如何降低以下代码的声纳认知复杂度？

Question

下面是我的代码，Sonar 抱怨它。要求做 15，现在是 19。重构此方法以将其认知复杂度从 19 降低到允许的 15。我尝试使用无法正常工作的枚举。它的 transactionId 作为 RequestParam。是否有任何eclipse工具可以提取这个或者我们需要手动提取。

@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;
    }

非常感谢任何帮助，枚举将不起作用。

Answer 1

将所有这些词典放在一个列表中。运行尝试从每个项目中提取密钥并将其添加到列表中的循环。

Answer 2

列出地图并遍历它们

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;

你可以进一步简化为

return Stream.of(
        manageProductOrderCISStore,
        manageAppointmentWWMStore,
        manageAppointmentHWMStore,
        manageAppointmentUnifyStore
        //the rest
    )
    .map(map -> map.get(transactionId))
    .filter(Objects::nonNull)
    .findFirst()
    .map(Collections::singletonList)
    .orElse(Collections.emptyList());

Answer 3

我看到了 3 种可能性：

遍历只有一个 if 的商店列表 - 如果你能有一个共同的 superclass/interface，它可能是最好的
拆分 ifs：假设在 10 之后，您调用 else {另一个方法与其余的 ifs}
告诉声纳接受这个（是的，你可以这样做）

我认为实际上可以在声纳中添加关于这个的例外，因为在我看来，另一个会使代码更难阅读。

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