instagram

Access-Control-Allow-Origin with instagram api

时光总嘲笑我的痴心妄想 提交于 2019-11-26 22:04:58
问题 I am trying to get my instagram feed with the following code $.ajax({ url: 'https://api.instagram.com/v1/users/xxxxxxx/media/recent/?access_token=xxxxxxxxxxx', error: function() { alert('error'); }, success: function(data) { alert('yes'); }, type: 'GET' }); The error I am getting is No 'Access-Control-Allow-Origin' header is present on the requested resource. Is there a work around? 回答1: Instagram API supports JSONP , so add &callback=? to the url and add dataType: "jsonp" to the $.ajax()

instagram jquery ajax type="GET, can't get around CORS

无人久伴 提交于 2019-11-26 22:02:11
问题 I have received this error Access-Control-Allow-Origin' header is present on the requested resource I am trying to make a $.ajax call to the instragram api. I am using the https://api.instagram.com/v1/tags/nofilter/media/recent?client_id=CLIENT-ID end point. I have put in my client-ID and I get the CORS error. It can't find my localhost:3000, I don't know what to do. the endpoint works in postman when I tested. So I know it has something to do with CORS but I don't know who to set other

Instagram embeds not working when adding embeds dynamically

て烟熏妆下的殇ゞ 提交于 2019-11-26 22:02:02
问题 Working on a blog that loads posts on an infinite scroller. Each blog post may or may not have Instagram embeds. I'm finding that the first one that shows on the page will get processed (regardless if it's in the initial page markup, or dynamically added), the following ones will not. Here is a simple JS Bin that illustrates the problem: http://jsbin.com/hilixi/1/edit?html,js,output The first Instagram embed is in the initial page markup. Another Instagram embed is added after page load,

Instagram API: How to get all user media?

China☆狼群 提交于 2019-11-26 21:28:24
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? You're right, the Instagram API will only return 20 images per call. So you'll have to use the pagination feature. If you're trying to use the API console. You'll want to first allow the API console to authenticate via your Instagram login. To do this you

Getting basic information from Instagram using PHP

有些话、适合烂在心里 提交于 2019-11-26 20:44:45
问题 $url = 'https://api.instagram.com/v1/users/XXXX?access_token=XXXX'; echo json_decode(file_get_contents($url))->{'followed_by'}; I am using this code and I do not understand what the issue is. I'm new to PHP so excuse the newbie mistake. I'm trying to get the "followed_by" to display on its own. I've managed to get Facebook's "like" and twitter's followers to display this way. 回答1: In case you need to grab follower count (or other fields) without logging in, Instagram is nice enough to put

How to open fb and instagram app by tapping on button in Swift

依然范特西╮ 提交于 2019-11-26 20:26:32
How can I open Facebook and Instagram app by tapping on a button in swift ? Some apps redirect to the Facebook app and open a specific page. How can I do the same thing? I found it: var url = NSURL(string: "itms://itunes.apple.com/de/app/x-gift/id839686104?mt=8&uo=4") if UIApplication.sharedApplication().canOpenURL(url!) { UIApplication.sharedApplication().openURL(url!) } but I have to know the app URL . Other examples were in ObjectiveC , which I don't know =/ jregnauld Take a look at these links, it can help you: https://instagram.com/developer/mobile-sharing/iphone-hooks/ http://wiki.akosma

Instagram API prevents redirection to iOS app upon successful login with 400 bad request error

冷暖自知 提交于 2019-11-26 20:26:17
问题 I'm trying to integrate Instagram login into my app to get the logged in user's details, I've registered my app on Instagram and got my client ID, then I added a new URL Type in the project settings with the identifier: igxxxx (where xxxx is my client id), i've also set the redirect URI to be ( xxxx://authorize ), once I successfully log in (using a webView) Instagram displays a white blank page with ( 400 Bad Request ) text written on top, however, when I change the redirect URI to anything

Instagram login programmatically

北城余情 提交于 2019-11-26 20:16:35
问题 I'm trying to programmatically login to Instagram via my own website because I want to retrieve direct messages from Instagram (this requires a login as it is not supported in the Instagram API (yet)). But the Instagram login page requires cookies to login. I keep getting the message that the page could not be loaded and that I might need to enable cookies. Is there any way possible to login programmatically via PHP to Instagram? This is what I have so far. $ch = curl_init('https://instagram

Instagram架构的分片和ID设计

旧时模样 提交于 2019-11-26 19:50:56
前言 每秒上传超过25张图和90个“喜欢”,在Instagram我们存了很多数据,为了确保把重要的数据都扔到内存里,达到快速响应用户的请求,我们已经开始把数据进行分片-换句话说,把数据放到更多的小桶子里,每个桶了装一部分数据。 我们的应用服务器跑的是 Django 和后端是 PostgreSQL ,在决定要分片后的第一个问题是,是否还继续用 PostgreSQL 作为主要数据仓库,或者换成别的?我们评估了一些NoSQL的解决方案,但最终决定最好的解决方案是:把数据分片到不同的 PostgreSQL 数据库。 在写数据到不同服务器之前,还需要解决一个问题,如何给在数据库里的每块数据都标识上唯一的标识(如,发布到我们系统的每张图)。单库好解决,就是用自增主键-但如果数据同时写到多个库就不行了,本博客将回答如果解决这个问题。 开始前,先列出系统的主要实现目标: 生成的ID可以按时间排序(如,一个图片列表的id,可以不用获取更多信息即可直接排序) ID最好是64位的(这样索引更小,存储的也更好,像Redis) 系统最好尽可能地只有部分是“可变因素”-很大部分原因为何在很少工程师的情况下 可以扩展 Instagram,就是因为我们相信简单好用! 现有的解决方案 很多类似的ID解决方案都有些问题,下面是一小部分例子: 在web应用层生成ID 这类方法把生成ID的任务都扔到应用层实现

canOpenUrl - This app is not allowed to query for scheme instragram

对着背影说爱祢 提交于 2019-11-26 19:07:05
问题 Im trying to add the Instagram url to my app in iOS9 however I am getting the following warning: -canOpenURL: failed for URL: "instragram://media?id=MEDIA_ID" - error: "This app is not allowed to query for scheme instragram" However, Ive added the following to the LSApplicationQueriesSchemes in my info.plist ; <key>LSApplicationQueriesSchemes</key> <array> <string>instagram</string> <string>instagram://media?id=MEDIA_ID</string>//this one seems to be the issue </array> Any help is greatly