So I want to slide down some content that is loaded with jquery .load function
I have this where I load the content and then Hide it
function froth(){
$("#area").hide();
$("#area").load("test.html #enquiry1");
$("#area").hide();
}
Here I want to make sure that when i click the button it loads the document and then once its loaded then it slides down the content.
$(document).ready(function() {
$('#t1').on('click', function(){
froth();
$('#area').ready(function(){$("#area").slideDown('slow');})
});
});
Any ideas on how to do this.
I've been looking at the API's and it looks like i will probably have to use some ajax function instead... but i wanted to see if there was any other way.
.load accepts a callback as parameter http://api.jquery.com/load/
$("#area").load("test.html #enquiry1", function() {alert('done')});
use the callback function of load to make sure the load content is loaded first.. and then do the animation
try this
$(document).ready(function() {
$('#t1').on('click', function(){
froth();
});
});
function froth()
{
$("#area").hide();
$("#area").load("test.html #enquiry1",function(){
$(this).slideDown('slow')
});
}
OR
$(document).ready(function() {
$('#t1').on('click', function(){
$("#area").hide().load("test.html #enquiry1",function(){
$(this).slideDown('slow')
});
});
});
Related
When I load Bootstrap popver content with ajax, the popover is not showing.
Javascript:
var id = 1;
$.post("load.php?pageid",
{
pageid:id;
},
function(data,status){
document.getElementById("body").innerHTML=data;
});
HTML response:
hover for popover
<script>
$(function ()
{ $("#example").popover();
});
</script>
When I place the HTML response above directly in the <body id="body"></body> code, the popover works fine. I dont understand what is going wrong here.
The problem is that you're setting up the popover inside of the function $(). Rather than
$(function ()
{ $("#example").popover();
});
It should be just
$("#example").popover();
Or perhaps better
$(document).ajaxComplete(function() {
$("#example").popover();
});
The reason being that any function inside of $() is run when the document is first finished loading, when the "document ready" event is fired. When that code is pasted inside of the original HTML, it works because it's present when the document finishes loading.
When it's the response from an AJAX call, it won't run inside of $(), because the "document ready" event already fired some time ago.
with the help of jQuery you can initialize all things that needs to be initialized using
$(document).ajaxSuccess(function () {
$("[data-toggle=popover]").popover();
$("[data-toggle=tooltip]").tooltip();
// any other code
});
inspired from Olaf Dietsche answer
<script>$(function () { $('[data-toggle="tooltip"]').tooltip()});</script>
Add this at the end of whatever you are loading in with ajax. You should already have this somewhere to opt-in to the tooltip, but put it again to re-initialize the tooltip.
I'm using a mix of .ready() and .load() to execute my desired function.
jQuery(document).ready(function($) {
$("img").load(function() {
// Function goes here
});
});
As you can see, this waits for the DOM to be ready, then on each <img> load, it executes the code.
If I only had one image to load this would be simple.
But the problem is -- what if I have 10 images to be loaded? The function will be called 10 times due to each image loading one by one, and that's not a very efficient way to go about it just to achieve what I want.
So here's the question -- is there a more efficient way to wait for all images to load, then execute the function once?
You could do something like this to avoid having your function run multiple times.
jQuery(document).ready(function($) {
var nrOfImages = $("img").length;
$("img").load(function() {
if(--nrOfImages == 0)
{
// Function goes here
}
});
});
jQuery(window).load(function() {
alert("page finished loading now.");
});
jQuery(window).load(...) will be triggered after all content on the page has been loaded. This different from jQuery(document).load(...) which is triggered after the DOM has been loaded. I think this will solve your issue.
If anybody wants to know, my final result was this:
(function($) {
$(window).load(function(){
// Function goes here
});
})(jQuery);
that's because
jQuery(window).load(function($) {});
isn't a jQuery object, as referenced in this question:
Calling jQuery on (window).load and passing variable for 'No Conflict' code
Here is basically what I want to do:
I want to slideUp() a div with my content. I want to .load() an external web page (on the same server) in that div and .slideDown() that div.
For now here is what I have:
$(document).ready(function() {
$('a').click(function(){
$('.content').slideUp('1000');
$('.content').hide().load('about.html');
$(".content").slideDown('1000');
});
});
Basically here's what I get: the div .content hides itself, loads the about.html page, and appears. But no slideUps or slideDowns.
Anyone has an idea?
Sorry if this is a noob question, this is the first real time I'm trying js/jquery.
Thanks in advance.
That's because those are asynchronous actions. You have to continue execution in callbacks:
$('a').click(function(){
$('.content').slideUp('1000', function() {
this.hide().load('about.html', function() {
this.slideDown('1000');
});
});
});
try:
$('a').click(function(){
var el = $('.content'), //cache content to avoid multiple calls
slidetime = 1000;
el.slideUp(slidetime,function(){ //slide up
el.hide().load('about.html',function(){ //afterwards, load
el.slideDown(slidetime); //afterwards, slide
});
});
});
you are passing string '1000' to slideUp or slideDown.
It should be number or string that it can take like 'slow'...
Try This
$(document).ready(function() {
$('a').click(function(){
$('.content').slideUp(1000);
$('.content').hide().load('index2.html');
$(".content").slideDown(1000);
});
});
Cheers :)
My current solution gets some content via AJAX, Inserts it into a DIV then hides it and fade it in.
The issue with that is the content contains images and an iframe, The solutions flickers and the fadeIn doesn't look smooth.
What i'd like to do is to only fadeIn after the iframe and pictures are fully loaded.
How to do that?
This will wait until all child content has finished loading:
$("#wrapperDivID").load(function (){
// do something, e.g.
$(this).fadeIn();
});
See .load().
In your specific case the following should work:
$('#content').load('/echo/html/',data, function() {
alert('fired');
});
I would change the duration for fadeIn so that it will take little long.
$("#content").load("getform.php"), function() {
$(this).fadeIn(1500); // or another bigger number based on your load time
});
If you know what all images are going to be in the page, Preload all those images in the page where you have the div "content", so that you will save some loading time.
If you use $.ajax to get the HTML you can wrap the resulting HTML like below:
$.ajax('some/path/to.html',{
'success':function(html){
var result = $(html).load(function(){
$('#content').fadeIn()
});
$('#content').html(result);
}
});
jsfiddle example
another way you can do this:
$( "#content" ).fadeOut(200, function() {
$(this).load( url, function() {
$(this).fadeIn(200);
});
});
change the fadeIn and Out times to whatever is good for you.
I would like to know a way that it will show a pre-load.gif in #page until the page is fully loaded.
How to do this?
$('#target').click(function() {
$('#page').html('first page load'); //will be replaced with .load
});
$('#target2').click(function() {
$('#page').html('second page load'); //will be replaced with .load
});
Please take a look into the working example
http://jsfiddle.net/EZj4R/1/
jQuery Load is a shorthand for ajax, and in the docs it says it accepts a complete function.
You could preload your loading gif before calling the load
$('#target').click(function() {
$('#page').html('<img src="loading.gif">').load('/some/url/');;
});
Check out this fiddle