这是在员工控制器功能样本这个问题如何防止CodeIgniter表单验证的代码重复?
function newStaff()
{
$data = array();
$data['departmentList'] = $this->department_model->list_department();
$data['branchList'] = $this->branch_model->list_branch();
$data['companyList'] = $this->company_model->list_company();
$this->load->view('staff/newstaff', $data);
}
function add_newStaff()
{
//when user submit the form, it will call this function
//if form validation false
if ($this->validation->run() == FALSE)
{
$data = array();
$data['departmentList'] = $this->department_model->list_department();
$data['branchList'] = $this->branch_model->list_branch();
$data['companyList'] = $this->company_model->list_company();
$this->load->view('staff/newstaff', $data);
}
else
{
//submit data into DB
}
}
从功能add_newStaff(),我需要的,如果表单验证返回false从数据库中加载回的所有数据。这可能很麻烦,因为我需要维护两份代码。我可以用来防止这种情况的任何提示?
谢谢。
哇谢谢。我以前怎么也想不起这种方法。 – cyberfly 2012-03-26 02:37:44