Zend Framework 2和jQuery模态对话框 [英] Zend Framework 2 & jquery modal dialog

问题描述

如何在jquery模式对话框中显示控制器动作?

How does one go about displaying a controller action inside of jquery modal dialog?

推荐答案

首先，您需要Javascript通过ajax加载url，这取决于您使用的是哪种模式，等等，这里有大量的库那里.我假设您使用的是基本的JQuery UI对话框模式.

Firstly you'll need your Javascript to load a url via ajax, this will depend on which kind of modal you are using etc, there's a ton of libraries out there. I will assume you are using the basic JQuery UI dialog Modal.

示例链接

<!-- this points to your action below.. -->
<a class="some-link" title="title here" href="mycontroller/test">testing</a>

示例Javascript(可在Google上找到快速示例，那里有许多示例.)

Example Javascript (quick example found on google, many examples out there..)

$(document).ready(function() {
    $('.some-link').each(function() {
        var $link = $(this);
        var $dialog = $('<div></div>')
            .load($link.attr('href'))
            .dialog({
                autoOpen: false,
                title: $link.attr('title'),
            });
    });
});

现在，当通过ajax请求提供模态的内容时，您需要确保您的操作不会呈现主布局.

Now you need to make sure your action doesn't render the main layout when providing the content for the modal via the ajax request.

这是一个非常简单的方法，通过用空视图替换ajax请求的基本布局来做到这一点.这不是最好的方法，但在这种情况下是最简单的方法；)

Here's a really simple method of doing that by replacing the base layout with an empty view for ajax requests. This isn't the best method but it's the simplest for this case ;)

示例动作

public function testAction()
{
    if($this->getRequest()->isXmlHttpRequest()) {
        $this->layout('application/layout/ajax-layout');
    }

    return new ViewModel(array()); // ..
}

application/layout/ajax-layout.phtml

application/layout/ajax-layout.phtml

<?php echo $this->content ?>

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