从手机差距建立应用程序退出的正确代码是什么？

Question

我使用了navigator.app.exitApp（）;在我的代码中。我尝试了通过防止后退按钮退出应用程序的多种方法。但这在我的应用程序中不起作用。

源代码：

$data = array(
      'vndr_name' => $this->input->post('vndr_name'),
      'vndr_phone' => $this->input->post('vndr_phone'),
      'vndr_mobile' => $this->input->post('vndr_mobile'),
      'vndr_gst' => $this->input->post('vndr_GST'),
      'vndr_pan' => $this->input->post('vndr_PAN'),
      'vndr_addressLine1' => $this->input->post('vndr_addr_line1'),
      'vndr_addressLine2' => $this->input->post('vndr_addr_line2'),
      'vndr_city' => $this->input->post('vndr_city'),
      'vndr_state' => $this->input->post('vndr_state'),
      'vndr_zipcode' => $this->input->post('vndr_ZIP'),
      'vndr_enrldate' => $this->input->post('enrlmnt_date')
    );
foreach(array_keys($data) as $h)
    $arr_cols[$h] = $this->db->escape($arr_cols[$h]);

$str_cols = implode(', ', $arr_cols);

$this->db->query("CALL sp_insert_vendor({$str_cols})");