我正在使用以下代码,该代码可以很好地加载与#spproject标记匹配的所有图像。我想要做的是一次加载9张照片,然后使用ajax加载更多或只链接到上一页/下一页。问题是我不太明白如何使用API,你能帮忙吗?
CODE:
<?PHP
function get_instagram($next=null,$width=160,$height=160){
if($next == null ) {
$url = 'https://api.instagram.com/v1/tags/spproject/media/recent?access_token=[TOKEN]&count=10';
}
else {
$url .= '&max_tag_id=' . $next;
}
//Also Perhaps you should cache the results as the instagram API is slow
$cache = './'.sha1($url).'.json';
//unlink($cache); // Clear the cache file if needed
if(file_exists($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));
}
$result = '<ul id="instagramPhotos">'.PHP_EOL;
if (is_array($jsonData->data))
{
foreach ($jsonData->data as $key=>$value)
{
$result .= '<li><div class="album">
<figure class="frame">
<a href="'.$value->link.'" target="_blank"><i><img src="'.$value->images->standard_resolution->url.'" alt="" width="'.$width.'" height="'.$height.'" name="'.$value->user->username.'"></i></a>
</figure>
<span class="count">#SPproject</span>
<a href="http://www.instagram.com/'.$value->user->username.'" target="_blank"><figcaption class="name">'.$value->user->username.'</figcaption></a>
</div></li>'.PHP_EOL;
}
}
$result .= '</ul>'.PHP_EOL;
if(isset($jsonData->pagination->next_max_tag_id)) {
$result .= '<div><a href="?next=' . $jsonData->pagination->next_max_tag_id . '">Next</a></div>';
}
return $result;
}
?>
<!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>#SPproject - A worldwide instagram idea</title>
<link rel="stylesheet" href="style.css">
<link rel="stylesheet" href="normalize.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
var totalPhotos = $('#instagramPhotos > li').size();
$('#result').text('Total tagged images: '+totalPhotos);
});
</script>
</head>
<body>
<div id="container">
<?=get_instagram(@$_GET['next']);?>
<div id="result"></div>
</div>
</body>
</html>
答案 0 :(得分:3)
Instagram返回的JSON对象包含一个分页变量,反过来又包含一个“next_url”变量。 next_url是您需要调用以获取下一页结果的API URL。
Here is a good tutorial与Instagram分页。 此外,未来的提示 - 不要在互联网上发布您的API访问代码...
下面的(修订版)代码应该是一个很好的起点。
<?PHP
function get_instagram($next=null,$width=160,$height=160){
$url = 'https://api.instagram.com/v1/tags/spproject/media/recent?access_token=[token]&count=9';
if($url !== null) {
$url .= '&max_tag_id=' . $next;
}
//Also Perhaps you should cache the results as the instagram API is slow
$cache = './'.sha1($url).'.json';
//unlink($cache); // Clear the cache file if needed
if(file_exists($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));
}
$result = '<ul id="instagramPhotos">'.PHP_EOL;
foreach ($jsonData->data as $key=>$value) {
$result .= '<li><div class="album">
<figure class="frame">
<a href="'.$value->link.'" target="_blank"><i><img src="'.$value->images->standard_resolution->url.'" alt="" width="'.$width.'" height="'.$height.'" name="'.$value->user->username.'"></i></a>
</figure>
<span class="count">#SPproject</span>
<a href="http://www.instagram.com/'.$value->user->username.'" target="_blank"><figcaption class="name">'.$value->user->username.'</figcaption></a>
</div></li>'.PHP_EOL;;
//$result .= '<li><a href="'.$value->link.'"><img src="'.$value->images->standard_resolution->url.'" alt="" width="'.$width.'" height="'.$height.'" name="'.$value->user->username.'" /></a></li>'.PHP_EOL;
}
$result .= '</ul>'.PHP_EOL;
if(isset($jsonData->pagination->next_max_tag_id)) {
$result .= '<div><a href="?next=' . $jsonData->pagination->next_max_tag_id . '">Next</a></div>';
}
return $result;
}
?>
<!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>#SPproject - A worldwide instagram idea</title>
<link rel="stylesheet" href="style.css">
<link rel="stylesheet" href="normalize.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
var totalPhotos = $('#instagramPhotos > li').size();
$('#result').text('Total tagged images: '+totalPhotos);
});
</script>
</head>
<body>
<div id="container">
<?=get_instagram(@$_GET['next']);?>
<div id="result"></div>
</div>
</body>
</html>