java-8 - java 8流按列表过​​滤

Question

我需要按正式或非官方姓名列表中存在的姓名列表中的名字和姓氏过滤条件，并返回员工详细信息列表。 下面是 pojo 类

public class EmployeeDetails {
    private NameDetails nameDetails;

    public NameDetails getNameDetails() {
        return nameDetails;
    }
    public void setNameDetails(NameDetails nameDetails) {
        this.nameDetails = nameDetails;
    }
}

public class NameDetails {
    private List<Name> officalName;
    private List<Name> unOfficialName;
    public List<Name> getOfficalName() {
        return officalName;
    }
    public void setOfficalName(List<Name> offiicalName) {
        this.officalName = offiicalName;
    }
    public List<Name> getUnOfficialName() {
        return unOfficialName;
    }
    public void setUnOfficialName(List<Name> unOfficialName) {
        this.unOfficialName = unOfficialName;
    }
}

public class Name {
    private String firstName;
    private String lastName;
    public String getFirstName() {
        return firstName;
    }
    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }
    public String getLastName() {
        return lastName;
    }
    public void setLastName(String lastName) {
        this.lastName = lastName;
    }
}

我尝试使用官方名称列表进行过滤，效果很好，但我也想过滤非官方名称中存在的名称。(或条件)

return employeeDetailList.stream().filter(employeeDetails->(employeeDetails.getNameDetails().getOffiicalName()).stream()
            .anyMatch(name-> (null!=name) && (firstName).equals(name.getFirstName()) && (lastName).equals(name.getLastName())))
            .collect(Collectors.toList());

帮我解决这个问题。谢谢

Answer 1

如果您无法覆盖equals/hashCode(如果可以的话会有更快的方法)，那么:

  employeeDetailList             
         .stream()
         .filter(x -> Stream.concat(
                          x.getNameDetails().getOfficalName().stream(),
                          x.getNameDetails().getUnOfficialName().stream()
                      )
                     .anyMatch(y -> Objects.equals(y.getFirstName(), firstName) &&
                                    Objects.equals(y.getLastName(), lastName)))
         .collect(Collectors.toList());