方法引用中的返回类型错误：无法将 Employee 转换为Optional<U>

Question

我正在尝试编写一个 lambda 函数来获取员工位置偏好并具有下面的代码示例。flatMap(this::buildEmployeeGeolocation) 但对于我的 lambda 函数，我在说 时遇到编译错误Bad return type in method reference: cannot convert com.abc.EmployeeGeolocation to java.util.Optional<U>。我在这里缺少什么？public Optional<EmployeeGeolocation> getEmployee(final SessionId sessionId) {    return Optional.ofNullable(employeePreferencesStore.getEmployeeAccountPreferences(sessionId))            .map(preferences -> preferences.getPreference(PreferenceKey.Location))            .filter(StringUtils::isNotBlank)            .map(this::readEmployeelocation)            .flatMap(this::buildEmployeeGeolocation);}private Optional<EncryptedGeolocation> readEmployeeLocation(@NonNull final String encryptedGeolocation) {    try {        return Optional.ofNullable(objectMapper.readValue(encryptedGeolocation, EmployeeGeolocation.class));    } catch (final IOException e) {        log.error("Error while reading the encrypted geolocation");        throw new RuntimeException(e);    }}private EmployeeGeolocation buildEmployeeGeolocation(@NonNull final EncryptedGeolocation unditheredEncryptedGeolocation) {    return EmployeeGeolocation.builder()            .latitude(10.0)            .longitude(10.0)            .accuracy(1.0)            .locationType(ADDRESS)            .build();}

Answer 1

看来您真正需要做的是交换map和flatMap。更改代码

.map(this::readEmployeeLocation) 
.flatMap(this::buildEmployeeGeolocation);

到

.flatMap(this::readEmployeeLocation) // since you already have an Optional<String>
.map(this::buildEmployeeGeolocation); // above results in Optional<EncryptedGeolocation>

重要提示：从代码中推断Optional.ofNullable(...).map(...).filter(StringUtils::isNotBlank)，它将导致Optional<String>直到此操作。