在MVC4站点中的排序或分页操作期间，如何使WebGrid而不是GET到POST? [英] How do I get my WebGrid to POST instead of GET during a sort or paging operation in my MVC4 site?

问题描述

我有一个相当简单的站点，具有搜索"局部视图和列表"局部视图.使用多种模型将它们汇总到索引"视图中.

I have a fairly simple site with a Search partial view and a Listing partial view. They're rolled up using multiple models into the Index view.

一切都很好.除非我单击网格列标题以排序或尝试分页到下一个数据列表，否则网格将恢复为空.如果我重新提交相同的搜索条件，则网格将重新填充所有正确排序或分页的适用数据.

Everything is fine. Except when I click the grid column headers to sort or attempt to page to the next listing of data, the grid comes back empty. If I re-submit the same search criteria, then the grid repopulates with all applicable data sorted or paged properly.

我已经将此行为归结为WebGrid将其分页和排序机制设置为GET而不是POST的事实.因此，显然我所有的模型数据都没有提交.

I've tracked this behavior down to the fact that the WebGrid sets up it's paging and sorting mechanisms as a GET instead of a POST. So obviously all my model data is left off the submission.

是否没有办法使WebGrid开机自检，所以数据标签也随之而来?对于WebGrid来说，不包含要分页或排序的数据的类似乎适得其反.

Isn't there a way to get the WebGrid to POST so the data tags along? Seems quite counterproductive for the WebGrid as a class to not include the data one wants to page or sort.

推荐答案

这可能不是最优雅的解决方案，但它可以起作用:

This may not be the most elegant solution, but it works:

在视图中将模型添加到您的会话中:

Add the model to your Session in the view:

Session.Add( "Model", Model );

然后，在控制器的Index GET Action(或任何GET Action)中，只需检查该值并调用POST Action:

Then, in the Index GET Action in your controller (or whatever the GET Action is), just check for the value and call the POST Action:

if ( Session[ "Model" ] != null )
    this.Index( Session[ "Model" ] as MyModel );

相应地清理您的会话.

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