对象数组的合并和去重

前提：默认对象中name为主键，即对象有相同name则需要合并为一个对象，且属性合并。

两种方法：原理版(Method 1) 和 精简版(Method 2)，直接上代码。

• Method 1

var songs = [
	{name:"李健",artist:"风吹麦浪",address:"春晚"}, 
	{name:"晴天",artist:"周杰伦",tag:"R&B"},
	{artist:"周杰伦",name:"晴天"}
];

var arr = [
	{name:"羽根",artist:"air"},
	{name:"晴天",artist:"周杰伦"}
]

for (let i = 0; i < arr.length; i++) {
    songs.push(arr[i])
}
console.log('songs:', songs);

function unique(songs){ 
    let result = {}; 
    let finalResult=[]; 
    for(let i=0;i<songs.length;i++){ 
    	result[songs[i].name]=Object.assign(result[songs[i].name]?result[songs[i].name]:{}, songs[i]);
 	} 
	console.log('result:', result);
    for(item in result){ 
    	finalResult.push(result[item]); 
    } 
	console.log('finalResult:', finalResult);
	return finalResult; 
}
console.log('unique:', unique(songs));

• Method 2

var songs = [
	{name:"李健",artist:"风吹麦浪",address:"春晚"}, 
	{name:"晴天",artist:"周杰伦",tag:"R&B"},
	{artist:"周杰伦",name:"晴天"}
];

var arr = [
	{name:"羽根",artist:"air"},
	{name:"晴天",artist:"周杰伦"}
]

this.songs = this.songs.concat(this.arr)
console.log('songs:', songs);
	
function unique2(arr) {
  const res = new Map();
  return arr.filter((arr) => !res.has(arr.name) && res.set(arr.name, 1))
}
console.log('unique2:', unique2(songs));