通过将字符串（来自数组）分配为对象的参数之一，将其他参数保留为空，将字符串数组转换为对象列表

Question

我有一个字符串数组

adrs[] = {

    "BD Jules Ferry",

    "Rue de Republique",

    "Avenue de Julin"};

我有一个名为“Address”的对象（为了简单起见，省略了 getter setter）

public class Address{

     adrsLine1 String;

     adrsLine2 String;

     adrsLine3 String;

     postalCode String;

}

我想通过将数组中的每个字符串分配为 的参数来将其转换adrs[]为。List<<Address>> addrsListadrsLine1Address

一旦转换后，我的每个地址对象addrsList就会像这样，

• addrsList.get(0)将有 [BD Jules Ferry, null, null, null]

• addrsList.get(1)将有 [Rue de Republique, null, null, null]

• addrsList.get(2)将有 [Avenue de Julin, null, null, null]

我使用Java 8，目前，我用它for loop来实现这一点，Java 8中有没有捷径或更有效的方法？

示例代码：

private String[] AdrsArray = {"BD Jules Ferry", "Rue de Republique", "Avenue de Julin"};

List<Address> addrsList = new ArrayList<>();

for (String adrsLine : AdrsArray){

    Address address = new ReturnAddress();

    address.setAdrsLine1(adrsLine);

    addrsList.add(address);

} 

输出：

Address address = addrsList.get(0); 

address.getAdrsLine1() //should have "BD Jules Ferry"

address.getAdrsLine2() //should have null

address.getAdrsLine3() //should have null

address.getAdrsPostalCode() //should have null

Answer 1

您可以使用streamsJava 8 中的 available 来简化您的代码，如下所示。List<Address> addrsList = Arrays.stream(adrs).map(adr -> {&nbsp; &nbsp; Address address = new ReturnAddress();&nbsp; &nbsp; address.setAdrsLine1(adr);&nbsp; &nbsp; return address;}).collect(Collectors.toList());

Answer 2

您可以使用流。我建议您在类中实现构造函数Address，例如Address(String adrsLine1) {&nbsp; this.adrsLine1 = adrsLine1;}然后你可以用这个简短的代码片段来做到这一点：List<Address> addrsList = Arrays.stream(adrs)&nbsp; &nbsp; .map(adr -> new Address(adr))&nbsp; &nbsp; .collect(Collectors.toCollection(ArrayList::new));如果您需要特定的列表类型，例如ArrayList，您应该使用Collectors.toCollection(ArrayList::new)，否则（如果您可以接受任何列表类型）您可以直接使用Collectors.toList()。

Answer 3

List<Address> addressList = Arrays.stream(AdrsArray)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .map(Address::new)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(Collectors.toList());您必须向接受的 Address 类添加一个构造函数adrsLine1。public class Address{&nbsp; &nbsp; &nbsp;private String adrsLine1;&nbsp; &nbsp; &nbsp;private String adrsLine2;&nbsp; &nbsp; &nbsp;private String adrsLine3;&nbsp; &nbsp; &nbsp;private String postalCode;&nbsp; &nbsp; &nbsp;Address(String address1) {&nbsp; &nbsp; &nbsp; &nbsp;adrsLine1 = address1;&nbsp; &nbsp; &nbsp;}}