当我将Jquery UI滑块元素克隆到另一个DIV时,当我移动克隆滑块移动原始滑块而不是克隆滑块时,它工作正常工作
HTML:
<div class="demo">
<p>
<label for="amount">Price range:</label>
<input type="text" id="amount" style="border:0; color:#f6931f; font-weight:bold;" />
</p>
<div class="tool-options">
<div id="slider-range" class="slider-range"></div>
</div>
</div>
<div class="clones"></div>
<button type="button" onclick="reset_slider('.slider-range')">Reset</button>
<br>
<br>
<button type="button" onclick="clone_elem();">Clone</button>
JS:
$(function() {
$( ".slider-range" ).slider({
range: true,
min: 0,
max: 500,
values: [ 75, 300 ],
slide: function( event, ui ) {
$( "#amount" ).val( "$" + ui.values[ 0 ] + " - $" + ui.values[ 1 ] );
}
});
$( "#amount" ).val( "$" + $( ".slider-range" ).slider( "values", 0 ) +
" - $" + $( ".slider-range" ).slider( "values", 1 ) );
});
var reset_slider = function( slider_selector ){
// Reset the sliders to their original min/max values
$( slider_selector ).each(function(){
var options = $(this).slider( 'option' );
$(this).slider( 'values', [ options.min, options.max ] );
});
};
var clone_elem = function(){
// alert('hi');
$(".tool-options").clone(true).appendTo(".clones");
};
如何将滑块移动动作附加到克隆元素?
答案 0 :(得分:0)
正如@ G.L.P评论中所提到的,我已经使用jquery删除了滑块形式的克隆元素,并且在我创建了一个滑块并将其附加到所需元素后,它现在正如我预期的那样运行良好
修改后的JS代码
$(function() {
$( ".slider-range" ).slider({
range: true,
min: 0,
max: 500,
values: [ 75, 300 ],
slide: function( event, ui ) {
$( "#amount" ).val( "$" + ui.values[ 0 ] + " - $" + ui.values[ 1 ] );
}
});
$( "#amount" ).val( "$" + $( ".slider-range" ).slider( "values", 0 ) +
" - $" + $( ".slider-range" ).slider( "values", 1 ) );
});
function CreateSlider(sliderdivID) {
$("." + sliderdivID).slider({
range: "min",
});
}
var clone_elem = function(){
// alert('hi');
var cloned= $(".tool-options").clone(true).appendTo(".clones");
$(cloned).find('.slider-range').remove();
var div = $("<div>", { class: "slider-range" });
$('.clones').after(div);
CreateSlider($(div).attr('class'));
};
最后一个箱子:Bin