與html鏈接標記和圖像的問題

-1

 <div class="col-lg-12"> 
     <h1 class="page-header">Anime!</h1> 
    </div> 

     <?php 

      include "config/database.php"; 

      $sql = "SELECT * FROM anime WHERE status = 'On Going' ORDER BY id"; 

      $query = mysql_query($sql); 

      if ($query > 0){ 

     ?> 
    <div class="container"> 
      <div class="description-plate"> 
       <?php 
        while 
         ($row = mysql_fetch_array($query)){ 
         $id = $row['id']; 
         $image = $row['image']; 
         $title = $row['title']; 
         $genre = $row['genre']; 
         $start = $row['start']; 
         $schedule = $row['schedule']; 
         $description = $row['description']; 

       ?> 
       <!--div class="caption-btm"> 
        <p style="margin-left:6px; margin-top:175px;">Start Airing From:</p> 
        <h5 style="margin-left:10px;"><?php echo $start; ?></h5> 
        <p style="margin-left:6px;">Airing Schedule:</p> 
        <h5 style="margin-left:10px;"><?php echo $schedule; ?></h5> 

       </div--> 
       <div class="thumbnail-fluid"> 
        <a href="<?php echo $row['image']; ?>"> 
        <div id="og-plate"> 
           <div><img src="admin/<?php echo $row['image']; ?>"></div> 
        <?php } ?> 
        </div> 
        </a> 
       </div> 
      </div> 
      <?php } ?> 
    </div>

所以當我試圖用php調用圖像時，標籤只出現在最後一張圖片上。我想要做的是在每張圖片上都貼上標籤。希望得到任何幫助，謝謝:)

來源

2014-12-26 sun station

好像你有額外的''後圖像div。 – dfsq

@dfsq多數民衆贊成是我發現 –

哎呀對不起，請再次檢查腳本。忘了粘貼完整的腳本，我的壞:) –

現在你正在經歷的循環（不知道爲什麼你正在使用while），每次創建

 <div class="thumbnail-fluid"> 
      <a href="<?php echo $row['image']; ?>"> 
      <div id="og-plate"> 
         <div><img src="admin/<?php echo $row['image']; ?>"></div> 
      <?php } ?> 
      </div> 
      </a> 
     </div>

你想要做的是建立一個HTML字符串在每次通過附加的下一個圖像標籤，更多的東西一樣

...

$myimages = '';  
    while // skipped details 
     $myimages .= ' <div class="thumbnail-fluid"> 
       <a href=". $row['image'] . '> 
       <div id="og-plate"> 
          <div><img src="admin/' . $row['image'] . '></div>' 

       . '</div> 
       </a> 
      </div>'; 
}

來源

2014-12-26 16:29:46 Elin

它最後出現圖像，因爲ORDER BY ID和T他的狀態='正在進行'可以返回一個圖像。你的html結構應該是這樣的。

<div class="col-lg-12"> 
    <h1 class="page-header">Anime!</h1> 
</div> 

<?php 
include "config/database.php"; 

$sql = "SELECT * FROM anime WHERE status = 'On Going' ORDER BY id"; 

$result = mysql_query($sql); 
$n = mysql_num_rows($result); 
if ($n > 0) { 
    ?> 
    <div class="container"> 
     <div class="description-plate"> 
      <?php 
      while ($row = mysql_fetch_array($result)) { 
       $id = $row['id']; 
       $image = $row['image']; 
       $title = $row['title']; 
       $genre = $row['genre']; 
       $start = $row['start']; 
       $schedule = $row['schedule']; 
       $description = $row['description']; 
       ?> 
       <div class="thumbnail-fluid"> 
        <a href="<?php echo $row['image']; ?>"> 
         <div id="og-plate"> 
          <div><img src="admin/<?php echo $row['image']; ?>"></div>      
         </div> 
        </a> 
       </div> 
      <?php } ?> 
     </div> 

    </div> 
<?php } ?>

來源

2014-12-26 16:44:50

與html鏈接標記和圖像的問題

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