如果php函数为true，则显示模式

Question

我有一个要在执行PHP函数时切换的模式。

模式代码

<div class="modal-normalise modal fade" tabindex="-1" role="dialog" id="free-modal">
    <div class="modal-dialog" role="document">
        <div class="modal-content">
            <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal" aria-label="Close"><i class="far fa-times-circle"></i></button>
                <br>
                <img class="img-center img-responsive" src="/wp-content/themes/x/assets/images/x@2x.png">
            </div>
            <div class="modal-body text-center">
                <p class="login-title head-title">test</p>
                <?php echo do_shortcode('[mepr-login-form]'); ?>
                <p class="subtitle fancy"><span>OR</span></p>
                <?php echo do_shortcode('[nextend_social_login]'); ?>
                <p class="login-title join-title">Not a member? Join now!</p>
                <a href="/join/" class="btn btn-fg">Join Now</a>
            </div>
        </div><!-- /.modal-content -->
    </div><!-- /.modal-dialog -->
</div><!-- /.modal -->

当前php代码

    // x
                    if (get_user_subscr_level() == "X") {

                        echo '<script type="text/javascript"> $("#free-modal").modal("show")</script>';

                    } elseif (get_user_subscr_level() != "X") {
                        echo '<script type="text/javascript"> $("#free-modal").modal("show")</script>';

                    };

在上述情况下，没有显示＃free-modal

尝试了echo js代码的多种变体