Question

163

Get the source.unsplash url of the picture

rated 0 times [ 170] [ 7] / answers: 1 / hits: 27354 / 5 Years ago, thu, october 10, 2019, 12:00:00

I am writing a website where I load this link (https://source.unsplash.com/random/1920x1080/?wallpaper,landscape) to get a random photo. The photo behind the link changes periodically and when its loaded of the site I only see this link https://source.unsplash.com/random/1920x1080/?wallpaper,landscape do you think there is a way to get the real source like this: https://images.unsplash.com/photo-1458571037713-913d8b481dc6?crop=entropy&cs=tinysrgb&fit=crop&fm=jpg&h=1080&ixid=eyJhcHBfaWQiOjF9&ixlib=rb-1.2.1&q=80&w=1920. When I open up the chrome devtools I can see the source in the Application tab but how can i access them?

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brennonr

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2 Years ago

answered 5 Years ago havenbilliec · Accepted Answer

You can use axios library to fetch the image and get the response url :

let randomURL = 'https://source.unsplash.com/random/1920x1080/?wallpaper,landscape';



axios.get(randomURL).then( data => {

  // the url of the random img

  console.log(data.request.responseURL);

});

<script src=https://cdnjs.cloudflare.com/ajax/libs/axios/0.19.0/axios.min.js integrity=sha256-S1J4GVHHDMiirir9qsXWc8ZWw74PHHafpsHp5PXtjTs= crossorigin=anonymous></script>

and with vanilla JS :

fetch(https://source.unsplash.com/random/1920x1080/?wallpaper,landscape).then( data => {

	console.log(data.url);                

});

and if you want to support old browsers :

request = new XMLHttpRequest();

request.open(GET, https://source.unsplash.com/random/1920x1080/?wallpaper,landscape, true);

request.send(null);

request.onreadystatechange = function() {

	if (request.readyState === 4) {

		if (request.status === 200) {

			console.log(request.responseURL);

		}

	}

}