In general I need to get all user media.
User has more than 250 photos.
I do /users/1/media/recent/?access_token=...&count=250
But it returns only 20 photos.
Maybe instagram has a limit for getting media. If it is, response has a pagination to solve it. But there are only max ID photo. How to know the first (min) ID photo to paginate it then?
It was a problem in Instagram Developer Console. max_id and min_id doesn't work there.
In June 2016 Instagram made most of the functionality of their API available only to applications that have passed a review process. They still however provide JSON data through the web interface, and you can add the parameter __a=1 to a URL to only include the JSON data.
max=
while :;do
c=$(curl -s "https://www.instagram.com/username/?__a=1&max_id=$max")
jq -r '.user.media.nodes[]?|.display_src'<<<"$c"
max=$(jq -r .user.media.page_info.end_cursor<<<"$c")
jq -e .user.media.page_info.has_next_page<<<"$c">/dev/null||break
done
Edit: As mentioned in the comment by alnorth29, the max_id parameter is now ignored. Instagram also changed the format of the response, and you need to perform additional requests to get the full-size URLs of images in the new-style posts with multiple images per post. You can now do something like this to list the full-size URLs of images on the first page of results:
c=$(curl -s "https://www.instagram.com/username/?__a=1")
jq -r '.graphql.user.edge_owner_to_timeline_media.edges[]?|.node|select(.__typename!="GraphSidecar").display_url'<<<"$c"
jq -r '.graphql.user.edge_owner_to_timeline_media.edges[]?|.node|select(.__typename=="GraphSidecar")|.shortcode'<<<"$c"|while read l;do
curl -s "https://www.instagram.com/p/$l?__a=1"|jq -r '.graphql.shortcode_media|.edge_sidecar_to_children.edges[]?.node|.display_url'
done
To make a list of the shortcodes of each post made by the user whose profile is opened in the frontmost tab in Safari, I use a script like this:
sjs(){ osascript -e'{on run{a}','tell app"safari"to do javascript a in document 1',end} -- "$1";}
while :;do
sjs 'o="";a=document.querySelectorAll(".v1Nh3 a");for(i=0;e=a[i];i++){o+=e.href+"\n"};o'>>/tmp/a
sjs 'window.scrollBy(0,window.innerHeight)'
sleep 1
done
I've solved this issue with the optional parameter count set to -1.
Use the best recursion function for getting all posts of users.
<?php
set_time_limit(0);
function getPost($url,$i)
{
static $posts=array();
$json=file_get_contents($url);
$data = json_decode($json);
$ins_links=array();
$page=$data->pagination;
$pagearray=json_decode(json_encode($page),true);
$pagecount=count($pagearray);
foreach( $data->data as $user_data )
{
$posts[$i++]=$user_data->link;
}
if($pagecount>0)
return getPost($page->next_url,$i);
else
return $posts;
}
$posts=getPost("https://api.instagram.com/v1/users/CLIENT-ACCOUNT-NUMBER/media/recent?client_id=CLIENT-ID&count=33",0);
print_r($posts);
?>
You can user pagination of Instagram PHP API: https://github.com/cosenary/Instagram-PHP-API/wiki/Using-Pagination
Something like that:
$Instagram = new MetzWeb\Instagram\Instagram(array(
"apiKey" => IG_APP_KEY,
"apiSecret" => IG_APP_SECRET,
"apiCallback" => IG_APP_CALLBACK
));
$Instagram->setSignedHeader(true);
$pictures = $Instagram->getUserMedia(123);
do {
foreach ($pictures->data as $picture_data):
echo '<img src="'.$picture_data->images->low_resolution->url.'">';
endforeach;
} while ($pictures = $instagram->pagination($pictures));
Use the next_url object to get the next 20 images.
In the JSON response there is an pagination array:
"pagination":{
"next_max_tag_id":"1411892342253728",
"deprecation_warning":"next_max_id and min_id are deprecated for this endpoint; use min_tag_id and max_tag_id instead",
"next_max_id":"1411892342253728",
"next_min_id":"1414849145899763",
"min_tag_id":"1414849145899763",
"next_url":"https:\/\/api.instagram.com\/v1\/tags\/lemonbarclub\/media\/recent?client_id=xxxxxxxxxxxxxxxxxx\u0026max_tag_id=1411892342253728"
}
This is the information on specific API call and the object next_url shows the URL to get the next 20 pictures so just take that URL and call it for the next 20 pictures.
For more information about the Instagram API check out this blogpost: Getting Friendly With Instagram’s API
What I had to do is (in Javascript) is go through all pages by using a recursive function. It's dangerouse as instagram users could have thousands of pictures i a part from that (so your have to controle it) I use this code: (count parameter I think , doesn't do much)
instagramLoadDashboard = function(hash)
{
code = hash.split('=')[1];
$('#instagram-pictures .images-list .container').html('').addClass('loading');
ts = Math.round((new Date()).getTime() / 1000);
url = 'https://api.instagram.com/v1/users/self/media/recent?count=200&min_timestamp=0&max_timestamp='+ts+'&access_token='+code;
instagramLoadMediaPage(url, function(){
galleryHTML = instagramLoadGallery(instagramData);
//console.log(galleryHTML);
$('#instagram-pictures .images-list .container').html(galleryHTML).removeClass('loading');
initImages('#instagram-pictures');
IGStatus = 'loaded';
});
};
instagramLoadMediaPage = function (url, callback)
{
$.ajax({
url : url,
dataType : 'jsonp',
cache : false,
success: function(response){
console.log(response);
if(response.code == '400')
{
alert(response.error_message);
return false;
}
if(response.pagination.next_url !== undefined) {
instagramData = instagramData.concat(response.data);
return instagramLoadMediaPage(response.pagination.next_url,callback);
}
instagramData = instagramData.concat(response.data);
callback.apply();
}
});
};
instagramLoadGallery = function(images)
{
galleryHTML ='<ul>';
for(var i=0;i<images.length;i++)
{
galleryHTML += '<li><img src="'+images[i].images.thumbnail.url+'" width="120" id="instagram-'+images[i].id+' data-type="instagram" data-source="'+images[i].images.standard_resolution.url+'" class="image"/></li>';
}
galleryHTML +='</ul>';
return galleryHTML;
};
There some stuff related to print out a gallery of picture.
Instagram developer console has provided the solution for it. https://www.instagram.com/developer/endpoints/
To use this in PHP, here is the code snippet,
/**
**
** Add this code snippet after your first curl call
** assume the response of the first call is stored in $userdata
** $access_token have your access token
*/
$maximumNumberOfPost = 33; // it can be 20, depends on your instagram application
$no_of_images = 50 // Enter the number of images you want
if ($no_of_images > $maximumNumberOfPost) {
$ImageArray = [];
$next_url = $userdata->pagination->next_url;
while ($no_of_images > $maximumNumberOfPost) {
$originalNumbersOfImage = $no_of_images;
$no_of_images = $no_of_images - $maximumNumberOfPost;
$next_url = str_replace("count=" . $originalNumbersOfImage, "count=" . $no_of_images, $next_url);
$chRepeat = curl_init();
curl_setopt_array($chRepeat, [
CURLOPT_URL => $next_url,
CURLOPT_HTTPHEADER => [
"Authorization: Bearer $access_token"
],
CURLOPT_RETURNTRANSFER => true
]);
$userRepeatdata = curl_exec($chRepeat);
curl_close($chRepeat);
if ($userRepeatdata) {
$userRepeatdata = json_decode($userRepeatdata);
$next_url = $userRepeatdata->pagination->next_url;
if (isset($userRepeatdata->data) && $userRepeatdata->data) {
$ImageArray = $userRepeatdata->data;
}
}
}
}
See http://instagram.com/developer/endpoints/ for information on pagination. You need to subsequentially step through the result pages, each time requesting the next part with the next_url that the result specifies in the pagination object.
Source: Stackoverflow.com