使用PHP获取Tumblr照片

Question

编辑:找到解决办法了。按照建议，我将脚本转换为使用JSON而不是XML。脚本如下，如果需要，应该在Wordpress安装中运行

    <?php
$offset = $_GET['offset'];
$ch = curl_init();
curl_setopt($ch,CURLOPT_URL,"http://blog.dgovil.com/api/read/json?num=10&start=".$offset);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
$result = curl_exec($ch);
curl_close($ch);
$result = str_replace("var tumblr_api_read = ","",$result);
$result = str_replace(';','',$result);
$result = str_replace('u00a0','&amp;nbsp;',$result);
$jsondata = json_decode($result,true);
$posts = $jsondata['posts'];

foreach($posts as $post){   ?>
<div class="tumblr_post post-<?php echo $post['type'] ?>">
    <?php if ($post['type'] == 'regular') { ?>
        <div class="post-title" id="post-<?php echo $post['id'];?>"><a href="<?php echo $post['url-with-slug']; ?>"><?php echo $post{'regular-title'}; ?></a></div>
    <?php echo $post{'regular-body'}; ?>
      <?php } ?>
    <?php if ($post['type'] == 'quote') {  ?>
        <?php echo $post{'quote-text'}; ?>
        <?php echo $post{'quote-source'}; ?>
      <?php } ?>

    <?php if ($post['type'] == 'photo') {  ?>
        <img src="<?php echo $post['photo-url-500'];?>">
        <?php echo $post{'photo-caption'}; ?>
        <?php echo $post{'photo-set'}; ?>
        <a href="<?php echo $post{'photo-url'}; ?>" class="fImage">View Full Size</a>
    <?php } ?>
    <?php if ($post['type'] == 'link') {  ?>
        <p><a href="<?php echo $post{'link-url'}; ?>"><?php echo $post{'link-text'}; ?></a>
        <?php echo $post{'link-description'}; ?>
      <?php } ?>
    <?php if ($post['type'] == 'conversation') {  ?>
        <?php echo $post{'conversation-text'}; ?>
      <?php } ?>

    <?php if ($post['type'] == 'video') {  ?>
        <!--<?php echo $post{'video-source'}; ?>-->
        <?php echo $post{'video-player'}; ?>
        <?php echo $post{'video-caption'}; ?>
      <?php } ?>
    <?php if ($post['type'] == 'conversation') {  ?>
        <?php echo $post{'audio-caption'}; ?>
        <?php echo $post{'audio-player'}; ?>
        <?php echo $post{'audio-plays'}; ?>
      <?php } ?>
<div id="post-date">
<?php echo date("jS D M, H:i",strtotime($post['date'])); ?>&nbsp; &nbsp;<a href="<?php echo $post['url-with-slug']; ?>">Read on Tumblr</a>
</div>
</div>

<?php   }?>

<div id="tumblr_down"></div>
<script type="text/javascript">
// load the api response
$.getJSON('http://tumblruptime.apigee.com/json?callback=?', function (response) {
    // check if an api method is broken
    if (!response.methods['/api/read'].up) {
        $('#tumblr_down').text('Tumblr seems to not want to work with me right now. You can still try and view the posts on blog.dgovil.com. Service provided by http://tumblruptime.icodeforlove.com/apiinfo');
    }
});
</script>
<div id="tumblr_postnav">
<a href="<?php echo '?offset='.($offset + 10) ?>" id="tumblr_next">Older posts</a>
<a href="<?php echo '?offset='.($offset - 10) ?>" id="tumblr_prev">Newer Posts</a>  
</div>

谢谢!

Answer 1

使用PHP解释JSON调用:

<?php
$ch = curl_init();
curl_setopt($ch,CURLOPT_URL,"http://username.tumblr.com/api/read/json");
curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
$result = curl_exec($ch);
curl_close($ch);
$result = str_replace("var tumblr_api_read = ","",$result);
$result = str_replace(';','',$result);
$result = str_replace('u00a0','&amp;nbsp;',$result);
$jsondata = json_decode($result,true);
$posts = $jsondata['posts'];
foreach($posts as $post){
echo $post['photo-url-500'].'</br>';
}
?>

你会想这样做b/c Tumblr的JSON响应似乎是非标准的。您可以将foreach循环更改为for循环或其他任何您想要的输出