php - 如何查询 mySQL 数据库的一个属性并将结果与​​表单变量进行比较？

Question

在我的控制页面中，我需要在我的 users6 表中查询所有用户名条目，并与我的 View 页面上的表单中的 $username 进行比较。其余代码没问题。我试过这个，但我似乎没有向 $myerror 变量添加任何内容，即使表中已经有一个表单用户名的条目。

  public function addPerson()
    {
      // GET AND SET POSTED DATA
      $username = $this->input->post('username');
      $password = $this->input->post('password');
      $accesslevel = $this->input->post('accesslevel');

      $myerror = "";
      // add the person to database with

        if (strlen($username) < 1)
        {
          $myerror = "The Username field is required.";
        }

        $query = $this->db->query("SELECT username FROM usersas6");
          foreach ($query->result() as $entry)
          {
              if ($username == $entry){
                $myerror .= "The Username is taken.";
              }
          }

        if(strlen($myerror) != 0){
          $this->TPL["myError"] = $myerror;
        }
        $this->getAllPerson();
        $this->template->show('Admin', $this->TPL);
    }

Answer 1

这样做

$myerror = array();
// add the person to database with

if (empty($username))
{
    $myerror[] = "The Username field is required.";
}
else
{
    $query = $this->db->query("SELECT username FROM usersas6 WHERE username = '$username' "); # search for specific Username
    $result = $query->num_rows(); # count rows 

    if (!empty($result)) # or  if ($result > 0) 
    {
        $myerror[] = "The Username is taken.";
    }   
}

$this->TPL["myError"] = $myerror;

FYI: It's better to use Model do DB queries.