cakePHP 使用 post 数据进行分页，无需会话、序列化或 post 即可获取

问题描述

我在控制器/索引中创建了一个带有 POST 操作的小型搜索和过滤器表单，它向自己发送要分页的条件和字段 ($this->paginate($conditions)).

I have created a small search and filter form with a POST action in controller/index, which POSTs to itself the conditions and fields to paginate ($this->paginate($conditions)).

这对第一页很好，但是在后续页面上，过滤条件丢失了.分页 passedArgs 很好地支持 GET 变量.

That is good for the first page, however on the subsequent pages the filter conditions are lost. Pagination passedArgs supports GET variables well.

是否有一种简单的方法可以将 POST 条件传递到其他分页页面?

Are there an un-complex ways to pass the POST conditions to the other paginated pages?

我看过的方法是通过会话传递 $conditions，这不是没有分配会话和取消设置会话再次提交表单的复杂性(对用户筛选条件).另一种方法是将 $conditions 作为序列化字符串传递，url_encode 作为 GET 参数.

The method I have looked at is to pass the $conditions via the session, which isn't without complexity of assigning the session and unsetting the session on submitting the form again (more refinements to the filter criteria by the user). The other method is passing the $conditions as serialized string with url_encode as a GET parameter.

是否有更好的蛋糕"方式来做到这一点，更像是 passedArgs.Sessions 和 url_encode 看起来不像蛋糕风格.

Is there a good 'cake' way to do this more like passedArgs. Sessions and url_encode do not look like cake style.

谢谢

推荐答案

你可以使用passedArgs.

You can use passedArgs.

在方法控制器中:

if ( empty($this->passedArgs['search']) ){
    $this->passedArgs['search'] = $this->data['YourModel']['search'];
}
if ( empty($this->data) ){
    $this->data['YourModel']['search'] = $this->passedArgs['search'];
}

在你看来:

$this->Paginator->options(array('url' => $this->passedArgs));

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