jQuery UI Widgets › Forums › Editors › ScrollBar, Slider, BulletChart, RangeSelector › jqxslider
This topic contains 3 replies, has 3 voices, and was last updated by Dimitar 9 years ago.
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Authorjqxslider Posts
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Hello sir,
i have used jqxslider in my code and i have define max=32 and min=1 value but i have need shows ticks and values in this order and position 1,2,4,8,16,32.
how can solve this problem.please tell me.
Thanks & Regards
Priyanka jainHello Priyanka jain,
This scenario is not supported by jqxSlider. However, you can at least hide the unneeded ticks as shown in the following example:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <link rel="stylesheet" href="../../jqwidgets/styles/jqx.base.css" type="text/css" /> <link rel="stylesheet" href="../../jqwidgets/styles/jqx.summer.css" type="text/css" /> <script type="text/javascript" src="../../scripts/jquery-1.10.2.min.js"></script> <script type="text/javascript" src="../../jqwidgets/jqxcore.js"></script> <script type="text/javascript" src="../../jqwidgets/jqxbuttons.js"></script> <script type="text/javascript" src="../../jqwidgets/jqxslider.js"></script> </head> <body> <div id='content'> <script type="text/javascript"> $(document).ready(function () { $("#jqxslider").jqxSlider({ theme: 'summer', value: 2, min: 1, max: 32, mode: "fixed", step: 1, ticksFrequency: 1 }); var upperTicks = $("#jqxslider .jqx-slider-tickscontainer:eq(0)").children(); var lowerTicks = $("#jqxslider .jqx-slider-tickscontainer:eq(1)").children(); for (var i = 0; i < upperTicks.length; i++) { if (i != 0 && i != 1 && i != 3 && i != 7 && i != 15 && i != 31) { $(upperTicks[i]).css("display", "none"); $(lowerTicks[i]).css("display", "none"); }; }; }); </script> <div id='jqxslider'> </div> </div> </body> </html>
Best Regards,
DimitarjQWidgets team
http://www.jqwidgets.com/Hi,
I wanted to use this solution to change the color and width of the ticks in position 0 & 10.
For some reason – even though the code is executed – and I get the “children” ticks it has no affect on my slider .Any idea? (I have several sliders in my page).
Thanks
Hello assaf.frank123,
Here is how to change the width and colour of the displayed ticks in the example:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <link rel="stylesheet" href="../../jqwidgets/styles/jqx.base.css" type="text/css" /> <link rel="stylesheet" href="../../jqwidgets/styles/jqx.summer.css" type="text/css" /> <script type="text/javascript" src="../../scripts/jquery-1.10.2.min.js"></script> <script type="text/javascript" src="../../jqwidgets/jqxcore.js"></script> <script type="text/javascript" src="../../jqwidgets/jqxbuttons.js"></script> <script type="text/javascript" src="../../jqwidgets/jqxslider.js"></script> </head> <body> <div id='content'> <script type="text/javascript"> $(document).ready(function () { $("#jqxslider").jqxSlider({ theme: 'summer', value: 2, min: 1, max: 32, mode: "fixed", step: 1, ticksFrequency: 1 }); var upperTicks = $("#jqxslider .jqx-slider-tickscontainer:eq(0)").children(); var lowerTicks = $("#jqxslider .jqx-slider-tickscontainer:eq(1)").children(); for (var i = 0; i < upperTicks.length; i++) { if (i != 0 && i != 1 && i != 3 && i != 7 && i != 15 && i != 31) { $(upperTicks[i]).css("display", "none"); $(lowerTicks[i]).css("display", "none"); } else { $(upperTicks[i]).css({ "width": 2, "background-color": "Blue" }); $(lowerTicks[i]).css({ "width": 2, "background-color": "Blue" }); } }; }); </script> <div id='jqxslider'> </div> </div> </body> </html>
Best Regards,
DimitarjQWidgets team
http://www.jqwidgets.com/ -
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