17

我想使用 PHP 获取用户的 Instagram 提要。我已经注册了 Instagram 开发者帐户并尝试提取用户的信息和照片，但响应不稳定。有时我会收到回复，有时我会不断收到错误消息：access_token 丢失。是否有通过用户名获取用户照片提要的可靠示例？

理想情况下，我希望它像这样简单：

$instagram = new Instagram();
$photos = $instagram->getPhotos("username-goes-here");

Instagram 是一个处理所有请求的类。任何帮助或方向表示赞赏。谢谢！

4

8 回答 8

57

尝试这个，

<?php

  function fetchData($url){
  $ch = curl_init();
  curl_setopt($ch, CURLOPT_URL, $url);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
  curl_setopt($ch, CURLOPT_TIMEOUT, 20);
  $result = curl_exec($ch);
  curl_close($ch); 
  return $result;
  }

  $result = fetchData("https://api.instagram.com/v1/users/ID-GOES-HERE/media/recent/?access_token=TOKEN-GOES-HERE");
  $result = json_decode($result);
  foreach ($result->data as $post) {
    // Do something with this data.
  }
?>

愿这对你有所帮助。

于 2012-07-10T10:13:44.667 回答
8

我这样做了：

<?php

  function fetchData($url){
  $ch = curl_init();
  curl_setopt($ch, CURLOPT_URL, $url);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
  curl_setopt($ch, CURLOPT_TIMEOUT, 20);
  $result = curl_exec($ch);
  curl_close($ch); 
  return $result;
  }

  $result = fetchData("https://api.instagram.com/v1/users/USER ID HERE/media/recent/?access_token=ACCES TOKEN HERE&count=14");


  $result = json_decode($result);
  foreach ($result->data as $post) {
     if(empty($post->caption->text)) {
       // Do Nothing
     }
     else {
        echo '<a class="instagram-unit" target="blank" href="'.$post->link.'">
        <img src="'.$post->images->low_resolution->url.'" alt="'.$post->caption->text.'" width="100%" height="auto" />
        <div class="instagram-desc">'.htmlentities($post->caption->text).' | '.htmlentities(date("F j, Y, g:i a", $post->caption->created_time)).'</div></a>';
     }

  }
?>
于 2013-08-23T16:03:58.007 回答
5

根据我在互联网上和这个页面上看到的内容，我在下面创建了一个 Instagram 类（非常简单，仅用于提取提要等）。

class Instagram {
    public static $result;
    public static $display_size = 'thumbnail'; // you can choose between "low_resolution", "thumbnail" and "standard_resolution"
    public static $access_token = "DEFAULTACCESSTOKEN"; // default access token, optional
    public static $count = 10;
    public static function fetch($url){
        $ch = curl_init();
        curl_setopt($ch, CURLOPT_URL, $url);
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
        curl_setopt($ch, CURLOPT_TIMEOUT, 20);
        $result = curl_exec($ch);
        curl_close($ch);
        return $result;
    }
    function __construct($Token=null){
        if(!empty($Token)){
            self::$access_token = $Token;

            // Remove from memory -- not sure if really needed.
            $Token = null;
            unset($Token);
        }
        self::$result = json_decode(self::fetch("https://api.instagram.com/v1/users/self/media/recent?count=" . self::$count . "&access_token=" . self::$access_token), true);
    }
}
$Instagram = new Instagram('ACCESSTOKENIFCHANGEDORNULLOREMPTY');
foreach ($Instagram::$result->data as $photo) {
    $img = $photo->images->{$Instagram::$display_size};
}
于 2014-08-30T07:42:42.563 回答
2

更新：2017 年 6 月 15 日 - Instagram 更改了终点，以下不再起作用。

由于没有获得批准的应用程序就不可能再获得随机用户提要，我已经想出了如何使用非官方 API 来获取它：

#!/bin/bash
instagram_user_id=25025320
count=12
csrftoken=$(curl --head -k https://www.instagram.com/ 2>&1 | grep -Po "^Set-Cookie: csrftoken=\K(.*?)(?=;)")
curl "https://www.instagram.com/query/" -H "cookie: csrftoken=$csrftoken;" -H "x-csrftoken: $csrftoken" -H "referer: https://www.instagram.com/" --data "q=ig_user($instagram_user_id)%20%7B%20media.after(0%2C%20$count)%20%7B%0A%20%20count%2C%0A%20%20nodes%20%7B%0A%20%20%20%20caption%2C%0A%20%20%20%20code%2C%0A%20%20%20%20comments%20%7B%0A%20%20%20%20%20%20count%0A%20%20%20%20%7D%2C%0A%20%20%20%20date%2C%0A%20%20%20%20dimensions%20%7B%0A%20%20%20%20%20%20height%2C%0A%20%20%20%20%20%20width%0A%20%20%20%20%7D%2C%0A%20%20%20%20display_src%2C%0A%20%20%20%20id%2C%0A%20%20%20%20is_video%2C%0A%20%20%20%20likes%20%7B%0A%20%20%20%20%20%20count%0A%20%20%20%20%7D%2C%0A%20%20%20%20owner%20%7B%0A%20%20%20%20%20%20id%2C%0A%20%20%20%20%20%20username%2C%0A%20%20%20%20%20%20full_name%2C%0A%20%20%20%20%20%20profile_pic_url%0A%20%20%20%20%7D%2C%0A%20%20%20%20thumbnail_src%2C%0A%20%20%20%20video_views%0A%20%20%7D%2C%0A%20%20page_info%0A%7D%0A%20%7D" -k

我稍后会用 PHP 改进这个答案，我也需要用 PHP 来做这个。

于 2016-06-04T17:19:34.743 回答
1

此函数在您的 App 类中，但可以是常规函数，并且无论如何它都会起作用。

<?php
    public function instagram(){
      $user = 'your user here';
      // you can get your token from here: https://instagram.pixelunion.net/
      $access_token = 'your access token here';
      $photo_count = 6;// you can choose the amount. 20 is the max per query

      $json_link = "https://api.instagram.com/v1/users/self/media/recent/?";
      $json_link .="access_token={$access_token}&count={$photo_count}";
      $json = file_get_contents($json_link);
      return json_decode($json);
    }

可以使用此工具交互式导航结果：http: //jsonviewer.stack.hu/

API 返回的 json 文件

在我的情况下，我使用刀片模板引擎（https://laravel.com/docs/5.8/blade）

所以模板将是

@foreach($instagram->data as $gram)
   <img src="{{$gram->images->thumbnail->url}}">
@endforeach

而已！

于 2019-10-22T02:08:12.950 回答
1

找到这篇中等文章：- https://medium.com/@bkwebster/how-to-get-instagram-api-access-token-and-fix-your-broken-feed-c8ad470e3f02

 <?php
                    $user_id=xxxxxx;//User ID is the first string of numbers before the first dot (.)
                    $count=2;
                    $width=100;
                    $height=100;
                    $url = 'https://api.instagram.com/v1/users/'.$user_id.'/media/recent/?access_token=xxxxxx.83c3b89.257e2fd9c2bd40c181a2a4fb9576628c&count='.$count;

                    // Also Perhaps you should cache the results as the instagram API is slow
                    $cache = './'.sha1($url).'.json';
                    if(file_exists($cache) && filemtime($cache) > time() - 60*60){
                        // If a cache file exists, and it is newer than 1 hour, use it
                        $jsonData = json_decode(file_get_contents($cache));
                    } else {
                        $jsonData = json_decode((file_get_contents($url)));
                        file_put_contents($cache,json_encode($jsonData));
                    }
                    foreach ($jsonData->data as $key=>$value) {
                        ?>
                        <ul class="w3_footer_grid_list1">
                            <li><label class="fa fa-instagram" aria-hidden="true"></label><a target="_blank" href="<?php echo $value->link;?>"><i><?php echo $value->caption->text; ?> </i></a><?php  ?>
                            </li>
                            <a target="_blank" href="<?php echo $value->link;?>">
                            <img src="<?php echo  $value->images->low_resolution->url;?>" alt="'.$value->caption->text.'" width="<?php echo $width;?>" height="<?php echo $height;?>" />
                            </a>
                        </ul>
                        <?php
                    }
                ?>
于 2019-06-08T05:19:07.177 回答
1

应该更新答案，因为现在唯一可能的方法是通过 Instagram Facebook API --> https://developers.facebook.com/docs/instagram/oembed

于 2021-03-22T21:26:41.197 回答
-1

试试这个原始形式的爬虫类型。

function feed_instagram($url = "https://www.instagram.com/titaniumheart_")
{  
    //$url ie https://www.instagram.com/titaniumheart_

    $dom = new DOMDocument();
    @$dom->loadHTMLFile($url);
    $f=$dom->saveHTML();  //load the url (crawl)

    $key="";    
    $swquote=0;     
    echo "<div>";

    for ($x=0;$x<strlen($f);$x++)
    {
       $c=substr($f,$x,1);
       //echo $c."-";
        if ($c==chr(34)) 
        {
            if($swquote==0)
            {
                $swquote=1; //to start get chars
            } else
            {
                $swquote=0;
                //echo $key;
                if($key=="code")
                {
                    //get the number of comments
                    $m=substr($f,$x+4,100);
                    $code= substr($m,0,strpos($m,chr(34)));
                    echo "code is ".$code;
                    echo "<br>";
                }               
                if($key=="comments")
                {
                    //get the number of comments
                    $m=substr($f,$x+12,20);
                    $comments= substr($m,0,strpos($m,"}"));
                    echo "number of comments is ".$comments;
                    echo "<br>";
                }
                if($key=="caption")
                {
                    //get the number of comments
                    $m=substr($f,$x+4,200);
                    $caption= substr($m,0,strpos($m,chr(34)));
                    echo "caption is ".$caption;
                    echo "<br>";
                }
                if($key=="likes")
                {
                    //get the number of comments
                    $m=substr($f,$x+12,20);
                    $likes= substr($m,0,strpos($m,"}"));
                    echo "number of likes is ".$likes;
                    echo "<br>";
                }
                if($key=="thumbnail_src")
                {
                    //get the number of comments
                    $m=substr($f,$x+4,200);
                    $src= substr($m,0,strpos($m,"?"));
                    echo "<br>image source is ".$src;
                    echo "<br>";                                    
                    echo "<a href=\"https://www.instagram.com/p/".$code."/\">";
                    echo "<img src=\"".$src."\">";
                    echo "</a><br>";                    
                }                               
               $key="";
        }

    }else
    {
        if($swquote==1)
        {
            $key.=$c;
        }
    }           
}
echo "</div>";
}

用法：https://www.instagram.com/titaniumheart_");?>

注意：您必须在 php.ini 上启用扩展“php_openssl”。

于 2016-09-21T10:45:44.760 回答