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Asked: 2026-05-10T23:10:32+00:00

This is in reference to the question previously asked The problem here is, each

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This is in reference to the question previously asked

The problem here is, each slider controls the other. It results in feedback.

How do I possibly stop it?

$(function() {     $('#slider').slider({ slide: moveSlider2 });     $('#slider1').slider({ slide: moveSlider1 });     function moveSlider2( e, ui )      {         $('#slider1').slider( 'moveTo', Math.round(ui.value) );     }      function moveSlider1( e, ui )      {         $('#slider').slider( 'moveTo', Math.round(ui.value) );     } });
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1 Answer

  1. This is sort of a hack, but works:

    $(function () {     var slider = $('#slider');     var slider1 = $('#slider1');     var sliderHandle = $('#slider').find('.ui-slider-handle');     var slider1Handle = $('#slider1').find('.ui-slider-handle');      slider.slider({ slide: moveSlider1 });     slider1.slider({ slide: moveSlider });      function moveSlider( e, ui ) {         sliderHandle.css('left', slider1Handle.css('left'));     }      function moveSlider1( e, ui ) {         slider1Handle.css('left', sliderHandle.css('left'));     } });

    Basically, you avoid the feedback by manipulating the css directly, not firing the slide event.

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