我要把头发拔出来了。谁能帮我把这个工作，我敢肯定这是愚蠢的…我已经得到了所有的PHP错误消失，但我不能得到图像显示。下面的代码…
<!DOCTYPE html>
<html>
<link rel="stylesheet" href="jquery.fancybox-1.3.4.css" type="text/css">
<script type='text/javascript' src='jquery.min.js'></script>
<script type='text/javascript' src='jquery.fancybox-1.3.4.pack.js'></script>
<script type="text/javascript">
$(function() {
$("a.group").fancybox({
'nextEffect' : 'fade',
'prevEffect' : 'fade',
'overlayOpacity' : 0.8,
'overlayColor' : '#000000',
'arrows' : false,
});
});
</script>
<?php
// Supply a user id and an access token
$userid = "1d458ab0c149424c812e664c32b48149";
$accessToken = "c195717e379f48c68df451cc3d60524a";
// Gets our data
function fetchData($url){
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_TIMEOUT, 20);
$result = curl_exec($ch);
curl_close($ch);
return $result;
}
// Pulls and parses data.
$result = fetchData("https://api.instagram.com/v1/users/{$userid}/media/recent/?access_token={$accessToken}");
$result = json_decode($result);
?>
<?php if(!empty($result->data)): ?>
<?php foreach ($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 } ?>
<?php endif ?>
</html>
你需要的是在不同的点添加一些检查，以找出从Instagram返回的内容并处理任何问题。在调试时，将var_dump()粘贴到所有地方可以帮助查看问题所在。
下面是一个带有一些额外检查的PHP节示例:
<?php
// Supply a user id and an access token
$userid = "USER ID";
$accessToken = "ACCESS TOKEN";
// Gets our data
function fetchData($url){
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_TIMEOUT, 20);
$result = curl_exec($ch);
$info = curl_getinfo($ch);
// Check a response was returned
if ($info['http_code'] == '404') {
echo ('Error: HTTP 404 returned, bad request');
die();
}
curl_close($ch);
return $result;
}
// Pulls and parses data.
$result = fetchData("https://api.instagram.com/v1/users/{$userid}/?access_token={$accessToken}");
$result = json_decode($result);
// Check the json_decode succeeded
if (empty($result)) {
echo "Error: JSON not returned from API";
die();
}
// Check no error was returned from Instagram
if ($result->meta->code != 200) {
echo "Error: ".$result->meta->error_message;
die();
}
?>
如果你打算用Instagram API做很多工作，你可能想看看一个库来完成大部分繁重的工作。这种似乎是目前最流行的。