过滤对象中的子列表并返回Java 8

问题描述

我需要一个雇员对象，该对象的地址应从该对象下面开始，其引脚为40001。

    Employee eObj= new Employee(1,"abc",10001,Arrays.asList(new Address("ad1","ad2","tel","40001"),new Address("ad1","tel1","40002")));

//需要帮助以编写下面的逻辑

Employee filteredEobj ={Logic}

//期望的响应

new Employee(1,"40001"));
How can we achieve this using java 8?

解决方法

流

假设您具有相应的吸气剂，并要求提供流变体：

Address address = eObj.getAddresses()
    .stream()
    .map(Employee::getAddress)
    .anyMatch(a -> a.getPostCode().equals("40001"))
    .orThrow(); // Whatever logic you want in case not found

Employee filteredEObj = new Employee(eObj.getFoo(),eObj.getBar(),eObj.getBaz(),List.of(address)); // i dont really know what those parameters are supposed to be

---

传统循环

还有更传统的方法：

Address matchAddress = null;
for (Address address : eObj.getAddresses()) {
    if (address.getPostCode().equals("40001")) {
        matchAddress = address;
        break;
    }
}

if (matchAddress == null) {
    // TODO case not found ...
}

Employee filteredEObj = new Employee(eObj.getFoo(),List.of(matchAddress));

---

只是流

最后是一路使用流的变体：

Employee filteredEObj = eObj.getAddresses()
    .stream()
    .map(Employee::getAddress)
    .filter(a -> a.getPostCode().equals("40001"))
    .limit(1)
    .map(a -> new Employee(eObj.getFoo(),List.of(a));
    .findAny()
    .orThrow(); // Whatever logic you want in case not found

---

多个匹配项

如果您对所有匹配的地址感兴趣，而不仅仅是一个，您可以简单地收集到一个列表中。例如：

List<Address> addresses = eObj.getAddresses()
    .stream()
    .map(Employee::getAddress)
    .filter(a -> a.getPostCode().equals("40001"))
    .collect(Collectors.toList());

// List is empty in case no matches

Employee filteredEObj = new Employee(eObj.getFoo(),addresses);

,

您可以先过滤地址，然后使用新地址的数据创建员工。

List<Address> filterdAddresses = eObj.getAdresses()
                            .stream()
                            .filter(a -> a.getPin().equals("40001"))
                            .collect(Collectors.toList());

Employee filteredEObj = new Employee(eObj.getId(),eObj.getName(),eObj.getEmpId(),filterdAddresses);

注意：我假设Employee的使用者正在学习，因为您尚未显示课程。

,

List<Employee> employees = empObjs.stream()
                                   .filter(e -> ("40001").equals(e.getPin())
                                   .collect(Collectors.toList());