I got a form, that send post request and show paginated results. There are problems, when i want to see pages number 2 and more, because there sended get request and controller doesn't see form to create query. Anyone know how to solve this problem?
2条回答 默认 最新
- douzhao7445 2013-12-15 06:14关注
I'm using symfony(1.4) and I dont know if there's a big difference between 2.x
so let me discuss about it...
creating url's you should use
<?php url_for('page/view?num='.$page_num) ?>
something like that, then you can now use the
request
of your module@app/{apps_name}/module/page/{actions.class.php} or {pageActions.class.php}
to your
view
methodpublic function executeView(sfWebRequest $request) { $page_num = $request->getParameter('num'); echo $page_num; }
you should get what the page number now.
one more thing, this only works in $_GET requests.
You should know how to use Routing, to configure atleast 3 parameters. It will help you to use $_POST requests.
解决 无用评论 打赏 举报
悬赏问题
- ¥15 使用ue5插件narrative时如何切换关卡也保存叙事任务记录
- ¥20 软件测试决策法疑问求解答
- ¥15 win11 23H2删除推荐的项目,支持注册表等
- ¥15 matlab 用yalmip搭建模型,cplex求解,线性化处理的方法
- ¥15 qt6.6.3 基于百度云的语音识别 不会改
- ¥15 关于#目标检测#的问题:大概就是类似后台自动检测某下架商品的库存,在他监测到该商品上架并且可以购买的瞬间点击立即购买下单
- ¥15 神经网络怎么把隐含层变量融合到损失函数中?
- ¥15 lingo18勾选global solver求解使用的算法
- ¥15 全部备份安卓app数据包括密码,可以复制到另一手机上运行
- ¥20 测距传感器数据手册i2c