仅使用本机JS，类似的方法将起作用：
a = [{ value:"4a55eff3-1e0d-4a81-9105-3ddd7521d642", display:"Jamsheer"}, { value:"644838b3-604d-4899-8b78-09e4799f586f", display:"Muhammed"}, { value:"b6ee537a-375c-45bd-b9d4-4dd84a75041d", display:"Ravi"}, { value:"e97339e1-939d-47ab-974c-1b68c9cfb536", display:"Ajmal"}, { value:"a63a6f77-c637-454e-abf2-dfb9b543af6c", display:"Ryan"}]b = [{ value:"4a55eff3-1e0d-4a81-9105-3ddd7521d642", display:"Jamsheer", $$hashKey:"008"}, { value:"644838b3-604d-4899-8b78-09e4799f586f", display:"Muhammed", $$hashKey:"009"}, { value:"b6ee537a-375c-45bd-b9d4-4dd84a75041d", display:"Ravi", $$hashKey:"00A"}, { value:"e97339e1-939d-47ab-974c-1b68c9cfb536", display:"Ajmal", $$hashKey:"00B"}]function comparer(otherArray){ return function(current){ return otherArray.filter(function(other){ return other.value == current.value && other.display == current.display }).length == 0; }}var onlyInA = a.filter(comparer(b));var onlyInB = b.filter(comparer(a));result = onlyInA.concat(onlyInB);console.log(result);
