switch 语句导致重复代码

Question

我是一名初级开发人员，希望更好地组织我的代码。

现在，我正在处理表单编号。我检索表单编号的哈希映射，并且基于表单编号，我需要调用不同的方法。每个方法都接受相同的参数，但做的事情不同。

例如：

var formDetails = new inferForms.buildFormsMap

for(form in formDetails){

switch(form.formNumber){
          case "A1345":
            getExclusionDetails(account, state, form, businessDealing)
            break
          case "B254":
            getExclusionDetails(account, state, form, businessDealing)
            break
          case "B297":
            getPartnershipDetails(account, state, form, businessDealing)
            break
          case "C397":
            getBrokerageDetails(account, state, form, businessDealing)
            break
          case "D972":
            getBrokerageDetails(account, state, form, businessDealing)
            break
          case "E192":
            getBrokerageDetails(account, state, form, businessDealing)
            break
          case "E299":
            getBrokerageDetails(account, state, form, businessDealing)
            break
          case "F254":
            getLocationDetails(account, state, form, businessDealing)
            break
          case "F795":
            getLocationDetails(account, state, form, businessDealing)
            break
          case "G642":
            getContractDetails(period, wcmJurisdiction, newForm, wcmBusiness, frm)
            break
          case "G979":
            getContractDetails(period, wcmJurisdiction, newForm, wcmBusiness, frm)
            break
   }
}

一些注意事项：

-这些方法是由另一个开发人员构建的。他辞职了，所以我承担了他的工作，并希望进行重构以使其变得更好。

-起点是表格编号的HashMap。我生成 HashMap，然后循环遍历它以根据 HashMap 中的每个表单编号收集详细信息。

-即使我要将方法转换为对象继承结构，我仍然需要一个 switch 语句来知道要实例化哪个子类，不是吗？ switch 语句会和上面的一样吗？

-其中一些 case 语句调用了完全相同的方法。有没有办法避免重复？

感谢您的所有帮助。我正在努力弄清楚如何更好地重新设计它。如果我可以提供更多详细信息，请告诉我。

Answer 1

至少某些情况下具有相同的主体 -> 使用 switch 语句失败

switch(form.formNumber){
      case "A1345": // fall through
      case "B254":
        getExclusionDetails(account, state, form, businessDealing)
        break;
      case "B297":
        getPartnershipDetails(account, state, form, businessDealing)
        break
      case "C397": // fall through
      case "D972": // fall through
      case "E192": // fall through
      case "E299":
        getBrokerageDetails(account, state, form, businessDealing)
        break
      case "F254": // fall through
      case "F795":
        getLocationDetails(account, state, form, businessDealing)
        break;
      case "G642": // fall through
      case "G979":
        getContractDetails(period, wcmJurisdiction, newForm, wcmBusiness, frm)
        break;
}

---

见 Holger 的评论：

在 JDK 14 及更高版本中，您可以使用允许多个标签的新语法，而不会失败。

switch(form.formNumber) { 
  case "A1345", "B254" -> getExclusionDetails(account, state, form, businessDealing); 
  case "B297" -> getPartnershipDetails(account, state, form, businessDealing); 
  case "C397", "D972", "E192", "E299" -> getBrokerageDetails(account, state, form, businessDealing); 
  case "F254", "F795" -> getLocationDetails(account, state, form, businessDealing); 
  case "G642", "G979" -> getContractDetails(period, wcmJurisdiction, newForm, wcmBusiness, frm); 
}

Answer 2

你可以把switch case换成if，else if，因为有多个条件相同的结果，这样会减少重复。

        var formDetails = new inferForms.buildFormsMap

        for(form in formDetails){
            var formNumber = form.formNumber
            if(formNumber.equals("A1345") || formNumber.equals("A1345")){
                getExclusionDetails(account, state, form, businessDealing)
            } else if (formNumber.equals("B297") || formNumber.equals("C397")) {
                getPartnershipDetails(account, state, form, businessDealing)
            } else if (formNumber.equals("D972") || formNumber.equals("E192")) {
                getBrokerageDetails(account, state, form, businessDealing)
            } else if (formNumber.equals("F254") || formNumber.equals("F795")) {
                getLocationDetails(account, state, form, businessDealing)
            } else if (formNumber.equals("G642") || formNumber.equals("G979")) {
                getContractDetails(period, wcmJurisdiction, newForm, wcmBusiness, frm)
            }
        }

Answer 3

对我来说，您的案例似乎是factory pattern 的理想人选。

首先定义一个抽象来收集不同的细节。

public interface DetailsManager {

  void gatherDetails(String account, String state, String form, String businessDealing);
}

继续具体实现。

public class ExclusionDetailsManager implements DetailsManager {

  @Override
  public void gatherDetails(String account, String state, String form, String businessDealing) {
    //do stuff
  }
}

public class PartnershipDetailsManager implements DetailsManager {

  @Override
  public void gatherDetails(String account, String state, String form, String businessDealing) {
    //do other stuff
  }
}

为避免 switch 和 if-else 语句，您可以在工厂中使用 Map。

public class DetailsManagerFactory {

  private final Map<String, Supplier<DetailsManager>> map;
  private final Supplier<DetailsManager> defaultSupplier;

  public DetailsManagerFactory(Map<String, Supplier<DetailsManager>> map) {
    this.map = map;
    this.defaultSupplier = DefaultDetailsManager::new;
  }

  public DetailsManager getManager(String formNumber) {
    return this.map.getOrDefault(formNumber, this.defaultSupplier).get();
  }

  private static final class DefaultDetailsManager implements DetailsManager {

    @Override
    public void gatherDetails(String account, String state, String form, String businessDealing) {
      //default manager doing nothing, just making sure not to cause NPE
    }
  }
}

DetailsManager 的创建被 Supplier 包裹而延迟。如果对象是重量级的，这将很有用——实例仅在需要时创建。如果不需要，您可以将地图的值更改为DetailsManager。

public class CachingDetailsManagerSupplier implements Supplier<DetailsManager> {

  private final Supplier<DetailsManager> managerSupplier;
  private DetailsManager cache;

  public CachingDetailsManagerSupplier(Supplier<DetailsManager> managerSupplier) {
    this.managerSupplier = managerSupplier;
  }

  @Override
  public DetailsManager get() {
    if (this.cache == null) {
      //init manager
      this.cache = this.managerSupplier.get();
    }
    return this.cache;
  }
}

此供应商会缓存创建的实例，但根据您的具体用例，这可能是可取的/不需要的。

例子

//init factory where appropriate
Supplier<DetailsManager> exclusionManagerSupplier = new CachingDetailsManagerSupplier(ExclusionDetailsManager::new);
Map<String, Supplier<DetailsManager>> map = new HashMap<>();
map.put("A1345", exclusionManagerSupplier);
map.put("B254", exclusionManagerSupplier);
map.put("B297", new CachingDetailsManagerSupplier(PartnershipDetailsManager::new));
DetailsManagerFactory factory = new DetailsManagerFactory(map);

//gather details
for (Object form : formDetails) {
  String formNumber = form.formNumber;
  DetailsManager manager = factory.getManager(formNumber);
  manager.gatherDetails(account, state, form, businessDealing);
}