Question

jquery timer script

Asked by: designersx

i am showing the images from database. i have to show only 4 images at a time. now i want to apply jquery timer of 3 seconds. after 3 seconds next images should be shown.

Make sure the area in which i am showing the images will remain fixed, so the previous image will get hidden and new images will be shown. this should be continued ....., so i want a script of that

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Asked On: 2009-08-17 at 07:45:35ID24658294
Topics: Jquery
,; PHP Scripting Language
,; JavaScript
Participating Experts: 3
Points: 500
Comments: 26

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Answers

by: designersxPosted on 2009-08-17 at 07:55:59ID: 25115018

i am using the below code which is very simple,
Timer concept is applied in this very well. it shows 1 image at a time in a div. but i want to show 4 images in a div. just do that for me, i will have my job done.

<html>
<head>
  <script src="jquery.js"></script>
  <script>
    function swapImages(){
      var $active = $('#myGallery .active');
      var $next = ($('#myGallery .active').next().length > 0) ? $('#myGallery .active').next() : $('#myGallery img:first');
      $active.fadeOut(function(){
      $active.removeClass('active');
      $next.fadeIn().addClass('active');
      });
    }
 
    $(document).ready(function(){
      // Run our swapImages() function every 5secs
      setInterval('swapImages()', 1000);
    });
  </script>
  <style>
    #myGallery{
      position:relative;
      width:400px; /* Set your image width */
      height:300px; /* Set your image height */
    }
    #myGallery img{
      display:none;
      position:absolute;
      top:0;
      left:0;
    }
    #myGallery img.active{
      display:block;
    }
  </style>
</head>
<body>
  <div id="myGallery">
    <img src="images/1.jpg" class="active" />
    <img src="images/2.jpg" />
    <img src="images/3.jpg" />
	<img src="images/4.jpg" />
	<img src="images/5.jpg" />
	<img src="images/6.jpg" />
	<img src="images/7.jpg" />
	<img src="images/8.jpg" />
	<img src="images/9.jpg" />
	<img src="images/10.jpg" />
  </div>
</body>
</html>

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by: designersxPosted on 2009-08-17 at 08:00:23ID: 25115069

you only need to modify the line 7 where it is written first. Because of this, only 1 image is showing in a div but i want to show 4 images in a div with spaces between the images.

by: profyaPosted on 2009-08-17 at 08:11:15ID: 25115195

Hi designerx,
quick and dirty, group your images in divs, each four in a div and then rotate on divs in stead of images.

by: designersxPosted on 2009-08-17 at 08:20:08ID: 25115295

r u saying me to take different divs??
don't get you clearly?

by: profyaPosted on 2009-08-17 at 08:25:57ID: 25115362

Something like:

<div id="myGallery">
	<div class="active">
		<img src="images/1.jpg" />
		<img src="images/2.jpg" />
		<img src="images/3.jpg" />
		<img src="images/4.jpg" />
	</div>
	<div>
		<img src="images/5.jpg" />
		<img src="images/6.jpg" />
		<img src="images/7.jpg" />
		<img src="images/8.jpg" />
	<div>
		<img src="images/9.jpg" />
		<img src="images/10.jpg" />
		<img src="images/1.jpg" />
		<img src="images/2.jpg" />
	<div>
</div>

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by: profyaPosted on 2009-08-17 at 08:26:31ID: 25115371

In stead of moving on images, you can move on divs.

by: designersxPosted on 2009-08-17 at 08:35:05ID: 25115468

doing like this but it does not show anything, why??

<div id="myGallery">
	<div class="active">
			<img src="images/1.jpg" />
			<img src="images/2.jpg" />
			<img src="images/3.jpg" />
			<img src="images/4.jpg" />
	</div>
	<div>
			<img src="images/5.jpg" />
			<img src="images/6.jpg" />
			<img src="images/7.jpg" />
			<img src="images/8.jpg" />
	</div>		
	<div>
			<img src="images/9.jpg" />
			<img src="images/10.jpg" />
			<img src="images/1.jpg" />
			<img src="images/2.jpg" />
	</div>
</div>

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by: designersxPosted on 2009-08-17 at 08:42:56ID: 25115561

you have not closed the divs and with that i don't see anything on the front end.

by: designersxPosted on 2009-08-17 at 08:47:14ID: 25115600

can u give me any other link if it does not working? means if u find it wrong.

by: hieloPosted on 2009-08-17 at 08:59:34ID: 25115728

try your ORIGINAL markup

<html>
<head>
  <script type="text/javascript" src="http://code.jquery.com/jquery-latest.js"></script>
  <script type="text/javascript" >
 
    function swapImages(){
 
		for( var i=0, limit = $("#myGallery>img").size(); i < limit; i=((i+4)%limit) )
		{				
			if( $("#myGallery > img:eq("+i+")").is(".active") )
			{
				$("#myGallery > img").slice(i,i+4).each(function(){ $(this).removeClass("active");});
				i= (i+4)%limit) ;
				if(i%4)i=0;
 
				$("#myGallery>img").slice(i,i+4).each(function(){$(this).addClass("active");});
				break;
			}
		}
    }
 
    $(document).ready(function(){
      // Run our swapImages() function every 5secs
      setInterval(swapImages, 1000);
    });
  </script>
  <style>
    #myGallery{
      position:relative;
      width:400px; /* Set your image width */
      height:300px; /* Set your image height */
    }
    #myGallery img{
      display:none;
      position:absolute;
      top:0;
      left:0;
    }
    #myGallery img.active{
      display:block;
    }
  </style>
</head>
<body>
  <div id="myGallery">
    <img src="images/1.jpg" class="active" />
    <img src="images/2.jpg"  class="active"/>
    <img src="images/3.jpg" class="active" />
     <img src="images/4.jpg"  class="active"/>
     <img src="images/5.jpg" />
     <img src="images/6.jpg" />
     <img src="images/7.jpg" />
     <img src="images/8.jpg" />
     <img src="images/9.jpg" />
     <img src="images/10.jpg" />
 
  </div>
  <div id="target"></div>
</body>
</html>

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by: designersxPosted on 2009-08-17 at 09:09:08ID: 25115822

still it shows me 1 image but i want to display 4 images with space between them and then timer should run.
this error is coming.

1.PNG
- 140 KB
1

by: dachusaPosted on 2009-08-17 at 09:13:05ID: 25115855

You need to change your css to get the below to work...

change the img in these to div

#myGallery img{
display:none;
position:absolute;
top:0;
left:0;
}
#myGallery img.active{
display:block;
}

<div id="myGallery">
        <div class="active">
                        <img src="images/1.jpg" />
                        <img src="images/2.jpg" />
                        <img src="images/3.jpg" />
                        <img src="images/4.jpg" />
        </div>
        <div>
                        <img src="images/5.jpg" />
                        <img src="images/6.jpg" />
                        <img src="images/7.jpg" />
                        <img src="images/8.jpg" />
        </div>          
        <div>
                        <img src="images/9.jpg" />
                        <img src="images/10.jpg" />
                        <img src="images/1.jpg" />
                        <img src="images/2.jpg" />
        </div>
</div>

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by: designersxPosted on 2009-08-17 at 09:20:18ID: 25115927

this is what i am using and i am not seeing any image on the front end. image path is right.

<html>
<head>
  <script type="text/javascript" src="jquery.js"></script>
  <script type="text/javascript" >
 
function swapImages(){
 
	for( var i=0, limit = $("#myGallery>img").size(); i < limit; i=((i+4)%limit) )
	{                               
		if( $("#myGallery > img:eq("+i+")").is(".active") )
		{
			$("#myGallery > img").slice(i,i+4).each(function(){ $(this).removeClass("active");});
			i= (i+4)%limit ;
			if(i%4)i=0;
 
			$("#myGallery>img").slice(i,i+4).each(function(){$(this).addClass("active");});
			break;
		}
	}
}
 
    $(document).ready(function(){
      // Run our swapImages() function every 5secs
      setInterval(swapImages, 1000);
    });
  </script>
  <style>
    #myGallery{
      position:relative;
      width:400px; /* Set your image width */
      height:300px; /* Set your image height */
    }
    #myGallery img{
      display:none;
      position:absolute;
      top:0;
      left:0;
    }
    #myGallery img.active{
      display:block;
    }
  </style>
</head>
<body>
  <div id="myGallery">
        <div class="active">
                        <img src="images/1.jpg" />
                        <img src="images/2.jpg" />
                        <img src="images/3.jpg" />
                        <img src="images/4.jpg" />
        </div>
        <div>
                        <img src="images/5.jpg" />
                        <img src="images/6.jpg" />
                        <img src="images/7.jpg" />
                        <img src="images/8.jpg" />
        </div>          
        <div>
                        <img src="images/9.jpg" />
                        <img src="images/10.jpg" />
                        <img src="images/1.jpg" />
                        <img src="images/2.jpg" />
        </div>
</div>
  <div id="target"></div>
</body>
</html>

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by: hieloPosted on 2009-08-17 at 09:27:33ID: 25115990

change:
i= (i+4)%limit) ;

to:
i= (i+4)%limit;

by: hieloPosted on 2009-08-17 at 09:29:49ID: 25116016

>>i am not seeing any image on the front end
the images are on top of each other. Try changing:

    #myGallery img{
      display:none;
      position:absolute;
      top:0;
      left:0;
    }
 
to:
    #myGallery img{
      display:none;
      /*
position:absolute;
      top:0;
      left:0;
*/
    }

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by: designersxPosted on 2009-08-17 at 09:40:30ID: 25116100

timer is not working sir and only 1 image is shown.

am i doing any mistake??

<html>
<head>
  <script type="text/javascript" src="jquery.js"></script>
  <script type="text/javascript" >
 
function swapImages(){
 
	for( var i=0, limit = $("#myGallery>img").size(); i < limit; i=((i+4)%limit) )
	{                               
		if( $("#myGallery > img:eq("+i+")").is(".active") )
		{
			$("#myGallery > img").slice(i,i+4).each(function(){ $(this).removeClass("active");});
			i= (i+4)%limit;
			if(i%4)i=0;
 
			$("#myGallery>img").slice(i,i+4).each(function(){$(this).addClass("active");});
			break;
		}
	}
}
 
    $(document).ready(function(){
      // Run our swapImages() function every 5secs
      setInterval(swapImages, 1000);
    });
  </script>
  <style>
    #myGallery{
      position:relative;
      width:400px; /* Set your image width */
      height:300px; /* Set your image height */
    }
     #myGallery img{
      
     
position:absolute;
      top:0;
      left:0;
 
    }
    #myGallery img.active{
      display:block;
    }
  </style>
</head>
<body>
  <div id="myGallery">
        <div class="active">
                        <img src="images/1.jpg" />
                        <img src="images/2.jpg" />
                        <img src="images/3.jpg" />
                        <img src="images/4.jpg" />
        </div>
        <div>
                        <img src="images/5.jpg" />
                        <img src="images/6.jpg" />
                        <img src="images/7.jpg" />
                        <img src="images/8.jpg" />
        </div>          
        <div>
                        <img src="images/9.jpg" />
                        <img src="images/10.jpg" />
                        <img src="images/1.jpg" />
                        <img src="images/2.jpg" />
        </div>
</div>
  <div id="target"></div>
</body>
</html>

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by: hieloPosted on 2009-08-17 at 09:49:17ID: 25116164

you removed the wrong css definitions. Read my previous post carefully.

by: designersxPosted on 2009-08-17 at 09:54:35ID: 25116191

oh sorry for that.

sir but you r saying me to do like this and it does not show no image because we have simple done display:none;

#myGallery img{
      display:none;
      /*
		position:absolute;
      top:0;
      left:0;
*/
    }

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by: hieloPosted on 2009-08-17 at 10:13:33ID: 25116322

try:

     #myGallery img{
      
display:none;     
    }
    #myGallery img.active{
      display:block !important;
    }

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by: designersxPosted on 2009-08-17 at 10:19:24ID: 25116375

this is the final code i am using, and does not display me any image sir.

<html>
<head>
  <script type="text/javascript" src="jquery.js"></script>
  <script type="text/javascript" >
 
function swapImages(){
for( var i=0, limit = $("#myGallery>img").size(); i < limit; i=((i+4)%limit) ){                               
	if( $("#myGallery > img:eq("+i+")").is(".active") ){
		$("#myGallery > img").slice(i,i+4).each(function(){ $(this).removeClass("active");});
		i= (i+4)%limit;
		if(i%4)i=0;
		$("#myGallery>img").slice(i,i+4).each(function(){$(this).addClass("active");});
		break;
	}
}
}
    $(document).ready(function(){
      // Run our swapImages() function every 5secs
      setInterval(swapImages, 3000);
    });
  </script>
  <style>
    #myGallery{
      position:relative;
      width:400px; /* Set your image width */
      height:300px; /* Set your image height */
    }
    #myGallery img{
		display:none;     
    }
    #myGallery img.active{
      display:block !important;
    }
  </style>
</head>
<body>
  <div id="myGallery">
        <div class="active">
                        <img src="images/1.jpg" />
                        <img src="images/2.jpg" />
                        <img src="images/3.jpg" />
                        <img src="images/4.jpg" />
        </div>
        <div>
                        <img src="images/5.jpg" />
                        <img src="images/6.jpg" />
                        <img src="images/7.jpg" />
                        <img src="images/8.jpg" />
        </div>          
        <div>
                        <img src="images/9.jpg" />
                        <img src="images/10.jpg" />
                        
        </div>
</div>
  <div id="target"></div>
</body>
</html>

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by: designersxPosted on 2009-08-17 at 10:20:16ID: 25116383

when i comment line 29, it shows all the images.

by: hieloPosted on 2009-08-17 at 10:21:45ID: 25116388

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"
 "http://www.w3.org/TR/REC-html40/loose.dtd">
 
<html>
<head>
  <style type="text/css">
    #myGallery{
      position:relative;
      width:400px; /* Set your image width */
      height:300px; /* Set your image height */
    }
    #myGallery img{
      display:none;
	 /*
      position:absolute;
      top:0;
      left:0;*/
    }
    #myGallery img.active{
      display:block !important;
    }
  </style>
  <script type="text/javascript" src="http://code.jquery.com/jquery-latest.js"></script>
  <script type="text/javascript" >
 
    function swapImages(){
 
		for( var i=0, limit = $("#myGallery>img").size(); i < limit; i=((i+4)%limit) )
		{				
			if( $("#myGallery > img:eq("+i+")").is(".active") )
			{
				$("#myGallery > img").slice(i,i+4).each(function(){ $(this).removeClass("active");});
				i= (i+4)%limit;
				if(i%4)i=0;
 
				$("#myGallery>img").slice(i,i+4).each(function(){$(this).addClass("active");});
				break;
			}
		}
    }
 
    $(document).ready(function(){
      // Run our swapImages() function every 5secs
      setInterval(swapImages, 5000);
    });
  </script>
 
</head>
<body>
  <div id="myGallery">
    <img src="http://images.experts-exchange.com/xp/images/addToKnowledgebase.png" class="active" />
    <img src="http://images.experts-exchange.com/xp/images/breadCrumbDelimiter.gif"  class="active"/>
    <img src="http://images.experts-exchange.com/xp/images/bulletArrow.gif" class="active" />
     <img src="http://images.experts-exchange.com/xp/images/cancelButton.png"  class="active"/>
     <img src="http://images.experts-exchange.com/xp/images/closeButtonTiny.gif" />
     <img src="http://images.experts-exchange.com/xp/images/compHOFgn.gif" />
     <img src="http://images.experts-exchange.com/xp/images/compMyAcc.gif" />
     <img src="http://images.experts-exchange.com/xp/images/compMyFav.gif" />
     <img src="http://images.experts-exchange.com/xp/images/expertTools.png" />
     <img src="http://images.experts-exchange.com/xp/images/hallOfFameButton.png" />
 
  </div>
  <div id="target"></div>
</body>
</html>

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Select all Open in new window

by: designersxPosted on 2009-08-17 at 10:28:58ID: 25116444

yes all working now.

i need to change the format only. showing me the images in 1 row , means it is showing me in 4 rows. ]
see the snapshot.
i wanted 4 images to be displayed in 2 rows.

after this i request you to see this link,

http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/Q_24658541.html#a25116161

1.PNG
- 230 KB
1

by: hieloPosted on 2009-08-17 at 10:54:26ID: 25116668

>>i wanted 4 images to be displayed in 2 rows.

use:

  <style type="text/css">
    #myGallery{
      position:relative;
      width:400px; /* Set your image width */
      height:300px; /* Set your image height */
    }
    #myGallery img{
      display:none;
    }
    #myGallery img.active{
      display:block !important;
    }
  </style>
 
and modify your document.ready to this:
 
    $(document).ready(function(){
   		$( "#myGallery > img:even").css({"float":"left","clear":"both"});
      // Run our swapImages() function every 5secs
      setInterval(swapImages, 5000);
    });

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Select all Open in new window

by: designersxPosted on 2009-08-17 at 10:58:52ID: 25116710

ok, can i have the space between the respective images like they have sticked to one another.

sir can u see at this link , i want to discuss with you.

http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/Q_24658541.html#a25116161

by: hieloPosted on 2009-08-17 at 11:15:09ID: 25116832

>>can i have the space between the respective images
change the 50px to whatever suits your need:
$( "#myGallery > img:even").css({"float":"left","clear":"both","margin-right":"50px"});

>>can u see at this link
Not today. Sorry. I need to focus my attention to something else.

20120131-EE-VQP-002

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