在字元串儲存引用中進行參數相關查找後，成員“名稱”未找到或不可見

我有這樣的程式碼，我想在我的結構中設置我的名字、姓氏和年齡，所以我想設置它們並讓它們與我的函式一起使用

這是我的程式碼：

pragma solidity >=0.4.0 <0.7.0;

contract NewHello{
   address owner;
   
   constructor()public{
       owner = msg.sender;
   }
   
   modifier onlyOwner (){
       require(msg.sender == owner);
       _;
   }
   
   struct user{
       string name;
       string surname;
       uint age;
   }
   
   user[] new_user;
   mapping(address => string) public users; // contains a name for every address
   
   function updateUser(address _address, string memory name,string memory surname,uint age) public onlyOwner {
       user memory new_user = user(name,surname,age);
       users[_address] = user(name,surname,age);
       
   }
   function getUser(address _address) public view returns(string memory name ,string memory surname,uint age) {
       //if(users[_address].age == 0) throw;
       return (users[_address].name,users[_address].surname,users[_address].age); 
   }
}

我得到 2 個錯誤，第一個是：

TypeError: Type struct NewHello.user memory is 
not implicitly convertible to expected type string storage ref. 
users[_address] = user(name,surname,age);
                 ^--------------------^

TypeError: Member "name" not found 
or not visible after argument-dependent lookup in string storage ref.

任何人都可以幫助我嗎？

使用如下映射：

mapping(address => user) public users

每個地址都會有一個使用者。如果您使用msg.sender而不是地址（函式參數），您可以改進這一點，因為您不需要提供地址，並且每個使用者都可以更改他的數據並查看它（onlyOwner在這種情況下您需要刪除修飾符） .

pragma solidity >=0.4.0 <0.7.0;

contract NewHello{ 

   address owner;

   constructor()public{
       owner = msg.sender;
   }

   modifier onlyOwner (){
       require(msg.sender == owner);
       _;
   }

   struct user{
       string name;
       string surname;
       uint age;
   }
   mapping(address => user) public users; // contains a user for every address

   function updateUser(address _address, string memory name, string memory surname, uint age) public onlyOwner {
       users[_address] = user(name, surname, age);
   }

   function getUser(address _address) public view returns(string memory name, string memory surname, uint age) {
        //if(users[_address].age == 0) throw;
       return (users[_address].name, users[_address].surname, users[_address].age);        
   }
}

引用自：https://ethereum.stackexchange.com/questions/72459