Question

1）如何访问paginator partial中的搜索$keyword以创建搜索友好网址？显然，将关键字传递为$ this-＆gt; view-＆gt; paginator-＆gt;关键字无效。
2）目前，搜索按钮的名称也作为参数发送。例如，当搜索“a”时，网址变为http://localhost/search/index/search/a/submit//page/2。有什么方法可以阻止这个？

搜索表单：

<form id="search" method="get" action="/search">
                        <input id="searchfield" onClick="this.value=''" type="text" name="search" value="Search wallpapers"/>
                        <input id="searchbutton" type="submit" value="" name="submit"/>
                    </form>

searchController中的动作：

public function indexAction()
    {        
        $keyword=$this->_request->getParam('search');        
        $alnumFilter=new Zend_Filter_Alnum();
        $dataModel=new Model_data();         

        $adapter=$dataModel->fetchPaginatorAdapter("title LIKE '%".$keyword."%'", '');
        $paginator=new Zend_Paginator($adapter);
        $paginator->setItemCountPerPage(18);
        $page=$this->_request->getParam('page', 1);
        $paginator->setCurrentPageNumber($page);

        $this->view->paginator=$paginator;
        $this->view->keyword=$keyword;

    }

index.phtml（查看）文件：

partialLoop（'wallpaper / table-row.phtml'，$ this-＆gt; paginator）;？＆gt;

    <div id="page-links">

        <?= $this->paginationControl($this->paginator,'Sliding','partials/search-pagination-control.phtml');?>       

    </div>

search-paginator-control.phtml文件：

if ($this->pageCount){
    $params=Zend_Controller_Front::getInstance()->getRequest()->getParams();


    if(isset($this->previous)){
    ?>
        <a href="<?php echo $this->url(array_merge($params,array('page'=>$this->previous)));?>">Previous</a>
    <?php
    }
    else{
        ?> Previous <?php
    }

    foreach($this->pagesInRange as $page){
        if($page !=$this->current){
            ?>
        <a href="<?echo $this->url(array_merge($params,array('page'=>$page)));?>"><?=$page;?></a>
        <?
        }
        else{
            echo $page;
        }
    }

    if(isset($this->next)){
    ?>
        <a href="<?php echo $this->url(array_merge($params,array('page'=>$this->next)));?>">Next</a>
    <?php
    }
    else{
        ?> Next <?php
    }
}

Answer 1

paginationControl视图助手接受第4个参数，即参数数组，因此在您的情况下，您可以执行以下操作：

<div id="page-links">

    <?= $this->paginationControl($this->paginator,'Sliding','partials/search-pagination-control.phtml', array('keyword' => $this->keyword));?>       

</div>

然后使用$ this-＆gt;关键字在您的分页内访问它。

访问paginator partial内的请求参数

1 个答案: