php - 我可以修复 Instagram 分页吗

Question

我的 Instagram 分页有问题。我需要媒体/喜欢的代码。我试过 $jsonurl = " https://api.instagram.com/v1/userers/self/media/liked?access_token= '".$data->access_token."'&count=2" 但不要返回任何东西。提前谢谢！

<?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

您的网址中有拼写错误：/useers/需要是/users/

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

需要是：

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