how to call function in jquery from javascript - javascript

<script type="text/javascript" src="jscripts/jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
alert("funciton");
$(function(){
$.fn.gotof(){
alert("I am calling form jquery");
}
});
});
</script>
<input type="button" onclick="dofunc();">
<script type="text/javascript">
function dofunc(){
gotof();
}
</script>
how do i call gotof() that is present in jquery
and below is the code written over jsfiddle

There are a few errors in your code. Fixed it should look like this:
$.fn.gotof = function() { // has to be defined as a function, does not need to be inside a nested document ready function
alert("I am calling form jquery");
};
$(document).ready(function() {
alert("function is ready to use now");
});
function dofunc() {
$.fn.gotof(); // can call it using $.fn.gotof(), but it should really be called properly via a selector $('div').gotof();
}
http://jsfiddle.net/pSJL4/8/

Check out http://docs.jquery.com/Plugins/Authoring - this should answer your questior. You also are not defining your function correctly; instead of
$.fn.gotof(){
alert("I am calling form jquery");
}
you need
$.fn.gotof = function(){
alert("I am calling from jquery");
};

Related

JQuery working on .html but not on .php

I really thought at first that my code is not working.. but then i tried to create a very simple click event in a clean page and my jquery works on .html but when i open the .php on my XAMPP server it does nothing when it's clicked but it alerts the first one.
index.php
<html>
<head>
<script src="js/jquery-3.2.0.min.js"></script>
</head>
<body>
<?php include "includes/widgets/login.php"; ?>
</body>
</html>
login.php
<script type="text/javascript">
$(document).ready(function() {
alert("Something");
t = clicked();
$('#login').click(function() {
alert("Login Alert!");
});
function clicked() {
alert("I was clicked!");
}
});
</script>
<input type="button" value="Login" id="login" onclick="clicked()">
place the function clicked(); outside from the document.ready();
Change login code to following and it will work.
<script type="text/javascript">
$(document).ready(function() {
alert("Something");
t = clicked();
$('#login').click(function() {
alert("Login Alert!");
});
});
function clicked() {
alert("I was clicked!");
}
</script>
<input type="button" value="Login" id="login" onclick="clicked()">
Uncaught ReferenceError: clicked is not defined at HTMLInputElement.onclick (index.php:22)
This problem will also solved. I don't know properly but i thought that browser can't find function inside jquery.
JQuery create function when DOM is ready.But Control onclick property was read by browser while creating a page but they can't find the function becuse it was not ready yet
So I think procedure may be like :
1) Browser read the page and find the function.
2) JQuery ready event will fire.
Place your function like this:
<script type="text/javascript">
$(document).ready(function() {
alert("Something");
t = clicked();
$('#login').click(function() {
alert("Login Alert!");
});
});
function clicked() {
alert("I was clicked!");
}
</script>

How can I all a JS function in a browser console if the JS function is included in the HTML?

HTML:
<script type="text/javascript" src="scripts/js/script.js"></script>
JQuery function here:
$(document).ready(function() {
function rnmtn(){
console.log("red");
}
});
How can I call rnmtn(); in the console without it returning undefined - the script is not linked to an html object - I need it to function on its own.
Make your function
function rnmtn(){
console.log("red");
}
and call it in console by typing
rnmtn()
you need to make the function accessible in the window scope,
such like
$(document).ready(function() {
window.rnmtn = function rnmtn(){
console.log("red");
}
});

Run Function in jQuery on Page Load and Change

I have a function in jQuery that I would like to run on Page Load and on Change. How would I do this? I currently only have the change part..
$('input[name=INPUTNAME]').change(function(){
....
});
You can pass a function to your input change listener and also call the same function on document ready.
$(function() {
yourFunction();
$("input").change(yourFunction);
});
function yourFunction() {
alert("foo bar");
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" />
Try
$(document).on("change", "input[name=INPUTNAME]", function(){})
Reference
http://api.jquery.com/on/
(function($) {
$(document).ready(yourFunctionName());
$('input[name=INPUTNAME]').on('change', yourFunctionName());
function yourFunctionName(){
//function body
}
})(jQuery);
Try this. Basically, we have a function and call it on input change and when the document is ready.

call javascript function on link click

I want to set a cookie value when user clicks on a link.
I am using the following code:
<script src="http://code.jquery.com/jquery-1.7.2.min.js"></script>
<script src="../../Scripts/jquery.cookie.js" type="text/javascript"></script>
<script type="text/javascript">
$(document).ready(function () {
function changeLang(lang) {
$.cookie('myCulture', lang);
window.location.reload();
return false;
}
});
</script>
HTML
<img src="../../Content/images/danishFlag.png" height="35px" width="35px"/>
<img src="../../Content/images/swedishFlag.png" height="35px" width="35px"/>
It looks like very simple code, but when I click on the link, there is an error in the browser.
It says.
ReferenceError: changeLang is not defined
changeLang("da-DK");
Where am I doing wrong??
you are defining the function inside the document ready scope, so it's not global therefore not available in the global scope
define it as a global simply by removing the var declaration or using window.changeLang = function
$(document).ready(function () {
changeLang = function(lang) {
document.cookie = 'myCulture' + lang;
window.location.reload();
return false;
}
});
define function outside of the jquery block
$(document).ready(function () {
});
function changeLang(lang) {
$.cookie('myCulture', lang);
window.location.reload();
return false;
}
Try moving the ChangeLang function to outside of the Document Ready section.
Also, if running in FireFox with Firebug you can experiment with executing the function directly from the Console.
Good luck
Please do not attach js functions in html like this.
You're using jquery so just do this:
$(document).ready(function () {
function changeLang() {
$.cookie('myCulture', $(this).data('lang'));
window.location.reload();
return false;
}
$('.link').on('click', changeLang);
});
and then in html:
...
$(document).ready(function () {
});
function changeLang(lang) {
document.cookie = 'myCulture=' + lang;
window.location.reload();
return false;
}
This will work for you.

how to execute jquery code one by one?

how to execute jquery code one by one?
I want first do "function 1", when finish the job. then do "function 2"...
Thanks.
<script type="text/javascript">
jQuery(document).ready(function(){
$(document).ready(function(){
//function 2, jqeury.ajax
});
$(document).ready(function(){
//function 3, jqeury.json
});
$('#col').load("home.html");
//function 4, jqeury.load
});
});
</script>
<script type="text/javascript">
jQuery(document).ready(function(){
//function 1, a jquery slider plungin
});
</script>
You don't need so many document ready calls. One will suffice
If you mean you want to call each method after one has received the response from the AJAX calls you are making, put the code in the callback;
$(document).ready(function(){
one(); //call the initial method
});
function one(){
$.get('url', {}, function(data){
two(); //call two() on callback
})
}
function two(){
$.getJSON('url', {}, function(data){
three(); //ditto
})
}
function three(){
$('#selector').load('url');
}
The docs
http://api.jquery.com/jQuery.get/
http://api.jquery.com/jQuery.getJSON/
http://api.jquery.com/load/
Instead of using more than one document.ready() use callback functions as below.
<script type="text/javascript">
function ajaxfun() {
//function 2, jqeury.ajax
//in ajax callback call the jsonfun();
}
function jsonfun(){
//function 3, jqeury.json
//after json function add the next function in it callback.
}
</script>
<script type="text/javascript">
jQuery(document).ready(function(){
//function 1, a jquery slider plungin
//Call ajaxfun() here to execute second.
});
</script>
It looks like you are doing three ajax calls. Since jQuery 1.5, we now have a Deferred object (technically a jqXHR object) returned from ajax calls, so you can chain them like this:
$(function() { // this is a document ready function. it's all you need.
$.ajax(/* your ajax specs */).done(function() {
$.getJSON('somepath').done(function() {
$('#container').load('etc.');
});
});
});
use setTimeout function
function f1(para) {
// ...
// do work;
setTimeout(f2, 10);
}
function f2() {
// ...
// do work
setTimeout(f3, 10);
}
<script type="text/javascript">
jQuery(document).ready(function(){
function2(){ //declare what function 2 will do here
//function 2, jqeury.ajax
}
function3(){
//function 3, jqeury.json
}
$('#col').load("home.html");
//function 4, jqeury.load
});
});
</script>
<script type="text/javascript">
jQuery(document).ready(function(){
//function 1, a jquery slider plungin
function2(); // execute function 2 after 1
function3();
});
</script>

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