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Javascript Fancybox没有';从数据库获取图像时无法工作(php)_Javascript_Php_Jquery_Html - Fatal编程技术网

Javascript Fancybox没有';从数据库获取图像时无法工作(php)

Javascript Fancybox没有';从数据库获取图像时无法工作(php),javascript,php,jquery,html,Javascript,Php,Jquery,Html,<?php $sqlInner=" select * from gallery_photos where series_id='".$row['id']."' "; $rr = mysql_query($sqlInner,$con); while($data=mysql_fetch_array($rr)) { echo"<a class='fancybox' rel='gallery' href='images/gallery_photos/".$data['p

<?php
 $sqlInner=" select * from gallery_photos where series_id='".$row['id']."' ";
 $rr = mysql_query($sqlInner,$con); 
 while($data=mysql_fetch_array($rr))    {
    echo"<a class='fancybox' rel='gallery' href='images/gallery_photos/".$data['photo_path']."'><img src='images/gallery_photos/".$data['photo_path']."'width='150' height='150' ></a>";
 }
?>


 <script type="text/javascript">
  $(document).ready(function() {
    $(".fancybox").fancybox({'type' : 'image'})
});
</script>
尝试删除“类型:图像”:

<?php
 $sqlInner=" select * from gallery_photos where series_id='".$row['id']."' ";
 $rr = mysql_query($sqlInner,$con); 
 while($data=mysql_fetch_array($rr))    {
    echo"<a class='fancybox' rel='gallery' href='images/gallery_photos/".$data['photo_path']."'><img src='images/gallery_photos/".$data['photo_path']."'width='150' height='150' ></a>";
 }
?>


 <script type="text/javascript">
  $(document).ready(function() {
    $(".fancybox").fancybox({'type' : 'image'})
});
</script>

$(文档).ready(函数(){
$(“.fancybox”).fancybox();
});

它可以处理静态图像吗?如果是,请尝试下面的代码

<?php
 $sqlInner=" select * from gallery_photos where series_id='".$row['id']."' ";
 $rr = mysql_query($sqlInner,$con); 
 while($data=mysql_fetch_array($rr))    {
    echo"<a class='fancybox' rel='gallery' href='images/gallery_photos/".$data['photo_path']."'><img src='images/gallery_photos/".$data['photo_path']."'width='150' height='150' ></a>";
 }
?>


 <script type="text/javascript">
  $(document).ready(function() {
    $(".fancybox").fancybox({'type' : 'image'})
});
</script>
<script type="text/javascript">
    $(document).ready(function() {
     $(".fancybox").fancybox({type : 'image'})
    });
</script>

$(文档).ready(函数(){
$(“.fancybox”).fancybox({type:'image'})
});

尝试将数据fancybox type=“image”属性添加到锚定标记。这对我有用。

请注意,自2012年以来,mysql扩展(提供mysql_uu函数)已被弃用,取而代之的是mysqli和PDO扩展。它的使用是非常不鼓励的。看,是的,它可以很好地处理静态图像。我也很累,但对我不起作用。
<?php
 $sqlInner=" select * from gallery_photos where series_id='".$row['id']."' ";
 $rr = mysql_query($sqlInner,$con); 
 while($data=mysql_fetch_array($rr))    {
    echo"<a class='fancybox' rel='gallery' href='images/gallery_photos/".$data['photo_path']."'><img src='images/gallery_photos/".$data['photo_path']."'width='150' height='150' ></a>";
 }
?>


 <script type="text/javascript">
  $(document).ready(function() {
    $(".fancybox").fancybox({'type' : 'image'})
});
</script>