2016-02-07 19 views
0

我不明白爲什麼我的photos未顯示。 photos div放置在echo語句之外時顯示正常,但在echo'd時它不會呈現。我的div的結構如下:Div不會出現在PHP echo語句中,但會呈現良好的外部回聲語句

userposts_panel 
    photos_div 
      photos 

下面是代碼(完整的)的所有div的存在:

<div class="userposts_panel"> 
     <div class="photos_div"> 
      <table> 
       <tr> 
      <?php 
      // get pictures from DB 
      $get_user_pics = mysqli_query ($connect, "SELECT * FROM user_photos WHERE uploaded_by='$username'"); // maybe $user 
      $check_rows = mysqli_num_rows ($get_user_pics); 
      while ($row = mysqli_fetch_assoc($get_user_pics)){ 
       $user_id = $row ['id']; 
       $photo_url = $row ['img_url']; 
       $phot_desc = $row ['photo_desc']; 
       $photo_uploaded_by = $row ['uploaded_by']; 
       $photo_uloaded_on= $row ['uploaded_on']; 
       $uid = $row ['uid']; 

       echo "<td> 
         <div class='photos'> 
          <img src='$photo_url' height='170px' width='170px'/> 
         </div> 
        </td>"; 
       } 
      ?> 
       </tr> 
      </table> 
     </div>   
    </div> 
</div> 

.userposts_panel CSS:

.userposts_panel{ 
    margin:0 auto; 
    height: auto; 
    text-align:left; 
    position:absolute; 
    top: 395px; 
    left:10px; 
    width: 800px; 
    background-color:#ffffff; 
    border-bottom: 1px solid #E1E3E4; 
    border-left: 1px solid #E1E3E4; 
    border-right: 1px solid #E1E3E4; 
    z-index:999; 
} 

。 photos_div CSS:

.photos_div{ 
    background-color: #FFFFFF; 
    width: 800px; 
    height: 500px; 
} 

.photos CSS:

.photos{ 
    position:absolute; 
    top:0; 
    left:0; 
    margin-top:11px; 
    margin-left:5px; 
    background-color: #ffffff; 
    border: 1px solid red; 
    padding: 3px; 
    width: 200px; 
    height:200px; 
    margin-right: 5px; 
    text-align:center; 
} 

我只是不明白爲什麼被回顯,但echo語句外呈現正常時,DIV沒有出現?任何幫助,將不勝感激。

+1

你至少讓你的​​元素中的串聯變量$ PHOTO_URL?或者是否也不會被迴應出來?如果是這種情況,我會懷疑你的查詢中有一個錯誤,導致while語句永不執行,因爲沒有行。 – Phil

+0

@菲爾你可能是對的。我嘗試了php塊外的'​​'標籤內的代碼,並出現了div。雖然我似乎無法弄清楚while循環出了什麼問題,但數據庫中的所有字段都是正確的。 – Freddy

+0

回顯您的check_rows變量並確保您有查詢返回的行。如果喬行檢查mysqli_error – Phil

回答

0

嘗試echo語句

<div class="userposts_panel"> 
    <div class="photos_div"> 
     <table> 
      <tr> 
     <?php 
     // get pictures from DB 
     $get_user_pics = mysqli_query ($connect, "SELECT * FROM user_photos WHERE uploaded_by='$username'"); // maybe $user 
     $check_rows = mysqli_num_rows ($get_user_pics); 
     while ($row = mysqli_fetch_assoc($get_user_pics)){ 
      $user_id = $row ['id']; 
      $photo_url = $row ['img_url']; 
      $phot_desc = $row ['photo_desc']; 
      $photo_uploaded_by = $row ['uploaded_by']; 
      $photo_uloaded_on= $row ['uploaded_on']; 
      $uid = $row ['uid']; 
      ?> 
      <td> 
        <div class='photos'> 
         <img src='<?php echo $photo_url; ?>' height='170px' width='170px'/> 
        </div> 
       </td> 
      <?php 
      } 
     ?> 
      </tr> 
     </table> 
    </div>   
</div> 

+0

你應該解釋一下「這」 – Michael

+0

@邁克爾「這」發表弗雷迪相同的代碼,我只是輕微增加至修復它 –

+0

@PeterWilson - 謝謝你的快速回復。但是,連接變量似乎對我無能爲力。我甚至嘗試過編輯完整的代碼,但無濟於事。 – Freddy