AJAX - Grab specific DOM elements from another html file?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I'm trying to use vanilla JS to get content to load without refreshing. When a menu option is clicked, I would like to get content from another html file, but only grab either the body or a particular class.
Is this doable without jQuery?
Thanks in advance!
var AJAX = function(page){
var call = new XMLHttpRequest();
call.open("GET", page);
call.send();
call.addEventListener("load", function(e){
console.log(e.target.responseText); // this shows all the HTML as text, but I only want to grab either the body or a particular class, and use it in an innerHTML method.
});
}
javascript ajax
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
add a comment |
I'm trying to use vanilla JS to get content to load without refreshing. When a menu option is clicked, I would like to get content from another html file, but only grab either the body or a particular class.
Is this doable without jQuery?
Thanks in advance!
var AJAX = function(page){
var call = new XMLHttpRequest();
call.open("GET", page);
call.send();
call.addEventListener("load", function(e){
console.log(e.target.responseText); // this shows all the HTML as text, but I only want to grab either the body or a particular class, and use it in an innerHTML method.
});
}
javascript ajax
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
add a comment |
I'm trying to use vanilla JS to get content to load without refreshing. When a menu option is clicked, I would like to get content from another html file, but only grab either the body or a particular class.
Is this doable without jQuery?
Thanks in advance!
var AJAX = function(page){
var call = new XMLHttpRequest();
call.open("GET", page);
call.send();
call.addEventListener("load", function(e){
console.log(e.target.responseText); // this shows all the HTML as text, but I only want to grab either the body or a particular class, and use it in an innerHTML method.
});
}
javascript ajax
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
I'm trying to use vanilla JS to get content to load without refreshing. When a menu option is clicked, I would like to get content from another html file, but only grab either the body or a particular class.
Is this doable without jQuery?
Thanks in advance!
var AJAX = function(page){
var call = new XMLHttpRequest();
call.open("GET", page);
call.send();
call.addEventListener("load", function(e){
console.log(e.target.responseText); // this shows all the HTML as text, but I only want to grab either the body or a particular class, and use it in an innerHTML method.
});
}
javascript ajax
javascript ajax
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
asked Nov 16 '18 at 23:03
giantqtipzgiantqtipz
303111
303111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
With XMLHttpRequest, it's pretty easy - with the responseType as document, the instiantiated XMLHttpRequest's .response property will contain the response document, which you can use ordinary DOM methods on. For example, the following code will select and log the text of the first element with a class name of foo from the response document:
function get(url){
var r = new XMLHttpRequest();
r.open('GET', url, true);
r.responseType = 'document';
r.onload = function () {
if (r.readyState !== 4 || r.status !== 200) return;
// also handle error statuses
// now, r.response is a document:
const doc = r.response;
console.log(doc.querySelector('.foo').textContent);
};
r.send();
}
If you use the more modern fetch method, you'll have to convert the response text into a document explicitly, possibly with DOMParser. For example:
function get(url){
fetch(url)
.then(res => res.text())
.then((text) => {
const doc = new DOMParser().parseFromString(text, 'text/html')
console.log(doc.querySelector('.foo').textContent);
});
}
One of the nice things about DOMParser is that you can use it to convert any (valid) HTML string into a document.
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
add a comment |
You can either:
- assign the content to a hidden div. And then query the DOM.
- parse the response as XML and access the tags.
For 1:
ajaxRequest.onreadystatechange = function() {
if(ajaxRequest.readyState == 4) {
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
For 2, this might help:
https://stackoverflow.com/a/13460821/1364747
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
With XMLHttpRequest, it's pretty easy - with the responseType as document, the instiantiated XMLHttpRequest's .response property will contain the response document, which you can use ordinary DOM methods on. For example, the following code will select and log the text of the first element with a class name of foo from the response document:
function get(url){
var r = new XMLHttpRequest();
r.open('GET', url, true);
r.responseType = 'document';
r.onload = function () {
if (r.readyState !== 4 || r.status !== 200) return;
// also handle error statuses
// now, r.response is a document:
const doc = r.response;
console.log(doc.querySelector('.foo').textContent);
};
r.send();
}
If you use the more modern fetch method, you'll have to convert the response text into a document explicitly, possibly with DOMParser. For example:
function get(url){
fetch(url)
.then(res => res.text())
.then((text) => {
const doc = new DOMParser().parseFromString(text, 'text/html')
console.log(doc.querySelector('.foo').textContent);
});
}
One of the nice things about DOMParser is that you can use it to convert any (valid) HTML string into a document.
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
add a comment |
With XMLHttpRequest, it's pretty easy - with the responseType as document, the instiantiated XMLHttpRequest's .response property will contain the response document, which you can use ordinary DOM methods on. For example, the following code will select and log the text of the first element with a class name of foo from the response document:
function get(url){
var r = new XMLHttpRequest();
r.open('GET', url, true);
r.responseType = 'document';
r.onload = function () {
if (r.readyState !== 4 || r.status !== 200) return;
// also handle error statuses
// now, r.response is a document:
const doc = r.response;
console.log(doc.querySelector('.foo').textContent);
};
r.send();
}
If you use the more modern fetch method, you'll have to convert the response text into a document explicitly, possibly with DOMParser. For example:
function get(url){
fetch(url)
.then(res => res.text())
.then((text) => {
const doc = new DOMParser().parseFromString(text, 'text/html')
console.log(doc.querySelector('.foo').textContent);
});
}
One of the nice things about DOMParser is that you can use it to convert any (valid) HTML string into a document.
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
add a comment |
With XMLHttpRequest, it's pretty easy - with the responseType as document, the instiantiated XMLHttpRequest's .response property will contain the response document, which you can use ordinary DOM methods on. For example, the following code will select and log the text of the first element with a class name of foo from the response document:
function get(url){
var r = new XMLHttpRequest();
r.open('GET', url, true);
r.responseType = 'document';
r.onload = function () {
if (r.readyState !== 4 || r.status !== 200) return;
// also handle error statuses
// now, r.response is a document:
const doc = r.response;
console.log(doc.querySelector('.foo').textContent);
};
r.send();
}
If you use the more modern fetch method, you'll have to convert the response text into a document explicitly, possibly with DOMParser. For example:
function get(url){
fetch(url)
.then(res => res.text())
.then((text) => {
const doc = new DOMParser().parseFromString(text, 'text/html')
console.log(doc.querySelector('.foo').textContent);
});
}
One of the nice things about DOMParser is that you can use it to convert any (valid) HTML string into a document.
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
With XMLHttpRequest, it's pretty easy - with the responseType as document, the instiantiated XMLHttpRequest's .response property will contain the response document, which you can use ordinary DOM methods on. For example, the following code will select and log the text of the first element with a class name of foo from the response document:
function get(url){
var r = new XMLHttpRequest();
r.open('GET', url, true);
r.responseType = 'document';
r.onload = function () {
if (r.readyState !== 4 || r.status !== 200) return;
// also handle error statuses
// now, r.response is a document:
const doc = r.response;
console.log(doc.querySelector('.foo').textContent);
};
r.send();
}
If you use the more modern fetch method, you'll have to convert the response text into a document explicitly, possibly with DOMParser. For example:
function get(url){
fetch(url)
.then(res => res.text())
.then((text) => {
const doc = new DOMParser().parseFromString(text, 'text/html')
console.log(doc.querySelector('.foo').textContent);
});
}
One of the nice things about DOMParser is that you can use it to convert any (valid) HTML string into a document.
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
answered Nov 16 '18 at 23:13
CertainPerformanceCertainPerformance
100k166291
100k166291
add a comment |
add a comment |
You can either:
- assign the content to a hidden div. And then query the DOM.
- parse the response as XML and access the tags.
For 1:
ajaxRequest.onreadystatechange = function() {
if(ajaxRequest.readyState == 4) {
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
For 2, this might help:
https://stackoverflow.com/a/13460821/1364747
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
add a comment |
You can either:
- assign the content to a hidden div. And then query the DOM.
- parse the response as XML and access the tags.
For 1:
ajaxRequest.onreadystatechange = function() {
if(ajaxRequest.readyState == 4) {
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
For 2, this might help:
https://stackoverflow.com/a/13460821/1364747
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
add a comment |
You can either:
- assign the content to a hidden div. And then query the DOM.
- parse the response as XML and access the tags.
For 1:
ajaxRequest.onreadystatechange = function() {
if(ajaxRequest.readyState == 4) {
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
For 2, this might help:
https://stackoverflow.com/a/13460821/1364747
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
You can either:
- assign the content to a hidden div. And then query the DOM.
- parse the response as XML and access the tags.
For 1:
ajaxRequest.onreadystatechange = function() {
if(ajaxRequest.readyState == 4) {
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
For 2, this might help:
https://stackoverflow.com/a/13460821/1364747
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
answered Nov 16 '18 at 23:15
TeddyTeddy
2,43411838
2,43411838
add a comment |
add a comment |
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