Question

任何人都可以使用ajax和代码点火器帮助我检查用户名是否在我的数据库中？我不能使用form_validation方法，因为我有模态窗口干扰检查。

目前我的控制器如下：

 function filename_exists(){
        $username = $this->input->post('username');
        $data['exists'] = $this->User_model->filename_exists($username);
    }

我的模特：

 function filename_exists($username)
 {

     $this->db->select('*'); 
     $this->db->from('users');
     $this->db->where('username', $username);
     $query = $this->db->get();
     if ($query->num_rows() == 0) {
         return true;
     } else {
         return false;
     }
 }

和我的ajax帖子：

function check_if_exists() {

     <?php $username = $this->input->post('username');
    ?>
     var username = '<?php echo $username ?>';
    var DataString=$("#form1").serialize();
     $.ajax({
     url: "<?php echo base_url(); ?>index.php/Files/filename_exists/",
     type: "post",
     data: DataString + '&username=' + username,
     success: function(response) {


            if (response == true) {
                $('#msg').html('<span style="color: green;">'+msg+"</span>");

            }
             else {

                $('#msg').html('<span style="color:red;">Value does not exist</span>');
            }


         }
     });


}

更新

 <form name = "form1" id = "form1" method ="post"> <!--action="<?php echo base_url()."index.php/Admin/create_user"; ?>"-->
    <?php echo validation_errors(); ?>
    <label for="userID" class = "labelForm">User ID:</label>
    <input type="text" id="userID" name="userID" class = "input2">
    <label for="first_name" class = "labelForm">First Name:</label>
    <input type="text" id="first_name" name="first_name" class = "input2">
    <label for="last_name" class = "labelForm">Last Name:</label>
    <input type="text" id="last_name" name="last_name" class = "input2">
    <label for="username" class = "labelForm">Username:</label>
    <input type="text" id="username" name="username" class = "input2" onblur="check_if_exists();">
    <label for="password" class = "labelForm">Password:</label>
    <input type="password" id="password" name="password" class = "input2" onblur="checkPasswords();">
    <label for="passconf" class = "labelForm">Password:</label>
    <input type="password" id="passconf" name="passconf" class = "input2" onblur="checkPasswords();">
    <label for="email" class = "labelForm">Email:</label>
    <input type="text" id="email" name="email" class = "input2">
  <button type="button"  id = "new_user_submit">Add New User</button>

Answer 1

试试这个

在Ajax中

function check_if_exists() {

var username = $("#username").val();

$.ajax(
    {
        type:"post",
        url: "<?php echo base_url(); ?>index.php/files/filename_exists",
        data:{ username:username},
        success:function(response)
        {
            if (response == true) 
            {
                $('#msg').html('<span style="color: green;">'+msg+"</span>");
            }
            else 
            {
                $('#msg').html('<span style="color:red;">Value does not exist</span>');
            }  
        }
    });
}

在控制器中

function filename_exists()
{
    $username = $this->input->post('username');
    $exists = $this->User_model->filename_exists($username);

    $count = count($exists);
    // echo $count 

    if (empty($count)) {
        return true;
    } else {
        return false;
    }
}

在模型中

function filename_exists($username)
{
    $this->db->select('*'); 
    $this->db->from('users');
    $this->db->where('username', $username);
    $query = $this->db->get();
    $result = $query->result_array();
    return $result
}

Answer 2

如果您只是想查看用户是否已经存在，则没有理由使用get（）函数进行查询。只需执行count_all_results（），如果找到用户，它将返回一个数字，否则返回0。

function filename_exists($username) {
  $this->db->where('username', $username);
  return $this->db->count_all_results('users'); 
}

如果你的数据库中存在用户名，那么所有这一切都会返回一个大于零的数字。

Answer 3

只是另一种结果相同的方法

<强> MODEL

function filename_exists($username) {
   $this->db->select()->from('users')->where('username', $username);
   $query = $this->db->get();

   return $query->first_row('array'); // returns first row if has record in db
}

<强> CONTROLLER

function filename_exists() {
   $username = $this->input->post('username');
   $user = $this->user_modal->filename_exists($username);

   return !empty($user); // not empty returns true else false
}

<强> AJAX

在check_if_exists功能中。

$.post('<?php echo base_url(); ?>index.php/files/filename_exists', { username: $('#username').val() }, function(response) {
    var html = '<span style="color:red;">Value does not exist</span>';
    if (response == true) {
       html = '<span style="color: green;">' + msg + '</span>';
    } 
    $('#msg').html(html);
});

使用codeigniter和ajax检查数据库中是否存在用户名

3 个答案: