另一个关于codeigniter的问题,这里有一些细节:
查看页面:
<?php if (isset($error)){echo $error; } ?>
<form action="<?php echo site_url('mem_posting/post');>" method="post">
<input type="text" name="fname">
some fields goes here...
</form>
控制器页面(mem_posting):
public function post_form()
{
$this->load->view('header');
$this->load->view(form_page);
}
public function post()
{
$post_data=array(
'mem_id'=>$this->input->post('mem_id'),
//other inputs...
)
$this->load->model('member_model');
if ($this->member_model->check_member($post_data)===true)
{
//row exist
// **i would like to load the same page but
// **with error message "like already exist".
}else{
$this->member_model->inset_member($post_data);
}
}
型号页面:
public function insert_member($post_data=array())
{
extract($post_data);
$this->db->where('member_id', $member_id);
$this->db->insert('membership', $post_data);
}
public function check_member($post_data=array())
{
extract($post_data);
$this->db->select('mem_id');
$this->db->where('mem_id', $mem_id);
$query = $this->db->get('membership');
if ($query->num_rows() > 0){
return true;
}else{
return false;
}
}
如您所见,视图页面包含表单,现在我想要实现的是回显像“已经存在”这样的错误,所以我不必在内部编写另一个 $this->load->view('post_form') if 语句。
先感谢您..