来自foreach look的图像名称

Question

我使用这部分代码来显示文件夹中的图像。问题是如何修改它以获得文件的名称作为标题。

<div id="tab1p" class="tab active">
                         <ul id="photoslist" class="photo_gallery_13">
                             <?php
                             $dirname = "images/kitchens/";
                             $images = glob($dirname."*.jpg");
                             $ignore = Array(".", "..");
                             foreach($images as $curimg){
                             if(!in_array($curimg, $ignore)) {};

                             ?>
                            <li><a rel="gallery-3" href="<?php echo "$curimg"?>" title="<?php echo "$curimg"?>" class="swipebox"><img src="<?php echo "$curimg"?>" alt="image"/></a></li>
                             <?php   }
                             ?>

                           <div class="clearleft"></div>
                          </ul>   
                      </div>

亲切的问候

帕特里克

Answer 1

也许不是正确的，但它是有效的。

<div id="tab1p" class="tab active">
                         <ul id="photoslist" class="photo_gallery_12">
                             <?php
                             $dirname = "images/kitchens/doors/zurfix/";
                             $images = glob($dirname."*.jpg");
                             $ignore = Array(".", "..");

                             foreach($images as $curimg){
                                 $name2 =pathinfo($curimg, PATHINFO_FILENAME);
                                 if(!in_array($curimg, $ignore)) {};

                                 ?>

                                 <li><a rel="gallery-3" href="<?php echo "$curimg"?>" title="<?php echo "$name2"?>" class="swipebox"><img src="<?php echo "$curimg"?>" alt="image"/></a></li>

                             <?php   }
                             ?>

                           <div class="clearleft"></div>
                          </ul>   
                      </div>