使用jQuery PHP CSS将删除图标放在图像的角落

Question

这里是将图像从数据库中的图像扔到php中的div中，在这里，我想通过单击删除图标来删除图像，该图标必须位于从db提取的每个图像的右上角。如何在图片的右上角找到删除图标？这就是我在对话框中显示图像的方式。

<?php        
$query = "SELECT * FROM files_images";    
$res = mysql_query($query);     
while ($row = mysql_fetch_array($res)) {
    echo "<div id='popimages'>";
    echo "<img title='$row[img]' id='$row[id]' style='width:100px;float:left; height:100px;margin-bottom:5px; margin-left:5px;border:2px solid #b06c1c;border-radius:10px;' src='http://localhost/PhpProject2/user_data/" . $row['filename'] . "'/>";
    echo "</div>";
}
?>

Answer 1

如果我理解正确，请尝试输出如下内容：

的HTML

  <div id='popimages'>
      <div class="image">
          <img class="btn-delete" onclick="alert('Do something!');" src="http://cdn1.iconfinder.com/data/icons/diagona/icon/16/101.png"/>
          <img title='$row[img]' id='$row[id]' style='width:100px;float:left; height:100px;margin-bottom:5px; margin-left:5px;border:2px solid #b06c1c;border-radius:10px;' src='http://i.i.com.com/cnwk.1d/i/tim/2011/03/16/Chrome-logo-2011-03-16.jpg'/>
      </div>
  </div>

CSS：

.image {
   width: 100px;
   height: 100px;
   position: relative;
}
.btn-delete {
   position: absolute;
   cursor: pointer;
   right: 2px;
   top: 2px;

   /* This was edited out because it was stupid. See fernholz's answer.
   left: 100%; 
   margin-left: -10px;
   margin-top: 2px; */
}

JSFiddle：http : //jsfiddle.net/TK7zB/