Roles角色合约(次要为前面的保险公司C提供环境)
// SPDX-License-Identifier: 3.0
pragma solidity ^0.8.20;
library Roles {
struct Role {
mapping (address => bool) bearer;
}
function has(Role storage role, address account) internal view returns (bool) {
require(account != address(0), "Roles: account is the zero address");
return role.bearer[account];
}
//查问增加的账户是否为空
function add(Role storage role, address account) internal {
require(!has(role, account), "Roles: account already has role");
role.bearer[account] = true;
}
//减少账户
function remove(Role storage role, address account) internal {
require(has(role, account), "Roles: account does not have role");
role.bearer[account] = false;
}
//删除账户
}
}.AirlineV.sol航空公司合约
// SPDX-License-Identifier: 3.0
pragma solidity ^0.8.20;
import "./Roles.sol";
contract AirlineV {
using Roles for Roles.Role;
Roles.Role private _airlineV;
首先申明两个事件(增删账户)
event airlineVAdded(address account);
event airlineVRemoved(address account);
初始化合约(构造函数)
constructor (address airlineV) {
_addairlineV(airlineV);
}
创立润饰器 onlyairlineV(限度拜访权限时调用)
modifier onlyairlineV() {
require( isairlineV(msg.sender),"airlineVRole: caller does not have the airlineV role");
_;
}
结构isairlineV函数判断是否具备airlineV角色
function isairlineV(address account)public view returns(bool){
return _airlineV.has(account);
}
结构两个增加航空公司的函数,用于合约外部调用(_addairlineV)和内部调用(addairlineV)
function _addairlineV(address account) internal {
_airlineV.add(account);
emit airlineVAdded(account);
}
结构一个合约外部删除函数(_removeairlineV)和用户自行退出函(renounceairlineV)
function addairlineV(address account) public onlyairlineV{
_addairlineV(account);
}
function _removeairlineV(address account) internal {
// _ai,.rlineV.remove(account);
emit airlineVRemoved(account);
}
function renounceairlineV() public {
_removeairlineV(msg.sender);
}
}演示:输出账号,点击transact
应用部署的账号进行减少操作
如果应用oxAb8…则会报错
isairlineV进行判断账号是否为航空公司身份
版权申明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协定,转载请附上原文出处链接和本申明。
原文链接:https://blog.csdn.net/2302_77339802/article/details/131834020
文章起源:CSDN 博主「发愣…」
文章原题目:《solidity案例详解(三)飞机治理合约》旨在流传区块链相干技术,如有侵权请与咱们分割删除。
发表回复