Question

美好的一天！我正在尝试将mysql语法转换为使用ajax的codeigniter，但我的头脑很难解决这个问题，可能是因为我是新的codigniter，..这里是我的代码

在我的模特中

雇员并网data.php

 <?php
 $servername = "localhost";
 $username = "root";
 $password = "";
 $dbname = "test";
 $conn = mysqli_connect($servername, $username, $password, $dbname) or   die("Connection failed: " . mysqli_connect_error());
 $requestData= $_REQUEST;
 $columns = array( 
 0 =>'employee_name', 
 1 => 'employee_salary',
 2=> 'employee_age',
 3 => 'id',
 );
 $sql = "SELECT id, employee_name, employee_salary, employee_age ";
 $sql.=" FROM employee";
 $query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get      employees");
  $totalData = mysqli_num_rows($query);
 $totalFiltered = $totalData; 
 $sql = "SELECT id, employee_name, employee_salary, employee_age ";
 $sql.=" FROM employee WHERE 1=1";
 if( !empty($requestData['search']['value']) ) {  
 $sql.=" AND ( employee_name LIKE '".$requestData['search']['value']."%' ";    
 $sql.=" OR employee_salary LIKE '".$requestData['search']['value']."%' ";
 $sql.=" OR employee_age LIKE '".$requestData['search']['value']."%' )";
 }
 $query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get  employees");
 $totalFiltered = mysqli_num_rows($query);
 $sql.=" ORDER BY ". $columns[$requestData['order'][0]['column']]."   ".$requestData['order'][0]['dir']."  LIMIT ".$requestData['start']." ,".$requestData['length']."   "; 
 $query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get   employees");
 $data = array();
 while( $row=mysqli_fetch_array($query) ) {  // preparing an array
 $nestedData=array(); 
 $nestedData[] = $row["id"];
 $nestedData[] = $row["employee_name"];
 $nestedData[] = $row["employee_salary"];
 $nestedData[] = $row["employee_age"];
 $data[] = $nestedData;
 }
 $json_data = array(
        "draw"            => intval( $requestData['draw'] ),
        "recordsTotal"    => intval( $totalData ),  // total number of  records
         "recordsFiltered" => intval( $totalFiltered ), // total number of  records after searching, if there is no searching then totalFiltered = totalData
         "data"            => $data   // total data array
         );
 echo json_encode($json_data);  // send data as json format
 ?>

这是我的看法的index.php

 <!DOCTYPE html>
 <html>
 <title>Datatable</title>
 <head>
    <link rel="stylesheet" type="text/css" href="css/jquery.dataTables.css">
    <script type="text/javascript" language="javascript" src="js/jquery.js">  </script>
     <script type="text/javascript" language="javascript"  src="js/jquery.dataTables.js"></script>
     <link rel="stylesheet" type="text/css" href="assets/css/table.css">
     <script type="text/javascript" language="javascript" >

        $(document).ready(function() {
            var dataTable = $('#employee-grid').DataTable( {
                "processing": true,
                "serverSide": true,
                "ajax":{
                    url :"employee-grid-data.php", // json datasource
                    type: "post",  // method  , by default get
                    error: function(){  // error handling
                        $(".employee-grid-error").html("");
                        $("#employee-grid").append('<tbody class="employee- grid-error"><tr><th colspan="3">No data found in the server</th></tr></tbody>');
                        $("#employee- grid_processing").css("display","none");

                    }
                }
            } );
        } );
     </script>
     <style>
        div.container {
            margin: 0 auto;
            max-width:760px;
        }
        div.header {
            margin: 100px auto;
            line-height:30px;
            max-width:760px;
        }
        body {
            background: #f7f7f7;
            color: #333;
            font: 90%/1.45em "Helvetica  Neue",HelveticaNeue,Verdana,Arial,Helvetica,sans-serif;
        }
     </style>
 </head>
 <body>
     <div class="header"><h1>DataTable</h1></div>
    <div class="container">
        <table style="color:black;" class="rwd-table" id="employee-grid"   cellpadding="0" cellspacing="0" border="0" class="display" width="100%">
                <thead>
                    <tr>
                        <th>ID</th>
                        <th>Employee name</th>
                        <th>Salary</th>
                        <th>Age</th>


                    </tr>
                </thead>
        </table>
    </div>
 </body>
 </html>

这是我正在尝试更改为模型区域的语法。

// getting total number records without any search
$this->db->select('id, employee_name, employee_salary, employee_age');
$this->db->from('employee');

我删除了数据库连接语法并应用于config / database.php但根本不工作。

Answer 1

您需要的大量信息：http://www.codeigniter.com/user_guide/database/index.html

如果您还没有完成教程，可以在此处找到它们：http://www.codeigniter.com/user_guide/tutorial/index.html

他们非常有助于弄清楚基础知识。

config/database.php

 $db['default'] = array(
    'hostname' => 'localhost',
    'username' => 'root',
    'password' => '',
    'database' => 'test',
    ...
);

model我称之为employee_model。 models/employee_model.php

class Employee_model extends CI_Model {

    public function __construct()
    {
            $this->load->database();  // or autoload it...
    }

    public function get_employee_grid() 
    {
        $result = StdClass();

        //$requestData= $_REQUEST;  
        // use $this->input->post('id') for POST Data
        // use $this->input->get('id') for GET Data
        // etc...

       $q = $this->db->select("id, employee_name, employee_salary, employee_age")->get('employee');
       $result->count = $q->num_rows();  //get's number of rows
       $result->rows = $q->result();   // object of resulting rows

    ...
       // when you need to search... use ->where()
       $this->db->$this->db->select("id, employee_name, employee_salary, employee_age")->where('id', $this->input->post('id'))->get('employee');

       return json_encode($result);
    }

从您的观点来看，url :"http://example.com/your_controller/employee_grid_data",

致电控制器：

class Your_controller extends CI_Controller {

    public function employee_grid_data()
    {
        $this->load->model('employee_model');

        $data = $this->employee_model->get_employee_grid();

        // if you expect ONLY JSON data, you can echo
        echo $data;

        // otherwise, you pass the variable to the view
    }
}

希望这会让你走上正确的道路......

将mysqli转换为codeigniter语法

1 个答案: