如何在 WebService (java) 上返回 List<T>

Question

我需要在 WebService 上返回不同的列表，目前我在一个封装文件中有超过 15 种不同的类型，但很难操作许多构造函数：

public class ResponseMessage implements java.io.Serializable {

    private Integer code;
    private String message;
    private List<Users> listUsers;
    private List<Products> listProducts;
    private List<Customers> listCustomers;
    private List<Suppliers> listSuppliers;
    private List<Reports> listReports;
    ...
    private Users userById;
    private Products productById;
    private Customers customerById;
    private Suppliers supplierById;
    ...

    public ResponseMessage() {
    }

    //My idea is something like that, not work
    public ResponseMessage(Integer code, String message, List<T> lstData) {
        this.code = code;
        this.message = message;
        this.lstData = lstData;
    }

    public ResponseMessage(Integer code, String message, T uniqueData) {
        this.code = code;
        this.message = message;
        this.uniqueData = uniqueData;
    }

    //Currently the constructor are this, work
    public ResponseMessage(Integer code, String message, List<Users> listUsers) {
        this.code = code;
        this.message = message;
        this.listUsers = listUsers;
    }

    public ResponseMessage(Integer code, String message, List<Users> listUsers, List<Customers> listCustomers) {
        this.code = code;
        this.message = message;
        this.listUsers = listUsers;
        this.listCustomers = listCustomers;
    }

    public ResponseMessage(Integer code, String message, List<Users> listUsers, List<Customers> listCustomers, List<Suppliers> listSuppliers) {
        this.code = code;
        this.message = message;
        this.listUsers = listUsers;
        this.listCustomers = listCustomers;
        this.listSuppliers = listSuppliers;
    }

    ...

    //Same history with unique result, work
    public ResponseMessage(Integer code, String message, Users userById) {
        this.code = code;
        this.message = message;
        this.userById = userById;
    }

    public ResponseMessage(Integer code, String message, Users userById, Products productById) {
        this.code = code;
        this.message = message;
        this.userById = userById;
        this.productById = productById;
    }

    //Getters and Setters
}

当我喜欢在 WebService 上返回构造函数时，我必须这样做，例如（工作）：

public ResponseMessage readAllSuppliers() {
   List<Suppliers> lstsuppliers = new ArrayList<Suppliers>();
   lstsuppliers = supplierDAO.getAllSuppliers();
   //ResponseMessage(code, message, user, customer, supplier list or unique supplier)
   ResponseMessage rm = new ResponseMessage(123, "reading suppliers", null, null, lstsuppliers);
   return rm;
}

但我认为你可以对任何人列表这样做：

public ResponseMessage readAllSuppliers() {
    List<Suppliers> lstsuppliers = new ArrayList<Suppliers>();
    lstsuppliers = supplierDAO.getAllSuppliers();
    //ResponseMessage(code, message, list or object data)
    ResponseMessage rm = new ResponseMessage(123, "reading suppliers", lstsuppliers);
    return rm;
}

最后，在 WebService 客户端上获取类似这样的信息数据：

public void getSuppliers() {
    WebServiceResponse wsr = new WebServiceResponse();
    ResponseMessage rm = wsr.readAllSuppliers();
    System.out.println("CODE: " + rm.getCode()); //CODE: 123
    System.out.println("MESSAGE: " + rm.getMessage()); //MESSAGE: reading suppliers
    for (Suppliers data : rm.getLstData()) {
       System.out.println("SUPPLIER INFO: " + data.getFullName()); 
    }
    //SUPPLIER INFO: Name1 Surname1
    //SUPPLIER INFO: Name2 Surname2
    //SUPPLIER INFO: Name3 Surname3
}

希望你能帮到我

Answer 1

你只需要在你的ResponseMessage 声明之后添加<T> 来告诉Java 你想使用泛型。希望这可以帮助。请注意，它希望每个响应仅返回一种类型。如果您需要发送多个类型，最好使用Map 或Types 到Lists，而不是@v1shnu 提到的lstData。

public class ResponseMessage<T> implements Serializable {

    private final List<T> lstData;
    private final Integer code;
    private final String message;

    public ResponseMessage(Integer code, String message, List<T> lstData) {
        this.code = code;
        this.message = message;
        this.lstData = lstData;
    }
}

您可以考虑为您的数据访问对象做同样的事情。

Answer 2

您可以创建一个HashMap，然后将其发送到ResponseMessage，如下所示：

Map<String,List<T>> listsMap = new HashMap<>();
listsMap.put("users",userDao.readAllUsers());
listsMap.put("customers",customerDao.readAllCustomers());
listsMap.put("suppliers",supplierDao.readAllSuppliers());

ResponseMessage rm = new ResponseMessage(123, "message", listsMap);

在您的ResponseMessage 类中，您可以添加如下构造函数：

public ResponseMessage (Integer code, String message,Map listsMap){
    this.code = code;
    this.message = message;
    this.listUsers = (List<Users>) listsMap.get("users");
    this.listCustomers = (List<Customers>) listsMap.get("customers");
    this.listSuppliers = (List<Suppliers>) listsMap.get("suppliers");
}