如何在单个Mono为空时合并两个Mono并创建Employee对象

我有两个Mono<String>类型的对象firstName和lastName，其中任意一个都可能为空（即返回Mono.empty()）。需要实现：

• 只要其中一个Mono有值，就创建对应字段填充的Employee对象
• 当两个Mono都为空时，创建一个空的Employee对象

当前方案的缺陷 #

之前尝试的两种写法，仅能在两个Mono同时不为空时成功构建Employee；只要其中一个为空，整个Mono就会返回空，无法满足需求：

第一种：使用Mono.zip的写法

public class Main {
    public static void main(String[] args) {

        Mono<String> firstName = Mono.just("Pavan");
        Mono<String> lastName = Mono.empty();

        Mono<Employee> map = Mono.zip(firstName, lastName)
                .flatMap(t -> Mono.just(Employee.builder()
                        .firstName(t.getT1())
                        .lastName(t.getT2())
                        .build()));

        System.out.println(map.block());

    }

    @Builder
    @ToString
    public static class Employee {
        private String firstName;
        private String lastName;
    }

}

第二种：嵌套flatMap+map的写法

Mono<Employee> employeeMono = firstName
        .flatMap(first -> lastName.map(last -> Employee.builder()
                .firstName(first)
                .lastName(last)
                .build()));

正确解决方案 #

核心思路是为每个可能为空的Mono设置默认值（比如null），确保组合操作总能拿到有效值，进而构建Employee对象。

方法一：defaultIfEmpty + Mono.zip

public class Main {
    public static void main(String[] args) {

        Mono<String> firstName = Mono.just("Pavan");
        Mono<String> lastName = Mono.empty();

        // 为空的Mono返回null作为默认值
        Mono<String> safeFirstName = firstName.defaultIfEmpty(null);
        Mono<String> safeLastName = lastName.defaultIfEmpty(null);

        Mono<Employee> employeeMono = Mono.zip(safeFirstName, safeLastName)
                .map(tuple -> Employee.builder()
                        .firstName(tuple.getT1())
                        .lastName(tuple.getT2())
                        .build());

        System.out.println(employeeMono.block());
        // 输出：Employee(firstName=Pavan, lastName=null)
    }

    @Builder
    @ToString
    public static class Employee {
        private String firstName;
        private String lastName;
    }
}

也可以简化为链式调用：

Mono<Employee> employeeMono = Mono.zip(
        firstName.defaultIfEmpty(null),
        lastName.defaultIfEmpty(null)
).map(tuple -> Employee.builder()
        .firstName(tuple.getT1())
        .lastName(tuple.getT2())
        .build());

方法二：Mono.combineLatest

和上面效果一致，写法更紧凑：

Mono<Employee> employeeMono = Mono.combineLatest(
        firstName.defaultIfEmpty(null),
        lastName.defaultIfEmpty(null),
        (first, last) -> Employee.builder()
                .firstName(first)
                .lastName(last)
                .build()
);

原方案失效原因 #

• Mono.zip要求所有传入的Mono都必须发出元素，若有一个是Mono.empty()，整个zip操作直接返回Mono.empty()，不会进入后续处理逻辑
• 嵌套flatMap写法中：若firstName为空，flatMap不会执行；若firstName有值但lastName为空，lastName.map(...)返回empty()，最终整个employeeMono也为空

内容的提问来源于stack exchange，提问作者nanpakal