2013-04-04 107 views
-3

我有画廊的格式,我想从数据库 获取图库相册图片下面是代码,请帮助我,我在PHP尝试这样的:如何在图片库获取从数据库图像

<script type="text/javascript"> 

           var countries=new ddtabcontent("countrytabs") 
           countries.setpersist(true) 
          countries.setselectedClassTarget("link") //"link" or "linkparent" 
          countries.init() 

         </script> 

         <!--For tab End--> 
         <!--<textarea_autoexpand>--> 

         <script type="text/javascript"> 
           $('#textarea1').autoresize(); 
           $('#textarea2').autoresize({ 
            animate: false, 
            buffer: 2, 
            onresize: function() { 
             $('#message').stop(true, true).hide() 
              .text('Resized to '+$(this).height()) 
              .fadeIn('slow', function() { 
               $(this).fadeOut(); 
              }); 
            } 
           }); 
         </script> 

         <!--</textarea _End>--> 
        </div> 
       </div> 
       <div class="comment"> 

        <!--Stack 1 --> 
        <div class="image_stack" style="margin-left: 235px"> 
         <img id="photo1" class="stackphotos" src="images/2.jpg"> 
         <img id="photo2" class="stackphotos" src="images/3.jpg"> 
         <img id="photo3" class="stackphotos" src="images/1.jpg"> 
        </div> 

        </div> 
       </div> 
      </div> 
+4

你好后

<?php mysql_connect('{YOUR-HOST-NAME}','{USER-NAME}','{PASSWORD}'); mysql_select_db('YOUR-DATABASE-NAME'?> 

基本queryetching图像哪里是你的数据库的查询? – 2013-04-04 08:21:27

+1

你的PHP在哪里? – JoDev 2013-04-04 08:21:59

回答

0

先连接数据库。从数据库连接

<?php 
    $query = mysql_query("SELECT {YOUR-IMAGE-FIELD-NAME} FROM {YOUR-TABLE-NAME}"); 
    $i = 1; 
    while ($row = mysql_fetch_array($query)) { 
?> 
<img id="photo<?php echo $i ?>" class="stackphotos" src="<?php echo $row['YOUR-IMAGE-FIELD-NAME']; ?>"> 

<?php $i++; } ?>