Question

我对instagram分页有问题。我需要媒体/喜欢的代码。我已尝试使用$ jsonurl =“https://api.instagram.com/v1/useers/self/media/liked?access_token=”。$ data-＆gt; access_token。“'＆amp; count = 2”但不返回任何内容。 Thx提前！

<?php 
session_start(); 
$data=$_SESSION['userdetails'];
echo $data->access_token;
if (isset($_REQUEST['next']) && isset($_SESSION['next'])) 
$jsonurl = $_SESSION['next']; 
else 
$jsonurl = "https://api.instagram.com/v1/useers/self/media/liked?access_token='".$data-    >access_token."'&count=2"; 
$json = file_get_contents($jsonurl,0,null,null); 
$response = json_decode($json, true); 

$_SESSION['next'] = $response['pagination']['next_url']; 
?> 
<div class="photo"> 
<a href="<?= $link ?>"><span></span><img src="<?= $thumbnail ?>" title="<?= $text ?>" alt="<?= $text ?>" /></a> 
    <div class="metaz">via <?= $author ?> at <?echo date("h:i:s A 
\o\\n d/m/Y",$date);?></div> 
    <?= $text ?>

// Button code here: 
<a href="?next">next</a>

Answer 1

网址中存在拼写错误：/useers/需要/users/

$jsonurl = "https://api.instagram.com/v1/useers/self/media/liked?access_token='".$data->access_token."'&count=2";

需要：

$jsonurl = "https://api.instagram.com/v1/users/self/media/liked?access_token='".$data->access_token."'&count=2";

Instagram分页我该如何修复

1 个答案: