我使用Instagram API和Fancybox Simplified来获取Instagram标签提要。说明:http://www.blueprintinteractive.com/blog/how-instagram-api-fancybox-simplified
这是要在网页上插入的PHP代码:
<?php
// Supply a user id and an access token
$userid = "MY_USER_ID";
$accessToken = "MY_ACCESS_TOKEN";
// Gets our data
function fetchData($url){
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_TIMEOUT, 18);
$result = curl_exec($ch);
curl_close($ch);
return $result;
}
// Pulls and parses data.
$result = fetchData("https://api.instagram.com/v1/tags/canon/media/recent?access_token=MY_ACCESS_TOKEN&count=18");
$result = json_decode($result);
?>
<?php foreach ($result->data as $post): ?>
<!-- Renders images. @Options (thumbnail,low_resoulution, high_resolution) -->
<a class="group" rel="group1" href="<?= $post->images->standard_resolution->url ?>"><img src="<?= $post->images->thumbnail->url ?>"></a>
<?php endforeach ?>
在结果网页上,我有一张带有特定网址的不合适的insta照片。
如何从网格Feed中删除此项?
答案 0 :(得分:0)
附加一些逻辑来删除您不想要的图像。我使用了一个简单的比较,只删除了一个项目,但您可以改为使用列表:
// Pulls and parses data.
$result = fetchData("https://api.instagram.com/v1/tags/canon/media/recent?access_token=MY_ACCESS_TOKEN&count=18");
$result = json_decode($result);
// Delete images according to certain specifications
foreach ($result as $id => $post) {
if ($post->images->thumbnail->url == 'unwanted_image.jpg') {
unset($result[$id]);
}
}
答案 1 :(得分:0)
(代表OP发布)。
以下是我修复它的方法:
<?php foreach ($result->data as $post): ?>
<!-- Renders images. @Options (thumbnail,low_resoulution, high_resolution) -->
<?php if ($post->images->thumbnail->url == 'http://distilleryimage11.s3.amazonaws.com/accb25dcbd9c11e3a46812e95f55900b_5.jpg') {
continue;
}
?>
<a class="group" rel="group1" href="<?= $post->images->standard_resolution->url ?>"><img src="<?= $post->images->thumbnail->url ?>"></a>
<?php endforeach ?>