我需要从instagram后链接抓取图像和视频(例如:https://www.instagram.com/p/BZiTZWXABMp)。当使用python的正常请求时,我在html响应中没有接收到图像url。请帮我拿一下。
发布于 2019-11-16 17:49:35
这很简单,只需在你的Instagram post链接末尾添加/media?size=l即可。
https://www.instagram.com/p/BZiTZWXABMp/media/?size=l
发布于 2017-09-27 18:46:57
您可以查看此url:https://api.instagram.com/oembed/?url=<your-url>
返回一个json,您可以获取它的thumbnail_url
属性
在您请求的帖子中,要查询的url是:https://api.instagram.com/oembed/?url=https://www.instagram.com/p/BZiTZWXABMp/
在那里您可以获得图像的url:https://instagram.fmad3-4.fna.fbcdn.net/t51.2885-15/s480x480/e35/21984643_995308220608235_7854693412774084608_n.jpg
发布于 2019-12-03 14:34:21
使用脚本
javascript:
var host = window.location.hostname;
var pathname = window.location.pathname;
var fullurl = host + pathname;
var https_host = "https://" + host + pathname;
var ps = fullurl.split("/")[1];
switch (host) {
case "www.instagram.com":
switch(ps) {
case "p":
var linkvid = document.getElementsByClassName("tWeCl")[0];
if(linkvid == undefined){
var length_photo = document.getElementsByClassName("FFVAD").length;
var linkins = length_photo - 1;
var link_photo = document.getElementsByClassName("FFVAD")[linkins].currentSrc;
window.open(link_photo, '_blank');
}else{
var linkvid = linkvid.currentSrc;
window.open(linkvid, '_blank');
}
case "stories":
var photo = document.getElementsByClassName('y-yJ5')[0].src;
if(photo == null){
var video = document.getElementsByClassName('y-yJ5')[2].currentSrc;
window.open(video, '_blank');
}
else{
window.open(photo , '_blank');
}
break;
default:
window.open("http://insd.ga", '_blank');
}
break;
}
https://stackoverflow.com/questions/46445788
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