Php smarty变量将不会输出
我正在尝试使用来自的paginator类 但由于某些原因,我无法在模板中的smarty变量中输出分页链接 我试过这个:Php smarty变量将不会输出,php,html,pagination,smarty,Php,Html,Pagination,Smarty,我正在尝试使用来自的paginator类 但由于某些原因,我无法在模板中的smarty变量中输出分页链接 我试过这个: function listadverts($where = null, $orderby = null, $limit = null) { $pages = new Paginator; $pages->items_total = $this->countadverts($where); $page
function listadverts($where = null, $orderby = null, $limit = null) {
$pages = new Paginator;
$pages->items_total = $this->countadverts($where);
$pages->mid_range = 9;
$pages->paginate();
$row = $this->db->dbh->query('SELECT ad.*, (SELECT img.image FROM '.$this->config->db_prefix.'_images AS img WHERE img.aid = ad.aid LIMIT 1) AS img FROM '.$this->config->db_prefix.'_adverts ad WHERE (ad.approved = 1) '. $where .' ORDER BY '.$orderby.' cr_date DESC '. $pages->limit .'');
$smarty = new Smarty();
$smarty->assign('paginate', $pages->display_pages());
return $row;
}
在我的模板中,我有
{$paginate}
功能
$pages->display_pages()
返回$this->return代码>
其中$this->returns包含此HTML
$this->return .= ($i == $this->current_page) ? "<a class=\"current\" href=\"#\">$i</a> ":"<a class=\"paginate\" href=\"$target?page=$i&ipp=$this->items_per_page\">$i</a> ";
解决方案:
return array(0 => $row, 1 => $pages->display_pages());
$res = $adverts->listadverts();
$app->view()->setData('adverts', $res[0]);
$app->view()->setData('paginate', $res[1]);
代码示例建议$smarty应该只存在于函数内部。试试全球$smarty;也可在外部访问,或在功能外部将数据插入$smarty
function listadverts($where = null, $orderby = null, $limit = null) {
$pages = new Paginator;
$pages->items_total = $this->countadverts($where);
$pages->mid_range = 9;
$pages->paginate();
$row = $this->db->dbh->query('SELECT ad.*, (SELECT img.image FROM '.$this->config->db_prefix.'_images AS img WHERE img.aid = ad.aid LIMIT 1) AS img FROM '.$this->config->db_prefix.'_adverts ad WHERE (ad.approved = 1) '. $where .' ORDER BY '.$orderby.' cr_date DESC '. $pages->limit .'');
return $pages->display_pages();
}
$smarty = new Smarty();
$smarty->assign('paginate', listadverts(... ));
代码示例建议$smarty应该只存在于函数内部。试试全球$smarty;也可在外部访问,或在功能外部将数据插入$smarty
function listadverts($where = null, $orderby = null, $limit = null) {
$pages = new Paginator;
$pages->items_total = $this->countadverts($where);
$pages->mid_range = 9;
$pages->paginate();
$row = $this->db->dbh->query('SELECT ad.*, (SELECT img.image FROM '.$this->config->db_prefix.'_images AS img WHERE img.aid = ad.aid LIMIT 1) AS img FROM '.$this->config->db_prefix.'_adverts ad WHERE (ad.approved = 1) '. $where .' ORDER BY '.$orderby.' cr_date DESC '. $pages->limit .'');
return $pages->display_pages();
}
$smarty = new Smarty();
$smarty->assign('paginate', listadverts(... ));
SQL查询用于什么?您将其分配给$row,但除了返回它之外,您似乎什么都不做。如果这段代码不需要知道$row,那么它应该在一个单独的函数中。此外,如果可以避免的话,字符串压缩也不被认为是构建查询的明智方法,尤其是在不使用转义函数以确保不会遭受SQL注入的情况下。代码示例建议$smarty应该只存在于函数中。尝试global$smarty代码>使在外部也可访问,或将数据插入到函数外部的$smarty中。。。未返回$pages对象。。。我的问题更新显示了我的解决方案。@Waygood-你会给出答案吗?您的评论让我看到了我在函数中返回的内容SQL查询的目的是什么?您将其分配给$row,但除了返回它之外,您似乎什么都不做。如果这段代码不需要知道$row,那么它应该在一个单独的函数中。此外,如果可以避免的话,字符串压缩也不被认为是构建查询的明智方法,尤其是在不使用转义函数以确保不会遭受SQL注入的情况下。代码示例建议$smarty应该只存在于函数中。尝试global$smarty代码>使在外部也可访问,或将数据插入到函数外部的$smarty中。。。未返回$pages对象。。。我的问题更新显示了我的解决方案。@Waygood-你会给出答案吗?你的评论让我看到了我在函数中返回的内容