I have a Social Media with posts, and I want to show the new with JS and AJAX. I tried to reload my page with AJAX and put it in a div element, but now the website is in this div element, so I have the full page twice. However, I only want to take the div element with AJAX. My code looks like this:
var timeout = setTimeout(loadDoc, 2000);
function loadDoc() {
const xhttp = new XMLHttpRequest();
xhttp.onload = function() {
document.getElementById("container").innerHTML = this.responseText;
}
xhttp.open("GET", "mywebsite.com");
xhttp.send();
timeout = setTimeout(loadDoc, 2000);
}<div id="container">
post2<br>
post1
</div>Kind regards, ComputerJoshi
There are two approaches you can take here:
The URL you are requesting is returning a bunch of data you don't want.
Change it (or create a new URL to request instead) so that you only get the data you want.
You could keep that data in HTML form, but it is more typical to present it in a structured format (JSON is popular) and then use that to update the DOM of the current page.
Convert the string of HTML into a DOM (e.g. with DOMParser), then use querySelector to find the parts of that DOM you want and then copy them into the current DOM.