(function( window ){
var Utils = {
//TODO 判断两个JSON是否相等
equals : function( param ,param2){
return JSON.stringify(param) === JSON.stringify(param2);
},
//TODO 获取当前项在数组出现的个数
getCountByItem : function( objects,item){
var count = 0;
for(var i = 0;i < objects.length;i++){
if( this.equals(objects[i] , item ) ){
count++;
}
}
return count;
},
//TODO 获取两个数组交集的项列表，支持List<object> 代码格式：[{}]
getRepeatItems : function( objects,objects2 ){
var tempObjects = objects.concat(objects2);
var repeatItems = [];
for(var i = 0;i < tempObjects.length;i++){
var itemCount = this.getCountByItem( tempObjects,tempObjects[i] );
if( itemCount > 1 ){
repeatItems.push( tempObjects[i] );
tempObjects[i] = null;
}
}
return repeatItems;
}
};
window.Utils = Utils;
})( window );
Javascript 获取两个数组交集(重复)的项列表，支持List<object>
原文：http://www.cnblogs.com/Ming2020/p/5061265.html