使用 CI CRUD 将数据从 DB 传递到更新表单
我正在尝试为 CRUD 活动编写一个紧凑的更新控制器。这是基本代码：
控制器：
function update($id)
{
$this->form_validation->set_rules('name','Name','required');
$this->form_validation->set_rules('age','Age','required|is_numeric');
$this->form_validation->set_rules('country','Country','');
$this->form_validation->set_error_delimiters('<br /><span class="error">', '</span>');
if ($this->form_validation->run() == FALSE) {
//Failed validation or first run
$data = $this->my_model->get_record($id);
$this->load->view('myform_view', $data);
} else {
//Validation success, update DB
}
}
视图：
<?php
$attributes = array('class' => '', 'id' => '');
echo form_open('my_form', $attributes); ?>
<p>
<label for="name">Name</label>
<?php echo form_error('name'); ?>
<br /><input id="name" type="text" name="name" value="<?php echo set_value('name'); ?>" />
</p>
<p>
<label for="age">Age</label>
<?php echo form_error('age'); ?>
<br /><input id="age" type="text" name="age" value="<?php echo set_value('age'); ?>" />
</p>
<p>
<label for="country">Country</label>
<?php echo form_error('country'); ?>
<br /><input id="country" type="text" name="country" value="<?php echo set_value('country'); ?>" />
</p>
<p>
<?php echo form_submit( 'submit', 'Submit'); ?>
</p>
<?php echo form_close(); ?>
这是基本结构，但是第一次运行表单时没有经过验证的数据。因此我必须从数据库中获取它。第一次运行时将其传递给视图的最佳方法是什么？然后，一旦提交表单，如果验证失败，那么我希望失败的数据显示不要再次从数据库重新加载。最好的方法是什么？
I'm trying to write a compact update controller for CRUD activity. Here is the basic code:
Controller:
function update($id)
{
$this->form_validation->set_rules('name','Name','required');
$this->form_validation->set_rules('age','Age','required|is_numeric');
$this->form_validation->set_rules('country','Country','');
$this->form_validation->set_error_delimiters('<br /><span class="error">', '</span>');
if ($this->form_validation->run() == FALSE) {
//Failed validation or first run
$data = $this->my_model->get_record($id);
$this->load->view('myform_view', $data);
} else {
//Validation success, update DB
}
}
View:
<?php
$attributes = array('class' => '', 'id' => '');
echo form_open('my_form', $attributes); ?>
<p>
<label for="name">Name</label>
<?php echo form_error('name'); ?>
<br /><input id="name" type="text" name="name" value="<?php echo set_value('name'); ?>" />
</p>
<p>
<label for="age">Age</label>
<?php echo form_error('age'); ?>
<br /><input id="age" type="text" name="age" value="<?php echo set_value('age'); ?>" />
</p>
<p>
<label for="country">Country</label>
<?php echo form_error('country'); ?>
<br /><input id="country" type="text" name="country" value="<?php echo set_value('country'); ?>" />
</p>
<p>
<?php echo form_submit( 'submit', 'Submit'); ?>
</p>
<?php echo form_close(); ?>
This is the basic structure, however the first time the form is run there is no validated data. Therefore I have to grab this from the DB. Whats the best way to pass this to the view on the first run? And then once the form has been submitted, if validation fails then I want the failed data to show not to reload from the DB again. Whats the best way to do this?
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评论（2）
您应该有另一种观看方式。然后根据“更新”方法提交您的表单。在那里，您可以像现在一样定义 form_validation 。
我问过类似的问题。请参阅此链接
You should have another method for the viewing aspect. Then submit your form against the "update" method. In there, you define the the form_validation as you have now.
I asked a similar question. See this link
首先抓取更新控制器中的数据进行编辑
，然后在视图文件中检查它
grab the data in update controller first for edit such as
and then check it in view file