p>";
    echo"</div>"; //close cases
   }
  ?>
  </div><!--end Families-->

  <script src="http://code.jquery.com/jquery-1.11.3.min.js"></script>
  <script src="js/jquery.showYtVideo.js"></script>

  <script>
   $(document).on("click",".show-modal",function(){
    //get video from data-attribute
    var videoId = $(this).attr("data-video-id");
    alert("Your video ID:"+videoId);

    //show video
    $.showYtVideo({
     videoId: videoId
    });
   });
  </script>
</body>

while循环只在弹出窗口中显示视频时获取最后一行

1 个答案: