I have a gallery that has back/next links below the main image, but I want them to be on each side of the image instead.
My current code looks like this:
Code:
<center>
<br class='clearit' />
<div id="main_image_wrapper">
[Code]...
I'm just not sure how to best do this. And I want the links centered to the height of the image.
I am trying to recreate this functionality on my website where you can drag a background image around and when you get ot the edges of the image it bounces back to the edge of that corrosponding side. have a look at the site in question - [url]
So far i have recreated the top left and right edges using
Code:
Here is how i thought the bottom would work
Code:
I am trying to create an Image menu for a site I'm working on that is rather simple in essence. When the user hovers over a button I have the menu will move left or right depending. I have not got a great deal into it yet as I have become stuck, as I'm new I figured it would be easier to troubleshoot if I build up the program bit by bit.
[Code]...
I still can't find a way to fade in/out on my main gallery pic. It seems pic is loaded before the fades start .
gallery there:
[URL]
I'm trying to make a simple page where I can move a div around by hovering the mouse over the different links. My problem is it wont stop once activated.
Code:
I have an image wrapped inside a link tag.<a href="somepage.html"><img id="content" src="img/some.gif" /></a>
I want this .click target to be the link: $('a').click(function(e){
Instead, the target returned is the image [HTMLImageElement].
I have tried using closest()and currentTarget:
But they all still return the image, not the link.
I came up with the idea of making the links in the back of the circle (i've got a total of 10 links), invisible and put a globe in the middle (as background) - making it look like the links in the back go behind the globe and comes back from behind it again. I've tried to do this myself with .this.style.invisibility="hidden";
[Code]...
I have an ajax call that sends some data to process.php and return back the image name. In process.php I returned the name as follows:
echo "image.png";
The ajax call is:
var $chart = $( '#chart' );
var $test = $( '#test' );
[Code]....
so from test div i get the path printed right: (./images/image.png), but I don't get the image to be desplayed
I have a script which uses an array of numbers to represent their position in a div.its 5 by 5 like so:
tileMap = [
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[code]....
I am currently building a simple photogallery. The photogallery contains a box with a few photo's in it, lined up horizontally. The box where the photo's are placed in is using an overflow so that there are only shown a few photo's. Below the box i have two buttons, left and right. The idea is simple if you click the left button the photo's must go 50 px to the left and visa versa goes with the right. I have coded this but the photo's unfortunately won't move! Have i done something wrong? You can see my code here:
View 2 Replies View RelatedI need an image to move from outside the viewed space, from somwhere on the page where users cannot hav acces, let's say from x position of -439px to 0px, so that the image looks like entering the window. And I need to do this after the user clicks a piece of text that is already on the screen. How can I do that ? In what tag should I include the image ? where should I put de event handler/ listener ? I know I need to change the CSS atributes but how. I tried this and it didn't work in Firefox nor in IE. THE HTML FILE
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
[code]....
I'm trying to make something a little like seen on Google Maps where you can search the location in text box and the map corresponding to the location will be dispalyed. Can any one send me the Javascript for the following in which my image is paced in an iframe and the upon entering the loaction in a textbox it image must move to that position witin the iframe.
View 1 Replies View RelatedI have been tasked with using javascript to move an image of a bus across the screen from right to left. I have found some code that almost works, but I cannot get the bus to finish offscreen. I have tried changing the x-position to a negative number, but it doesn't seem to work. If there are any javascript gurus out there, take a look at the sample code at :[url]
How to modify the existing code to achieve the desired effect.
How to make a cloud like the websites below to move ?[URL]
View 5 Replies View RelatedManaged to get stuck over something that should be very simple.
I'm wanting to write the src attribute of an img tag on the client-side.
Doctype is XHTML 1.0 Transitional
In the head, I'm loading a js function that generates the src path. I want to call it and have it re-write the src in an img element with a specific id.
I have used draggable, and yes, the image can be moved by the user inside the div.when using the containtment:'parent' the image can not get out of the parent div,BUT I need some more functionality than that:In my app the image can be zoomed out / in When zooming out the image can grow more than the parent container then I need:
1. the image can keep growing with out affecting the dimensions of the container
2. the draggable function should still be usable so the user can see the rest of the image by dragging it.
Basically it's the same function that we use when we zoom in in photoshop and then we can move the image to see those areas that get out of the visible scope.
I'm trying to move an image with the following JS function, but it only seems to work when I remove the XHTML doctype from the top of the document.
Here's the javascript:
Code:
I read something about this on another forum and I think it has something to do with the declaration of obj.style.top, but I can't find a solution. I've tried creating a new variable that contains "document.getElementByID(obj)" and then using that variable to change the top of the image, but it still doesn't work.
The idea is move the foot to room no.1 .. but i want to see the foot moving on the red line to the room :(
Like this pic : [url]
I'm trying to write a function to determine whether or not an image exists.
Most people recommend setting the onload event handler. That works like a
charm except that side-effective actions seem to be impossible. I can, as
nearly as I can tell, run alert and opera.postError (I test on IE, firefox,
and opera by the way). I've tried using the 'this' while in the event
handler, setting this.name, test.name, window.document.getElementById(test's
id).name. I just can't seem to find any means of using side effects inside
the event handler.
Below are several examples of things I've tried so far:
I'd like to see and use a javascript to resize an image.
I've creating a site and I'd like that only this image will be resizable automatically based on resolutions (800x600, 1024x768, 1280x1024, ecc...)...
I'm trying to flip an image horizontally on the client side. Is it possible to rotate or change the orientation of an image by using JavaScript or CSS2?
View 3 Replies View RelatedI am trying to move an image down 20px each time the button is clicked, but I can only get it to move down once. I see what I am doing wrong, but I'm not sure how to fix it. I don't know how to store the updated position of the image in a variable.
View 5 Replies View RelatedI have an assignment doing something simple with javascript, but I'm lost because I can't figure out what I am doing wrong. This is the assignment. I'm using the code provided and modifying it as stated, but I cannot get the image to move. My textbook doesn't provide answers (it honestly has just 2 chapters devoted to javascript) that I can find. Here is my code
<?xml version="1.0"?>
<!DOCTYPE html PUBLIC "-W3CDTD XHTML 1.0 TransitionalEN"
"http:www.w3.org/TR/xhtmll/DTD/xhtmll-transitional.dtd">
[code]....
This is meant to be a very simple animation. Everything seems to be correct, but it doesn't seem to run. Basically it is a system where the image is 50px taller than the box it is being displayed in. When you hover over the image, the image is meant to move up via jquery to show the bottom (previously hidden) 50px and cut off the top 50px.
Here is my Site: [URL]
Here is my HTML:
Code:
<div id="container"><div>
<a class="none" href="<?php the_permalink() ?>"><img src="/themes/smo/portfolio/<?php echo get_post_meta($post->ID, 'portfolio_img',true).".gif"; ?>" alt="" /></a>
</div></div>
Here is the Jquery:
Code:
$(function(){
$("div.container div a").hover(function(){
$("img", this).stop().animate({top:"-51px"},{queue:false,duration:200});
}, function() {
$("img", this).stop().animate({top:"0px"},{queue:false,duration:200});
});
I think its something to do with the container class.
I have been using the JQuery Cycle plugin on my website and love it! It has so many options and seems customizable to do basically anything. My question is about the previous and next buttons. At the moment I am using next:'#next2, .pics img', prev:'#prev2', to make previous and next arrows underneath the image as well as use the image itself a "next" button. I would like to remove the arrows underneath the image and just use left and right halves of the image as a previous and next button. Previous being the left half and next being the right half of course. I've been searching for this being posted before and have not found any discussions on it. Also, the JQuery Slideviewer does what I'm trying to accomplish: JQuery Slideviewer Demo
if(zeroLeft >= pictWidth/2) { var uiprev = ui.parent().prev().find("a"); (jQuery(uiprev).length != 0)? uiprev.trigger("click") : ui.parent().parent().find("a:last").trigger("click"); } else { var uinext = ui.parent().next().find("a"); (jQuery(uinext).length != 0)? uinext.trigger("click") : ui.parent().parent().find("a:first").trigger("click"); }
Is there a way to incorporate this type of button in JQuery Cycle? It seems easy enough to make the image a next button I feel like dividing it between the two shouldn't be that much more difficult.
I didn't want to write iPhone there.have any idea on how to do this? But most important, it IS possible, right?
View 5 Replies View Related