As I work on a website, I incorporate a link () to display a random photo. However, the photo refreshes periodically and upon loading the site, all that is visible is this default link: . Is there a method to extract the actual source like so: "". Although the Chrome devtools reveal the source in the Application tab, how can it be accessed?
If you're looking to fetch an image and retrieve the response URL, you can utilize the axios library:
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>For those preferring vanilla JS:
fetch("https://source.unsplash.com/random/1920x1080/?wallpaper,landscape").then( data => {
console.log(data.url);
});In case you need to cater to older browser versions:
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);
}
}
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