javascript - 如果找到具有正确属性的obj，则js返回true

Question

let userScopes = [
  {
    scope: 'user',
    actions: ['create', 'read', 'update'],
  },
];

function checked(scope, actions) {
  return !!~userScopes.find(
    _scope =>
      _scope.scope === scope && _scope.actions.lastIndexOf(actions) !== -1
  );
}

console.log(checked("user","create"));//true
console.log(checked("users","create"));//false
console.log(checked("user","creat"));//false
console.log(checked("make","create"));//false

我必须确保如果找到具有以下属性的对象，则它返回 true，否则返回 false。

但是它行不通，有什么建议吗？

Answer 1

您可以使用 Array#some 查看和 Array#includes .

function checked(scope, action) {
    return userScopes.some(o => o.scope === scope && o.actions.includes(action));
}

let userScopes = [{ scope: 'user', actions: ['create', 'read', 'update'] }];

console.log(checked("user", "create"));  //  true
console.log(checked("users", "create")); // false
console.log(checked("user", "creat"));   // false
console.log(checked("make", "create"));  // false