Instagram post URL

Question

我有一个非常简单的脚本可以帮助我在我的网站上显示 IG 图片并且效果很好，但理想情况下我希望一旦点击图片就会 link 到实际 post页面而不是图像源。有什么办法可以调整吗？

PS：我知道如何去掉 link 中的 fancybox - 但我不知道如何以及在何处检索实际的 post id / link 来自.

这是我当前的代码：

    <?php

    function rudr_instagram_api_curl_connect( $api_url ){
        $connection_c = curl_init(); // initializing
        curl_setopt( $connection_c, CURLOPT_URL, $api_url ); // API URL to connect
        curl_setopt( $connection_c, CURLOPT_RETURNTRANSFER, 1 ); // return the result, do not print
        curl_setopt( $connection_c, CURLOPT_TIMEOUT, 20 );
        $json_return = curl_exec( $connection_c ); // connect and get json data
        curl_close( $connection_c ); // close connection
        return json_decode( $json_return ); // decode and return
    }

    $access_token = 'my access token ';
    $username = 'my username';
    $user_search = rudr_instagram_api_curl_connect("https://api.instagram.com/v1/users/search?q=" . $username . "&access_token=" . $access_token);
    // $user_search is an array of objects of all found users
    // we need only the object of the most relevant user - $user_search->data[0]
    // $user_search->data[0]->id - User ID
    // $user_search->data[0]->first_name - User First name
    // $user_search->data[0]->last_name - User Last name
    // $user_search->data[0]->profile_picture - User Profile Picture URL
    $limit = 8;
    $i = 0;
    // $user_search->data[0]->username - Username

    $user_id = $user_search->data[0]->id; // or use string 'self' to get your own media
    $return = rudr_instagram_api_curl_connect("https://api.instagram.com/v1/users/" . $user_id . "/media/recent?access_token=" . $access_token);

    //var_dump( $return ); // if you want to display everything the function returns

    foreach ($return->data as $post) {
        echo '<a href="' . $post->images->standard_resolution->url . '" class="fancybox"><img src="' . $post->images->standard_resolution->url . '" /></a>';
    }
    ?>

Answer 1

实际上你 link 打算 $post->images->standard_resolution->url

现在，如果我们看一下 API returns (https://www.instagram.com/developer/endpoints/users/)

},
        "link": "http://instagr.am/p/BWrVZ/",
        "user": {
            "username": "kevin",
            "profile_picture": "http://distillery.s3.amazonaws.com/profiles/profile_3_75sq_1295574122.jpg",
            "id": "3"
        },
        "created_time": "1296710327",
        "images": {
            "low_resolution": {
                "url": "http://distillery.s3.amazonaws.com/media/2011/02/02/6ea7baea55774c5e81e7e3e1f6e791a7_6.jpg",
                "width": 306,
                "height": 306
            },
            "thumbnail": {
                "url": "http://distillery.s3.amazonaws.com/media/2011/02/02/6ea7baea55774c5e81e7e3e1f6e791a7_5.jpg",
                "width": 150,
                "height": 150
            },
            "standard_resolution": {
                "url": "http://distillery.s3.amazonaws.com/media/2011/02/02/6ea7baea55774c5e81e7e3e1f6e791a7_7.jpg",
                "width": 612,
                "height": 612
            }

如您所见，在顶部有 link - 您所要做的就是将 href 更改为此 link 而不是图像 URL.

所以改变：

 echo '<a href="' . $post->images->standard_resolution->url . '" class="fancybox"><img src="' . $post->images->standard_resolution->url . '" /></a>';

至：

 echo '<a href="' . $post->link . '" class="fancybox"><img src="' . $post->images->standard_resolution->url . '" /></a>';

它应该可以工作。希望它有所帮助。

Instagram post URL

Instagram post URL

php

instagram-api