使用浏览器后退按钮获取Ajax先前的请求

Question

I'm facing a situation with Ajax request on back button.

I created a form which posts values and returns results via ajax request according the given filters and loads on a specific div. When I click on any of the link on page, it opens the new url as expected. On clicking on browser back button it opens the previous form page with default values. How can I enable browser state functionality so that I have results with last posted values with browser back button. Note that type of ajax is POST .

One solution I got is that to modify this form type to GET instead of POST, but this would take much time to do changes in server side code.

var page_url = $(this).attr('href');
page_url = page_url.split(':');
var page = page_url['1'];

$form = $('#form);

        $.ajax({
        type: 'POST',
        data: $form.serialize(),
        url: webroot + 'controller /action/page:'+page
        }).done(function (result){
})

I want to know the possible solution.

Answer 1

You should set cache to false:

$.ajax({
    dataType: "json",
    url: url,
    cache: false,
    success: function (json) {...}
});

Source https://stackoverflow.com/a/25230377

Answer 2

您可以使用Jquery .unload()函数或尝试使用cookies您可以从此处读取它。 如何使用jQuery设置/取消设置cookie？

Answer 3

Once you submitting that time dont go for new page, just hide the form elements and display new content in other div. Once you click on back button just show previously hidden div and hide current showing div