如何让我的弹出窗口与动态类一起工作

时间:2017-05-11 11:55:14

标签: jquery

我正在尝试创建一个弹出窗口,根据类别弹出相应的模态。弹出窗口有一个动态类。所以,如果我有一个叫做的课程 popup_1然后它应该显示popup_modal_1

@foreach($single_portfolio->portfolio as $portfolio)
    <h2>{!! $portfolio->title !!}</h2>

    <?php
        $thumb = $portfolio->image;
    ?>


    <div class="popup popup_{!! $portfolio->id !!}">
        <img src="{!! asset("portfolio_images/thumbs/$thumb") !!}">
    </div>

    <div id="myModal" class="popup_modal popup_modal_{!! $portfolio->id !!}">
        <div class="modal-content">
            <span class="close">
                &times;
            </span>

            <div class="portfolio_popup">
                <div class="portfolio_wrapper">
                    <?php
                        $fullImage = $portfolio->image2;
                        $fullImg = $portfolio->image2;
                    ?>

                    <div>
                        <img src="{!! asset("portfolio_images/$img") !!}">
                    </div>
                </div>

                <div class="portfolio_nav">
                    <div>
                        <img src="{!! asset("portfolio_images/$thumbImg") !!}">
                    </div>
                </div>
            </div>
        </div>
    </div>
@endforeach

我的JS

$('.popup').click(function(){
    $('.popup_modal').show();
});

1 个答案:

答案 0 :(得分:1)

您可以尝试:

$('.popup').click(function(){
    $(this).next('.popup_modal').show(); //assuming that associated popup_modal will be immediate to your popup link
});