I'm making a simple game in javascript and I would like to .append() a box. To serve as a bullet. But I'm stuck. This is what I have
var existingdiv1 = document.getElementById('bullet');
and
$("#test").click(function() {
$("div").append([existingdiv1]);
});
It wont create additional "divs" when I press the button "#test".
You will have to select the existing div (I guess this is the bullet?). Then append it.
Here's and example:
Working Demo
Javascript:
$("#test").click(function(){
$("#appendToThis").append($('#bullet').html());
});
Html:
<input id="test" type="button" value="click" />
<div id="appendToThis"></div>
<div id="bullet"><div>BANG</div></div>
You will see the word "bang" be appended everytime you click. You can remove it by using the empty() method on the test div.
From the .append documentation:
If an element selected this way is inserted into a single location elsewhere in the DOM, it will be moved into the target (not cloned).
It seems like you want to clone [docs] the element first:
$("div").append($(existingdiv1).clone());
// or simpler
$("div").append($('#bullet').clone());
Note though that if you have multiple div elements on your page, $("div") will select all of them and the element you pass to .append will cloned automatically (as stated in the documentation).
You should append the element only to a single div.
I believe you wanted to add a new div to the existing div, so your code has been reversed the append order. and you need wrap the existing div by $('#bullet')
Try this
$('#bullet').append('<div></div>');
You can use .clone()
$("#test").click(function(){
$("#appendToThis").append($('#bullet').clone());
});
Related
I have problem with html() function of JQuery, which doesn't return updated css of the divs.
My divs are encapsulated in parent div. In JQuery, according to some logic, I add or remove css classes on divs through addClass() or removeClass().
I also change text content of some divs.
Later I want to use html() function to get whole html of updated parent html() and copy somewhere else.
html() function() get initial html and doesn't get ant changes :(
I founded this topic, but there is described solution to update values of form:
similar problem
Is there any similar solution to update css classes and div text contents, so html() function will pick up changes to them?
that because you save the html() to var, then you call the var instead using the new one. see the example
$(function(){
var first = $('p').html();
console.log(first); //first
$('p').html('second');
console.log(first); //first
console.log($('p').html()); //second
$('p').html('third');
console.log(first); //first
console.log($('p').html()); //third
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p>first</p>
I'm probably being especially dense about this, but I can't get an element to return using prev(). My basic HTML structure is:
<div>
<table></table>
</div>
<input type="button">
Where when I press the button, I want to get the previous element (the div element). To achieve this my button has a function attached to it with
var nearestDiv = $(this).prev();
When I've checked the contents of nearestDiv in the console it appears to be some kind of JQuery object rather than a HTML div. I've tried popping .val() at the end of .prev() but this comes back empty. How can I get the div element?
Note that my button is generated on the fly and doesn't have anything which identifies it.
you need to use jquery get function, to get a native html object and not the jquery wrapper:
$("input").on("click",function(){
console.log("jquery wrapper:",$(this).prev());
console.log("native html div object:",$(this).prev().get(0));
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>
<table></table>
</div>
<input type="button">
If your html structure is same as you provided in the question, it will definitely return the div element. Note that there is no val() method for div element, you need to either use .html() or .text() inorder to get the contents.
$("input[type='button']").click(function () {
var div = $(this).prev();
alert(div.html());
alert(div.text());
});
Fiddle
You need to give .text() or .html() for standard HTML Elements. So your code should be:
var nearestDiv = $(this).prev().html();
var nearestDiv = $(this).prev().text();
So I try to select a div within another div. My html goes like this:
<div id="Stage_game_page1"><div id="cube0">[...]</div><div id="cube1">[...]</div></div>
I want to select my #cube0 within my Stage_game_page specifically, with jQuery or JS.
The goal of the selection is to use it in an loop.
I tried :
var count =$("#Stage_game_page").children().length;
for(i=0; i<count;i++){
$("#Stage_game_page")$("#cube"+i)[...]
}
I don't understand what I'm doing wrong.
var count =$("#Stage_game_page").children().length;
for(i=0; i<count;i++){
$("#cube"+i);
}
This is sufficient to select the "#cube0"/"#cube1"/"#cube2" etc. especially since ids are always unique. To answer the question $("#cube0", "#Stage_game_page")... that is how you select a div in another div
The id attribute should only be used once! I see above that you're using id="cube0" twice. If you want your divs to be recognized in multiple instances, use a class instead (the . instead of the #). Using the same id twice will probably break your script.
I believe for your html, you could use id "cube0", "cube1", etc., as long as you're ok with entering them manually. That should work for the loop you'd like to use.
Loops through each div that starts with the id cube inside Stage_game_page1
$("#Stage_game_page1 > div[id^='cube']").each(function () {
alert($(this).html());
});
JSFiddle
Child Selctor
Starts with Selector
use each() for loop.
$('#Stage_game_page1').children().each(function(index) {
// your code here with index starts from 0
});
or this using jquery attribute starts with selector
$('#Stage_game_page1').find('[id^="cube"]').each(function(index) {
// your code here
});
You need to use .find() or .children() or the like.
The correct jQuery usage would be
$("#Stage_game_page").find('#cube'+i)
to find a div with that id inside the container #stage_game_page
You have duplicate cube0 in your html code..
and i think the look should contain something like that:
$("#cube"+i)[...]
One another solution is:
$("#Stage_game_page1 div[id='cube0']")
I have a textarea and button that I need to replace with new ones. I use replaceWith in jquery to achieve this but it seems that I'm doing it wrongly.
This is my javascript:
<script>
$(document).ready(function () {
$(document).on('click', 'div', function(){
$('textarea, button').replaceWith('<textarea>New</textarea><button>Old</button>');
});
});
</script>
My HTML:
<textarea>Old</textarea>
<button>Old</button>
<div>Replace</div>
Clicking the Replace div should replace both the Old text area and the button with the new ones but for some reason it leads to displaying 2 text areas and 2 buttons.
Tried using $('textarea', 'button') but this does nothing at all.
If per your comments elsewhere you cannot split the two elements apart for text purposes, then alternatively you should ensure that both existing elements share a common parent (e.g. a <div>) and then replace the contents of that parent:
<div id="parent">
<textarea>Old</textarea>
<button>Old</button>
</div>
$('#parent').empty().append(newContent);
Alternatively if you cannot change the downloaded HTML, then within the event handler if you can assume that there are no other matching elements between the "replace" div and the original content:
$(this).prevAll('button').first().remove();
$(this).prevAll('textarea').first().remove();
$(this).before(newContent);
You should separate the two out, to avoid trying to replace both in the same statement.
$('textarea').replaceWith('<textarea>New</textarea>');
$('button').replaceWith('<button>Old2</button>');
I've been using jQuery for a while but this is a new one. A simplified example:
HTML
<div class='custom'></div>
<div class='custom'></div>
<div class='custom'></div>
jQuery:
var $customElems = $('.custom'),
$spanOuter = $('<span class="outer"/>'),
$spanInner = $('<span class="inner"/>');
$customElems.each( function() {
$(this).wrap($spanOuter).after($spanInner);
});
JSFiddle:
http://jsfiddle.net/a3ZK8/
I would have expected the 'inner' span to be added to all three elements in the selection but it gets always inserted into the last one only (no matter how many). I tried it with .before(), with and without the chaining, same result. What am I missing??
The problem is you are using a reference to a jQuery object.
Hence you keep moving the object reference around within each iteration.
If you have no events attached or no need for the span to be a jQuery object then just pass the parameter as a HTML string literal instead of an object reference
Cloning a jQuery object that doesn't need to be a jQuery object in the first place is just redundant processing and unnecessary overhead.
Change your jQuery object to a string similar to this:
spanInnerString = '<span class="inner"/>';
and your method like this:
$(this).wrap($spanOuter).after(spanInner);
The result is:
<span class="outer"><div class="custom"></div><span class="inner"></span></span>
<span class="outer"><div class="custom"></div><span class="inner"></span></span>
<span class="outer"><div class="custom"></div><span class="inner"></span></span>
DEMO - Passing parameter as HTML string
Off course, the same goes for the outer span. Don't create jQuery objects unless you have to.
If you must use a jQuery object because you want to attach events to the span or similar, than cloning is the way to go, though make sure you use clone(true, true) then to also clone the attached events.
You need to clone the element. Otherwise, after() will relocate the same element 3 times, which results in it being attached to only the last looped element.
$customElems.each(function () {
$(this).wrap($spanOuter).after($spanInner.clone());
});
Demo: Fiddle
You might ask, "Why would wrap() work?" That's because 'wrap()' internally clones the element.
You're moving the same span from place to place. If you acted on all three divs at once, jquery will instead clone the span.
http://jsfiddle.net/a3ZK8/1/
var $customElems = $('.custom'),
$spanOuter = $('<span class="outer"/>'),
$spanInner = $('<span class="inner"/>');
$customElems.wrap($spanOuter).after($spanInner);
From the documentation for .after:
Important: If there is more than one target element, cloned
copies of the inserted element will be created for each target except
for the last one.
which means the last element will always get the original, while all other selected elements will get a clone. That's why when you acted on one element at a time, it simply moved the same span around.