如何 return 提交到动态页面

Question

我正在尝试 return 到一个页面，该页面是根据存储在数据库中的数据动态创建的。但是，当我尝试 return 到主动态页面时，它不起作用。它应该提交重定向到共享轨道的院子 ID 的页面。相反，它只是给我一条失败消息

控制器

public function update_track($yard_id)
{
    $data = array(
        'track_id'=>$this->input->post('track_id'),
        'train_id'=>$this->input->post('train_id'),
        'train_direction'=>$this->input->post('train_direction'),
        'train_length' =>$this->input->post('train_length'),
        'train_hpt'=>$this->input->post('train_hpt'),
        'train_tons'=>$this->input->post('train_tons'),
        'train_status'=>$this->input->post('train_status'),
        );
        $this->load->model('atisyard_model');
        $this->atisyard_model->update_track($data);
        if($this->db->affected_rows()>0)
        {
        redirect('/atis/'.$value->yard_id);
        }
        else
        {
            echo 'failure';
        }
}

查看

    <?=form_open('atis/update_track/'); foreach ($record as $value) {?>
<h1>Edit Track:</h1>
<b><p>Track Name: </b><?php echo $value->track_no;?></br>
<b>Track Length: </b><?php echo $value->track_length;?>'</br></br></p>
<table cellpadding="5" border="0" align="center">
<input type="hidden" name="track_id" title="track_id" value="<?php echo $value->track_id;?>">
    <tr>
        <td>Lead Engine No. / Symbol / Work Order:</td>
        <td><input type="text" name="train_id" id="train_id" size="28" maxlength="28" value="<?php echo $value->train_id;?>"></td>
    </tr>
    <tr>
        <td>Train Direction:</td>
        <td><?=form_dropdown('train_direction', $train_direction, $value->train_direction);?></td>
    </tr>
        <tr><td>Train Length:</td>
        <td><input type="text" name="train_length" id="train_length" size="4" maxlength="6" value="<?php echo $value->train_length;?>"></td>
    </tr>
    <tr><td>HP/T:</td>
        <td><input type="text" name="train_hpt" id="train_hpt" size="2" maxlength="5" value="<?php echo $value->train_hpt;?>"></td>
    </tr>
    <tr><td>Train Tons:</td>
        <td><input type="text" name="train_tons" id="train_tons" size="4" maxlength="6" value="<?php echo $value->train_tons;?>"></td>
    </tr>
        <td>Train Status:</td>
        <td> <?=form_dropdown('train_status', $train_status, $value->train_status);?></td>
    </tr>
    <tr>
    <tr><td><button id="submitbtn" >Submit</button></td></tr>
</table>
<a href="<?=site_url("atis/yard/{$value->yard_id}");?>">Back to Yard</a>
<?php } echo form_close();?>

Answer 1

如果我理解正确并且你将 $yard_id 传递给你的 update_track() 函数，那么不应该这样：

redirect('/atis/'.$value->yard_id);

公正 :

redirect('/atis/'.$yard_id);

因为我看不到你从哪里获得 $value。

Answer 2

再次查看您的代码，将 yard_id 添加为表单上的隐藏字段是否更好：

<input type="hidden" name="yard_id" name ="<?=$value->yard_id; ?>" />

然后将您的重定向更改为：

redirect('/atis/'.$this->input->post('yard_id'));

如何 return 提交到动态页面

How to return to a dynamic page on submit

php

codeigniter-3