鍍金池/ 問(wèn)答/HTML/ 求es6,se7方法合并對(duì)象

求es6,se7方法合并對(duì)象

objA = {
"D組": {"title":"D組","val":1,"unit":"bar"},
"C組": {"title":"C組","val":1,"unit":"bar"},
"B組": {"title":"B組","val":1,"unit":"bar"},
"A組": {"title":"A組","val":1,"unit":"bar"},
"右上土壓": {"title":"右上土壓","val":1,"unit":"bar"},
"右下土壓": {"title":"右下土壓","val":1,"unit":"bar"},
"上部土壓": {"title":"上部土壓","val":1,"unit":"bar"},
"左上土壓": {"title":"左上土壓","val":1,"unit":"bar"},
"左下土壓": {"title":"左下土壓","val":1,"unit":"bar"}
};

objB = {
"D組": 0,
"C組": 0,
"B組": 0,
"A組": 0,
"右上土壓": 0,
"右下土壓": 0,
"上部土壓": 0,
"左上土壓": 0,
"左下土壓": 0
};

如何用es6方法將后者的值賦值給前者的val?
真實(shí)的數(shù)據(jù)結(jié)構(gòu)比這個(gè)要復(fù)雜，求大神給個(gè)思路。

回答
編輯回答
薄荷綠
for (let i in objA) {
  objA[i].val = objB[i];
}
2018年3月29日 09:48
編輯回答
尛曖昧
Object.keys(objB).forEach(key => objA[key].val = objB[key])
2018年4月12日 03:11