如何降低声纳对以下代码的认知复杂度？
如何降低声纳对以下代码的认知复杂度？
提问于 2021-08-04 15:11:52
下面是我的代码，Sonar在抱怨。要求制作15，现在是19。重构该方法，将其认知复杂度从19降低到允许的15。我试过了，但这不管用。并将transactionId作为RequestParam。是否有eclipse工具来提取这个，或者我们需要手动进行。
@GetMapping(value = "display",produces = APPLICATION_JSON_VALUE)
@ResponseBody
public List<RspNotification> display(@RequestParam(name = Constants.TRANSACTION_ID) final String transactionId)
{
List<RspNotification> list = new ArrayList<>();
if (manageProductOrderCISStore.get(transactionId) != null)
{
list.add(manageProductOrderCISStore.get(transactionId));
}
else if (manageAppointmentWWMStore.get(transactionId) != null)
{
list.add(manageAppointmentWWMStore.get(transactionId));
}
else if (manageAppointmentHWMStore.get(transactionId) != null)
{
list.add(manageAppointmentHWMStore.get(transactionId));
}
else if (manageAppointmentUnifyStore.get(transactionId) != null)
{
list.add(manageAppointmentUnifyStore.get(transactionId));
}
else if (manageAppointmentUnifyXmlStore.get(transactionId) != null)
{
list.add(manageAppointmentUnifyXmlStore.get(transactionId));
}
else if (manageProductOrderFLSStore.get(transactionId) != null)
{
list.add(manageProductOrderFLSStore.get(transactionId));
}
else if (manageProductOrderHFSStore.get(transactionId) != null)
{
list.add(manageProductOrderHFSStore.get(transactionId));
}
else if (fulfilmentQuoteEEStore.get(transactionId) != null)
{
list.add(fulfilmentQuoteEEStore.get(transactionId));
}
else if (manageInventoryHFSStore.get(transactionId) != null)
{
list.add(manageInventoryHFSStore.get(transactionId));
}
else if (manageInventoryFLSStore.get(transactionId) != null)
{
list.add(manageInventoryFLSStore.get(transactionId));
}
else if (manageBillingDocumentStore.get(transactionId) != null)
{
list.add(manageBillingDocumentStore.get(transactionId));
}
else if (serviceHealthStore.get(transactionId) != null)
{
list.add(serviceHealthStore.get(transactionId));
}
else if (fulfilmentOrderEEStore.get(transactionId) != null)
{
list.add(fulfilmentOrderEEStore.get(transactionId));
}
else if (manageWorkOrderUnifyStore.get(transactionId) != null)
{
list.add(manageWorkOrderUnifyStore.get(transactionId));
}
else if (manageDiagnosticAssuranceStore.get(transactionId) != null)
{
list.add(manageDiagnosticAssuranceStore.get(transactionId));
}
else if (materialsSupplyUnifyStore.get(transactionId) != null)
{
list.add(materialsSupplyUnifyStore.get(transactionId));
}
///////// more if statement
return list;
}任何帮助都是非常感谢的，而枚举是行不通的。
Stack Overflow用户
回答已采纳
发布于 2021-08-04 15:19:32
做一个地图的列表并循环它们。
List<Map<String, RspNotification>> maps = Arrays.asList(
manageProductOrderCISStore,
manageAppointmentWWMStore,
manageAppointmentHWMStore,
manageAppointmentUnifyStore
//the rest
);
List<RspNotification> list = new ArrayList<>();
for (Map<String, RspNotification> map : maps) {
RspNotification notification = map.get(transactionId);
if (notification != null) {
list.add(notification);
return;
}
}
return list;您可以进一步简化为
return Stream.of(
manageProductOrderCISStore,
manageAppointmentWWMStore,
manageAppointmentHWMStore,
manageAppointmentUnifyStore
//the rest
)
.map(map -> map.get(transactionId))
.filter(Objects::nonNull)
.findFirst()
.map(Collections::singletonList)
.orElse(Collections.emptyList());页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/68653763
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