javascript – 如何通过检查多个值来过滤数组/对象

我正在玩数组试图了解更多,因为我最近很喜欢与他们合作.
我得到这种情况,我想搜索一个数组,并将其元素值与包含某些所选过滤器值的另一个数组进行比较.

例如,如果我选择3个过滤器,我以后要在新数组中写入匹配 – 只有匹配所有3个过滤器的匹配.

为了更容易理解,我在http://jsfiddle.net/easwee/x8U4v/36/设置了一个例子

代码是：

var workItems =   [
    { "id": 2616,"category": ".category-copy .category-beauty .category-fashion"},//this is a match
    { "id": 1505,"category": ".category-beauty"},// NOT
    { "id": 1500,"category": ".category-beauty .category-fashion"},// NOT
    { "id": 692,"category": ".category-stills .category-retouching"},// NOT
    { "id": 593,"category": ".category-beauty .category-capture .category-fashion .category-product .category-stills .category-stills-retouching "},// NOT
    { "id": 636,"category": ".category-beauty .category-copy .category-fashion"},//this is a match
    { "id": 547,"category": ".category-fashion .category-lifestyle .category-stills .category-stills-retouching "},// NOT
    { "id": 588,"category": ".category-capture .category-recent-work .category-copy .category-beauty .category-fashion"} //this is a match
];

var filtersArray = [".category-beauty",".category-fashion",".category-copy"];

var i;
for (i = 0; i < filtersArray.length; ++i) {
    var searchString = filtersArraY[i];
    console.log('Searching for: ' + searchString);
    var filtered = $(workItems).filter(function(){
        return this.category.indexOf(searchString);
    });    
}   
console.log('Filtered results: ' + JSON.Stringify(filtered,null,4));

我也试过

filtered = $.grep(workItems,function(element,indeX){
    return element.category.indexOf(filtersArraY[i]); 
},truE);

但它只匹配第一个过滤器,只有当它在workItems.category的开头

我已经尝试了许多不同的解决方案,但不能真正使这项工作.@R_693_10675@用什么功能来返回所需的结果？

var filtered = workItems.filter(function(element) {
   // Create an array using `.split()` method
   var cats = element.category.split(' ');

   // Filter the returned array based on specified filters
   // If the length of the returned filtered array is equal to
   // length of the filters array the element should be returned  
   return cats.filter(function(cat) {
       return filtersArray.indexOf(cat) > -1;
   }).length === filtersArray.length;
});

var filtered = $(workItems).filter(function(i,element) {
   var cats = element.category.split(' ');

    return $(cats).filter(function(_,cat) {
       return $.inArray(cat,filtersArray) > -1;
    }).length === filtersArray.length;
});