简化 if else 条件以降低 Java 中的认知复杂度

Question

有没有办法简化这个java函数？

为了代码的可维护性，需要简化。

public void pushDocument(ESDocumentType esDocumentType, Object data, String documentId, long userId, long organizationId) {
    boolean proceed = false;

    if (esDocumentType.equals(ESDocumentType.XMLACTIVITY)) {
        proceed = Constants.ELASTIC_LOGGING_ENABLED && Constants.ELASTIC_XMLACTIVITY_ENABLED || Constants.SQS_LOGGING_ENABLED;
    }

    else if (esDocumentType.equals(ESDocumentType.XMLREQRES)) {
        proceed = Constants.ELASTIC_LOGGING_ENABLED && Constants.ELASTIC_XMLREQRES_ENABLED || Constants.SQS_LOGGING_ENABLED;
    }

    else if (esDocumentType.equals(ESDocumentType.ORDERHISTORY)) {
        proceed = Constants.ELASTIC_LOGGING_ENABLED && Constants.ELASTIC_ORDERHISTORY_ENABLED || Constants.SQS_LOGGING_ENABLED;
    }

    else if (esDocumentType.equals(ESDocumentType.SINGIN)) {
        proceed = Constants.ELASTIC_LOGGING_ENABLED && Constants.ELASTIC_SIGNIN_ENABLED || Constants.SQS_LOGGING_ENABLED;
    } else if (esDocumentType.equals(ESDocumentType.GOOGLESEARCH)) {
        proceed = Constants.ELASTIC_LOGGING_ENABLED && Constants.ELASTIC_GOOGLESEARCH_ENABLED || Constants.SQS_LOGGING_ENABLED;
    }

    if (proceed) {
        LogThread logThread = new LogThread();
        logThread.pushDocument(esDocumentType, data, documentId, userId, organizationId);
    }
}

Answer 1

我不知道你的确切用例，你必须自己稍微完善一下，但这样的东西可能会起作用。

List<ESDocumentType> enabled; // fill this based on your "Constant.ELASTIC_<BLAH>_ENABLED" constants in the constructor

public void pushDocument(ESDocumentType type, other parameters) {
    boolean proceed = (Constants.ELASTIC_LOGGING_ENABLED && enabled.contains(type)) || Constants.SQS_LOGGING_ENABLED;
    
    if (proceed) {
        LogThread logThread = new LogThread();
        logThread.pushDocument(esDocumentType, data, documentId, userId, organizationId);
    }
}