从零到一构建短链接系统（八）

1.git上传远程仓库（现在才想起来）

git init

git add .

git commit -m "first commit"

git remote add origin OLiyscxm/shortlink

git push -u origin "master"

2.开发全局异常拦截器之后就可以简化UserController

拦截器可以替代为空判断

3.common目录下创建目录serialize，再创建类

public class PhoneDesensitizationSerializer extends JsonSerializer<String> {@Overridepublic void serialize(String phone, JsonGenerator jsonGenerator, SerializerProvider serializerProvider)throws IOException {String phoneDesensitization = DesensitizedUtil.mobilePhone(phone);jsonGenerator.writeString(phoneDesensitization);}
}

4.UserRespDTO添加json序列化注释

5.resp目录下定义一个UserActualRespDTO类，这个类就不要json序列化注释了，再修改UserController，这时就可以选择是否查看真实用户手机号了

@RestController
@RequiredArgsConstructor
public class UserController {private final UserService userService;/*** * 根据用户名查询用户信息* */@GetMapping("/api/shortlink/v1/user/{username}")public Result<UserRespDTO> getUserByUsername(@PathVariable("username") String username) {return Results.success(userService.getUserByUsername(username));}@GetMapping("/api/shortlink/v1/actual/user/{username}")public Result<UserActualRespDTO> getActualUserByUsername(@PathVariable("username") String username) {return Results.success(BeanUtil.toBean(userService.getUserByUsername(username), UserActualRespDTO.class));}
}

6.新增查询用户名是否存在功能

UserService中添加

/**
* 查询用户名是否存在
* @param username 用户名
* @return 存在返回True，不存在返回False
*/Boolean hasUsername(String username);

UserServiceImpl中添加

@Override
public Boolean hasUsername(String username) {LambdaQueryWrapper<UserDO> queryWrapper = Wrappers.lambdaQuery(UserDO.class).eq(UserDO::getUsername, username);UserDO userDO = baseMapper.selectOne(queryWrapper);return userDO==null;
}

UserController中添加

/**
*查询用户名是否存在
*/@GetMapping("/api/shortlink/v1/user/has-username")
public Result<Boolean> hasUsername(@RequestParam("username") String username){return Results.success(userService.hasUsername(username));
}

7.data一栏OLi存在，返回false