i,m trying to make a map who show me as position A and a target adress as point B.I have made it so i can choose adress a and adress b from a dropdown but i want to automaticly load my position as possition A then choose position B from a dropdownlist. How can i do this ?
#navigation li is the parent element, which is positioned relative.The ul element above that is also position relative. I previously tested a click function and was able to confirm I was getting the correct position back, so now I just need to set the CSS property correctly for all of those links.The reason I want to do this is I have a set of links that appear over a photo of a city skyline. When you hover over those items, I want them to be given a background image that is a blurred and lightened version of the same photo so it needs to line up (sort of like the tabs are made of frosted glass).
I'm new to javascript and am not sure why this works in firefox and not chrome. I am trying to create a script that keeps an object fixed horizontally while bing positioned absolute vertically. if I replace the toPP variable in document.getElementById('fire').style.top = toPP; with say '50px' it will move the element down 50 pxs, but how I have it currently it doesn't do anything in chrome
<script type="text/javascript" > window.onscroll = function() { if( window.XMLHttpRequest ) { var x = 0 -document.documentElement.scrollTop; var toP = String(x); var toPP = toP + "px";
I have a HTML file that has a lot of content. In it I have a <div id="d1">somecontent</div> block. Would it be possible for me to know the position of this d1 (absolute or relative) displayed on the page / window if not setting it first ?
I have a div inside another div, I want to keep inner div centered vertically at the visible portion of outer div. Say the outer div is 3000px in height, most of the monitors can not display all of it, only a portion will be seen, I want the inner div to be centered in the view port, if users scroll down the browser window, they will see another portion of outer div but the inner div should still be centered vertically.
I want to do something but here is an example. (That explain much more) : Here you have a toolbar called content-header. It change from position:static to position:fixed if it leave the screen. (whan you scroll down) I want to do the same thing with my first <tr> : That way on a my long table you will never loose the title of each columns. I tried to copy this website by using his stuff but it didn't work.
I'm using a div with contenteditable=true, and I need to know where the caret is for operations like splitting the html contents of that div where the caret is. Ultimately, if a piece of text is selected, I'll want to split that html content on the selection start / end.jCaret only seems to work for textareas.Is this possible in a div?
Do you know how to calculate position of image to move it to vertical center of screen? I mean especially the general problem of scrolling. The image doesn't change its absolute position, but when user scrolls or views a page with images, so he sees how the image gets more and more close to the top edge of screen, till the distance is 0. And now imagine the image is positioned absolute, 1000 from top. And on the screen it is at the position 0. How can I calculate its position from the top of screen?
I am trying to build this interface where there are icons and onmouseover, it displays menu choices for the corresponding. Instead of fixing the position of the menu box, is there a way to specify the left and top of that, relative to the icon that is being hovered? Here is part of my code:
I want to use the .css of jquery to take lets say 5 buttons css position absolute left value. I want increase this value. So it will move the buttons to the right. I have a button that when clicked that it will run this jquery code. Where the 5 buttons moves to the right. I want to do this all at ones. In otherwords I want to put $("#1_button,#2_button,#3_button").css("left": function (value) { code to increase left position value}); I want to do something like that. Is it possible? and how would one do it. what I want it to do is grab all the buttons left position value and then add numbers to it to increase all buttons left position shifting all to the right. So if I add 10px I want all buttons to increase by 10px causing it to move 10px to the right.
I'm using jquery.animate() to scroll a certain DIV out of my screen at the top. Now I want that same DIV to fade out when it reaches a position which is 150px from the top (so before actually sliding off the screen). How do I achieve this? The code I have for the animation itself is as follows:
JQuery - Event triggered by div position Is it possible to use JQuery to trigger an event from the position of a div in relation to the browser window. I have this demo pageHEREthat contains three div's positioned below each other.
When I scroll down the page I want an alert to say "Div one in view", "Div two in view" etc. I know more than one div could be visible in the browser window but I wanted it to alert the one in the centre of the window. I was thinking it would be best to fire an event when the div is a set distance from the top of the browser window - I don't know where to start with it, or if using the distance from the top of the browser is the best way.
I'm making a dynamic div in a page but I'm having some doubts:-I have an input text field and i need to get it's position (left andtop) to make the div appear near this input. $.(':INPUTNAME').css('left') returns me zero like 'top' does.Is there a way to get left/top position from a static input text field
I need to change the position for a group as reaction for a mouse movement so I wrote this: this.dragged.attr('transform').baseVal.getItem(0).setTranslate(x, y) But this does not integrate well with the animation this.dragged.animate({svgTransform:'translate('+(-x)+' '+(-y)+')'},delay)
I'm using the following code to get tooltips to appear for elements within my SVG DOM:
function tooltip_event_for_svg_node(svgDoc, nodeID) { var node_obj = svgDoc.getElementById(nodeID); $(node_obj).tooltip({ delay: 0, track: true, [Code]...
The SVG object is defined as follows:<object id="svg_obj" data="svg_graph.svg" type="image/svg+xml" /> and there are <div> objects which correspond appropriately to each of the href attribute values on the elements within the SVG source (svg_graph.svg). The tooltips show up as expected--except their coordinates are extremely offset up and to the left from the pointer, instead of sitting right and below the pointer as expected. I discovered that when I removed HTML from the page source, this distance shrunk. When I removed everything except the <object> (and the appropriate js) from the HTML, the tooltip was in the right location. When I used tooltip in other (non-SVG) areas of the page, I didn't have this issue.
I have this table on my html page: <table id="t_menu_options" width="125" border="1" cellspacing="0" bordercolor="#F9F9F9"> <tr><td colspan="2"><font size="-2" face="Verdana"><strong>Options...</strong></font></td> </tr><tr><td width="15%">></td> <span id="edit"><td><font size="-2" face="Verdana"><span id="et">Edit</span></font></td></span> </tr><tr><td width="15%">></td> <span id="delete"><td><font size="-2" face="Verdana"><span id="dt">Delete</span></font></td></span> </tr></table>
I start the page by hiding the above html by using $('#t_menu_options').hide(); And that works.
Now upon some user mouseclick event, I want to show the html and move it to a position where the user clicked: function showTaskOptionsMenu(e) { alert(e.pageX+" "+e.pageY); $('#t_menu_options').css({'left':e.pageX+'px','top':e.pageY+'px'}).show(); } The alert command works as expected. however, the html is simply showed at the bottom of the page and is not moved to the desired location.
Using Jquery, and I'm trying to make a element positioned either from the top or bottom based on where it is in a grid.
Code JavaScript: if (position.top+width > container_height) { var position_type = "bottom"; } else { var position_type = "top"; }
And then apply it like so Code Javascript: .css({position_type:position.top+padding,"left":position.left+7,"width":width,"height":width}) This isn't working.
I have this photo gallery that I'm setting up where I've used some jquery to pop a magnified view of an image when the mouse is hovered over the thumbnail. As it is right now, it always positions the magnified image with the top-left corner at the x and y position of the mouse cursor. What I need it to do is if the image is going to be displayed beyond the viewport, that it would change its positioning to the corner that will show the whole image.