Question

我想知道是否有人可以帮助我。我有一个form_dropdown，它填充了数据库中的选项。我想显示用户在插入记录时选择的默认值，但无法在不将其添加到数据库的情况下弄清楚如何执行此操作这是模型

function get_type()
{
    $results = $this->db->select('t_id, t_name')->from('account_type')->get()->result();

    $t_id = array('-SELECT-');
    $t_name = array('-SELECT-');

    for ($i = 0; $i < count($results); $i++)
    {
        array_push($t_id, $results[$i]->t_id);
        array_push($t_name, $results[$i]->t_name);
    }
    return $type_result = array_combine($t_id, $t_name);
}
function get_account_record($a_id)
{
    $this->db->where('a_id', $a_id);
    $this->db->from('account_info');
    $query = $this->db->get();
    return $query->result();
}

这是控制器

function update($a_id)
{
    $data['a_id'] = $a_id;

    $data['type'] = $this->sf_model->get_type();
    $data['view'] = $this->sf_model->get_account_record($a_id);

    $this->form_validation->set_rules('a_name', 'Account Name', 'trim|required|xss_clean|callback_alpha_only_space');
    $this->form_validation->set_rules('a_web', 'Website', 'trim|required|xss_clean');
    if ($this->form_validation->run() == FALSE)
    {
        $this->load->view('viewUpdate', $data);
    }
    else
    {
        $data = array(
            'a_id' => $this->input->post('a_id'),
            'a_name' => $this->input->post('a_name'),
            'a_website' => $this->input->post('a_web'),
            't_id' => $this->input->post('a_type'),
            'a_billingStreet' => $this->input->post('a_billingStreet'),
            'a_billingCountry' => $this->input->post('a_billingCountry'),
            'a_mobile' => $this->input->post('a_mobile')
        );

        $this->db->where('a_id',$_POST['a_id']);
        $this->db->update('account_info', $data); 
       redirect('salesforce' . $a_id);
    }

}

这是仅下拉功能TYPE的视图

    <?php
    $attributes = 'class = "form-control" id = "a_type"';
    echo form_dropdown('a_type',$type,set_value('a_type'),$attributes);?>
    <span class="text-danger"><?php echo form_error('a_type'); ?></span>
?>

Answer 1

请考虑以下事项。从这一点可以明显看出，我不是PHP编码器，但它应该给你一个想法......

    <?php

    /*  DROP TABLE IF EXISTS colours;

        CREATE TABLE colours(colour VARCHAR(12) NOT NULL PRIMARY KEY);

        INSERT INTO colours VALUES
        ('Red'),('Orange'),('Yellow'),('Green'),('Blue'),('Indigo'),('Violet');

        DROP TABLE IF EXISTS users;

        CREATE TABLE users
        (username VARCHAR(12) NOT NULL PRIMARY KEY
        ,colour VARCHAR(12) NOT NULL
        );

        INSERT INTO users VALUES
        ('Adam','Orange'),('Bob','Green'),('Charlie','Red'),('Dan','Yellow');


      */

include('path/to/connection/stateme.nts');
$query = "
        SELECT c.*
             , CASE WHEN u.colour = c.colour THEN 1 ELSE 0 END selected
          FROM colours c
          LEFT
          JOIN users u
            ON u.colour = c.colour
           AND u.username = 'Adam';";

$result = mysqli_query($db,$query);

$options = '';

 while($row = mysqli_fetch_assoc($result)){
 if ($row['selected'] == 1){
  $selected = 'selected';
  } else {
  $selected = '';
  }
    $options .= "<option $selected >{$row['colour']}\n";

 } // end of while loop
?>

<select><? echo "\n$options"; ?> </select>

输出：

<select>
<option  >Blue
<option  >Green
<option  >Indigo
<option selected >Orange
<option  >Red
<option  >Violet
<option  >Yellow
</select>

CodeIgniter - 在更新记录时从数据库向form_dropdown添加默认值

1 个答案: