如何在Solidity中返回手机号与余额的映射格式数据？

现有Solidity代码如下：

//Owner to Phone
mapping(address => uint256[]) public phones;
// Phone to Balance
mapping(uint256 => uint256) public balance;

function register(uint256 phone, uint256 Balance) public {
    _mint(msg.sender, phone);
    phones[msg.sender].push(phone);
    balance[phone] = Balance;
}
 
function details(address owner) public view returns(uint256[] memory){
    return (phones[owner]);
}

当前调用details函数仅返回已铸造的手机号数组，希望返回类似{"9222111888":"150","9093164641":"550"}这样的手机号与对应余额的映射格式数据，请问该如何实现？

Solidity本身不支持直接返回JSON格式的数据，因为EVM（以太坊虚拟机）的返回值只能是基础类型、数组或自定义结构体。我们可以通过以下两种方式实现需求：

方式1：返回自定义结构体数组 #

定义一个包含手机号和余额的结构体，让details函数返回该结构体的数组，前端拿到数据后可轻松组装成所需的JSON格式。

修改后的代码如下：

//Owner to Phone
mapping(address => uint256[]) public phones;
// Phone to Balance
mapping(uint256 => uint256) public balance;

// 定义存储手机号与余额的结构体
struct PhoneBalance {
    uint256 phone;
    uint256 balance;
}

function register(uint256 phone, uint256 Balance) public {
    _mint(msg.sender, phone);
    phones[msg.sender].push(phone);
    balance[phone] = Balance;
}
 
function details(address owner) public view returns(PhoneBalance[] memory) {
    uint256[] memory userPhones = phones[owner];
    PhoneBalance[] memory result = new PhoneBalance[](userPhones.length);
    
    for (uint256 i = 0; i < userPhones.length; i++) {
        uint256 phone = userPhones[i];
        result[i] = PhoneBalance(phone, balance[phone]);
    }
    
    return result;
}

方式2：返回两个平行数组 #

如果不想定义结构体，也可以同时返回手机号数组和对应的余额数组，前端再将两个数组按索引对应组装成映射：

//Owner to Phone
mapping(address => uint256[]) public phones;
// Phone to Balance
mapping(uint256 => uint256) public balance;

function register(uint256 phone, uint256 Balance) public {
    _mint(msg.sender, phone);
    phones[msg.sender].push(phone);
    balance[phone] = Balance;
}
 
function details(address owner) public view returns(uint256[] memory, uint256[] memory) {
    uint256[] memory userPhones = phones[owner];
    uint256[] memory userBalances = new uint256[](userPhones.length);
    
    for (uint256 i = 0; i < userPhones.length; i++) {
        userBalances[i] = balance[userPhones[i]];
    }
    
    return (userPhones, userBalances);
}

说明 #

两种方式的返回结果都需要前端做简单处理：

• 对于结构体数组，遍历数组将每个元素的phone作为键、balance作为值，构建成目标JSON对象。
• 对于平行数组，按索引一一对应，将userPhones[i]作为键、userBalances[i]作为值来组装。

内容的提问来源于stack exchange，提问作者monish nagre