+1 vote
in Other by
I need help with the following question:

So I want to place text in jquery using one of the following (open for suggestions): http://fancybox.net/home http://www.shadowbox-js.com/index.html http:// jacklmoore. kom/colorbox/example1/

But what I want to do is: if the next button is pressed the website should smoothly scroll to a predefined location:

something along the lines of:

$(document).ready(function() {

function filterPath(string) {

return string

.replace(/^\//,'')

.replace(/(index|default).[a-zA-Z]{3,4}$/,'')

.replace(/\/$/,'');

}

var locationPath = filterPath(location.pathname);

var scrollElem = scrollableElement('html', 'body');

$('a[href*=#]').each(function() {

var thisPath = filterPath(this.pathname) || locationPath;

if (  locationPath == thisPath

&& (location.hostname == this.hostname || !this.hostname)

&& this.hash.replace(/#/,'') ) {

  var $target = $(this.hash), target = this.hash;

  if (target) {

    var targetOffset = $target.offset().top;

    $(this).click(function(event) {

      event.preventDefault();

      $(scrollElem).animate({scrollTop: targetOffset}, 400, function() {

        location.hash = target;

      });

    });

  }

}

});

// use the first element that is "scrollable"

function scrollableElement(els) {

for (var i = 0, argLength = arguments.length; i <argLength; i++) {

  var el = arguments[i],

      $scrollElement = $(el);

  if ($scrollElement.scrollTop()> 0) {

    return el;

  } else {

    $scrollElement.scrollTop(1);

    var isScrollable = $scrollElement.scrollTop()> 0;

    $scrollElement.scrollTop(0);

    if (isScrollable) {

      return el;

    }

  }

 }

return [];

}

});

JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

1 Answer

0 votes
by
Have a look at this:

--> scoling

http://fancybox.net/api

Related questions

+1 vote
asked Feb 1, 2022 in Other by DavidAnderson
+1 vote
asked Feb 1, 2022 in Other by DavidAnderson
...