如果会话数据存在，则在视图中输出 [英] IF Session data exists, output in view

问题描述

因此，我试图将会话的变量值输出到视图中（仅当它存在时）。但是我似乎无法正常工作。没错没有。我想念什么？以为我现在已经收拾好了-猜不出来...

So I'm trying to output a session's variable value into a view (only if it exists). But I can't seem to get it to work. No error. Nothing. What am I missing? Thought I had this down packed by now - guess not...

在控制器中...仅供参考，$ data已分配给视图（即 $ this->加载-> view（'view'，$ data）;

In controller...FYI, $data is assigned to the view (i.e. $this -> load -> view('view', $data);

$data['campaign_name'] = $this -> session -> userdata('campaign_name');

以下是我要输出的php代码段。因此，简而言之，如果该会话存在，则将其输出。否则，不执行任何操作。

Here's my php snippet in the view that I'm trying to output. So in short, if the session exists, output it. If not, do nothing.

<input type="text" name="campaign_name" class="wizardInput nameField" value="<? if (isset($campaign_name)) ;?> ">

有人吗？

编辑好吧，我应该提到我正在尝试将会话值输出为FORM值。上面修改了视图代码。表单提交的好像是值存在-甚至将值一起发送。它在文本输入中不可见...

EDIT Okay, I should have mentioned that i'm trying to output the session value into a FORM value. Modified view code above. The form submits as though the value is there - and even sends the value along. However, it's not visible in the text input...

推荐答案

您可以轻松地执行此操作在您的视图中：

you can easly do this in your view:

if($this->session->userdata('campaign_name')){
 // do somenthing cause it exist
}

然后如果要创建会话数据作为输入的值，请执行以下操作：

then if you want to make session data as the value of an input do this:

<input type="text" name="campaign_name" class="wizardInput nameField" value="<?php echo $this->session->userdata('campaign_name') ?>">

您无需控制会话用户数据是否存在，原因是不存在，它不打印任何内容，导致 userdata（）方法返回 false ！

you don't need to control if session userdata exist cause if it not exist it doesn't prints anything, cause userdata() method returns false !

然后，您无需通过 $ data []数组传递会话数据，因为可以从任何地方检索会话数据（模型/控制器/视图/挂钩/库/帮助器等）

Then you don't need to pass session data trough the $data[] array, cause session data can be retrieved from anywhere (model/controllers/views/hooks/libraries/helpers and so on)

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