我有三组复选框,允许我根据数据属性过滤多个div。这很好但不是我需要的。
如果你访问下面的jsfiddle,你会看到一个有效的测试。
我需要做的是:
如果在同一类别中选择了多个选项,例如'小'和'中',它应该增加结果,即显示所有'小'和'中'结果,(这就是它目前所做的)。
在不同类别中选择多个选项时,例如'中等'和'北美'它应该减少结果,即在'北美'显示所有'中等'花朵
我如何修复/修改我的代码以使其正常工作? (我希望这是有道理的。)
以下是我正在使用的jquery:
$('.flowers-wrap, .continents-wrap').delegate('input[type=checkbox]', 'change', function() {
var $lis = $('.flowers > div'),
$checked = $('input:checked');
if ($checked.length) {
var selector = '';
$($checked).each(function(index, element){
if(selector === '') {
selector += "[data-category~='" + element.id + "']";
} else {
selector += ",[data-category~='" + element.id + "']";
}
});
$lis.hide();
console.log(selector);
$('.flowers > div').filter(selector).show();
} else {
$lis.show();
}
});
请参阅我设置的jsfiddle,以证明我遇到的问题 - http://jsfiddle.net/n3EmN/
答案 0 :(得分:1)
非常感谢Pho3nixHun的帮助:)
这是我自己的答案:(jsfiddle here)。
$(document).ready(function() {
var byProperty = [], byColor = [], byLocation = [];
$("input[name=fl-colour]").on( "change", function() {
if (this.checked) byProperty.push("[data-category~='" + $(this).attr("value") + "']");
else removeA(byProperty, "[data-category~='" + $(this).attr("value") + "']");
});
$("input[name=fl-size]").on( "change", function() {
if (this.checked) byColor.push("[data-category~='" + $(this).attr("value") + "']");
else removeA(byColor, "[data-category~='" + $(this).attr("value") + "']");
});
$("input[name=fl-cont]").on( "change", function() {
if (this.checked) byLocation.push("[data-category~='" + $(this).attr("value") + "']");
else removeA(byLocation, "[data-category~='" + $(this).attr("value") + "']");
});
$("input").on( "change", function() {
var str = "Include items \n";
var selector = '', cselector = '', nselector = '';
var $lis = $('.flowers > div'),
$checked = $('input:checked');
if ($checked.length) {
if (byProperty.length) {
if (str == "Include items \n") {
str += " " + "with (" + byProperty.join(',') + ")\n";
$($('input[name=fl-colour]:checked')).each(function(index, byProperty){
if(selector === '') {
selector += "[data-category~='" + byProperty.id + "']";
} else {
selector += ",[data-category~='" + byProperty.id + "']";
}
});
} else {
str += " AND " + "with (" + byProperty.join(' OR ') + ")\n";
$($('input[name=fl-size]:checked')).each(function(index, byProperty){
selector += "[data-category~='" + byProperty.id + "']";
});
}
}
if (byColor.length) {
if (str == "Include items \n") {
str += " " + "with (" + byColor.join(' OR ') + ")\n";
$($('input[name=fl-size]:checked')).each(function(index, byColor){
if(selector === '') {
selector += "[data-category~='" + byColor.id + "']";
} else {
selector += ",[data-category~='" + byColor.id + "']";
}
});
} else {
str += " AND " + "with (" + byColor.join(' OR ') + ")\n";
$($('input[name=fl-size]:checked')).each(function(index, byColor){
if(cselector === '') {
cselector += "[data-category~='" + byColor.id + "']";
} else {
cselector += ",[data-category~='" + byColor.id + "']";
}
});
}
}
if (byLocation.length) {
if (str == "Include items \n") {
str += " " + "with (" + byLocation.join(' OR ') + ")\n";
$($('input[name=fl-cont]:checked')).each(function(index, byLocation){
if(selector === '') {
selector += "[data-category~='" + byLocation.id + "']";
} else {
selector += ",[data-category~='" + byLocation.id + "']";
}
});
} else {
str += " AND " + "with (" + byLocation.join(' OR ') + ")\n";
$($('input[name=fl-cont]:checked')).each(function(index, byLocation){
if(nselector === '') {
nselector += "[data-category~='" + byLocation.id + "']";
} else {
nselector += ",[data-category~='" + byLocation.id + "']";
}
});
}
}
$lis.hide();
console.log(selector);
console.log(cselector);
console.log(nselector);
if (cselector === '' && nselector === '') {
$('.flowers > div').filter(selector).show();
} else if (cselector === '') {
$('.flowers > div').filter(selector).filter(nselector).show();
} else if (nselector === '') {
$('.flowers > div').filter(selector).filter(cselector).show();
} else {
$('.flowers > div').filter(selector).filter(cselector).filter(nselector).show();
}
} else {
$lis.show();
}
$("#result").html(str);
});
function removeA(arr) {
var what, a = arguments, L = a.length, ax;
while (L > 1 && arr.length) {
what = a[--L];
while ((ax= arr.indexOf(what)) !== -1) {
arr.splice(ax, 1);
}
}
return arr;
}
});
答案 1 :(得分:0)
您应该在过滤元素之前隐藏元素并在之后显示它们,在.hide()
之前添加.filter(selector)
。像这样:
$('.flowers > div').hide().filter(selector).show();
这应隐藏所有元素,然后显示所选元素。
<强> DEMO 强>