【发布时间】:2019-02-06 13:17:25
【问题描述】:
尝试将我的 Spring Boot 应用程序迁移到 WebFlux,我开始转换 api 层,同时保持存储库不变(即数据库访问是同步和阻塞的)。我遇到了如何从 Mono/Flux 类型获取数据并将它们转发到存储库的问题。
考虑以下
@POST
@Path("/register")
public String register( String body ) throws Exception
{
ObjectMapper objectMapper = json();
User user = objectMapper.readValue( body, User.class );
int random = getRandomNumber( 111111, 999999 );
String uuid = null;
//first, check if user already did registration from that phone
UserDbRecord userDbRecord = UserDAO.getInstance().getUserByPhone( user.phone );
if( userDbRecord != null )
{
logger.info( "register. User already exist with phone: " + user.phone + ", id: " + userDbRecord.getId() );
uuid = userDbRecord.getToken();
}
else
{
uuid = UUID.randomUUID().toString();
}
SMS.send( user.phone, random );
Auth auth = new Auth();
auth.token = uuid;
return objectMapper.writeValueAsString( auth );
}
因此尝试执行以下操作:
public Mono<ServerResponse> register( ServerRequest request )
{
Mono<User> user = request.bodyToMono( User.class );
Mono<UserDbRecord> userDbRecord = user.flatMap( u -> Mono.just( userRepository.findByPhone( u.phone ) ) );
int random = getRandomNumber( 111111, 999999 );
String uuid = null;
//first, check if user already did registration from that phone
//now what???
if( userDbRecord != null )
{
logger.info( "register. User already exist with phone: " + userDbRecord.getPhone() + ", id: " + userDbRecord.getId() );
uuid = userDbRecord.getToken();
}
else
{
uuid = UUID.randomUUID().toString();
}
SMS.send( user.phone, random );
Auth auth = new Auth();
auth.token = uuid;
return ok().contentType( APPLICATION_JSON ).syncBody( auth );
}
检查 userDbRecord Mono 是否为空以及从中提取电话属性的最佳方法是什么?
【问题讨论】:
标签: spring-boot spring-webflux