如何正确地将磁通包装在单对象中

Question

我有一个网络服务，可以返回学生和注册班级的详细信息。

{
  "name": "student-name",
  "classes": [
    {
      "className": "reactor-101",
      "day": "Tuesday"
    },
    {
      "className": "reactor-102",
      "day": "Friday"
    }
  ]
}

该类的DTO如下所示：

public class Student {
        private String name;
        private Flux<StudentClass> classes;
        @Data
        @AllArgsConstructor
        @JsonInclude(JsonInclude.Include.NON_DEFAULT)
        public static class StudentClass {
            private String className;
            private String day;
        }
    }

Flux<StudentClass> studentClassFlux = studentClassRepository.getStudentClass(studentName);

return Mono.just(new Student(studentName, studentClassFlux));

{
  "name": "student-name",
  "classes": {
    "prefetch": 32,
    "scanAvailable": true
  }
}

List<StudentClass> studentClassList = studentClassRepository.getStudentClass(studentName)..toStream().collect(Collectors.toList());
return Mono.just(new Student(studentName, studentClassList)); // Change the Student#classes from flux to list

Answer 1

当相互包装时，反应类型并不意味着要序列化。在这种特殊情况下，您可能希望student对象包含列表 。您可以这样实现：

public Mono<Student> findStudent(String studentName) {

    return studentClassRepository
             .getStudentClass(studentName)
             .collectList()
             .map(studentClasses -> new Student(studentName, studentClasses));
}