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简化 if else 条件以降低 Java 中的认知复杂度

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) …
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java sonarqube

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