Question

我正在尝试从表“花名册”中读取数据，该名册基于登录的人从该表中获取数据，如果有人对如何执行此操作有任何想法，我已经尝试了一些方法，但是我认为他们是对的

这是我的页面，我希望将数据加载到

<?php
include('session.php');
?>
<!DOCTYPE html>
<html>
<head>
<meta charset='utf-8' />
<link href='fullcalendar.min.css' rel='stylesheet' />
<link href='fullcalendar.print.min.css' rel='stylesheet' media='print' />
<script src='js/moment.min.js'></script>
<script src='js/jquery.min.js'></script>
<script src='fullcalendar.min.js'></script>
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<meta name="description" content="">
<meta name="author" content="">
<script>

$(document).ready(function () {
    var calendar = $('#calendar').fullCalendar({
		header: {
			left: 'prev,next today',
			center: 'title',
			right: 'month,basicWeek,basicDay'
		},
		navLinks: true, // can click day/week names to navigate views
		editable: true,
		eventLimit: true,
        events: "load_Emp_Roster.php",
        displayEventTime: true,
        eventRender: function (event, element, view) {
            if (event.allDay === 'true') {
                event.allDay = true;
            } else {
                event.allDay = false;
            }
        },
       selectable: true,
        selectHelper: true,
       select: function (start, end, allDay) {
      
            calendar.fullCalendar('unselect');
        }, 
        
    

    });
});

以下是我的会话文件

<?php
   include('config1.php');
   session_start();
   
   $user_check = $_SESSION['login_user'];
   
   $ses_sql = mysqli_query($db,"select * from emp where email = '$user_check' ");
   
   $row = mysqli_fetch_array($ses_sql);
   
   $login_session = $row['name'];
   
   if(!isset($_SESSION['login_user'])){
      header("location:login.php");
   }
?>

下面是我的文件，我在其中从包含已登录用户名的名册表中获取数据，我认为他是我出了错的地方，但我会提出任何建议

<?php
include('session.php');


$connect = new PDO('mysql:host=localhost;dbname=paperlesspub', 'root', '');

$data = array();

$query = "SELECT * FROM roster WHERE Name = '$user_check'";

$statement = $connect->prepare($query);

$statement->execute();

$result = $statement->fetchAll();

foreach($result as $row)
{
 $data[] = array(
  'id'   => $row["reqID"],
  'title'   => $row["name"],
  'start'   => $row["Start"],
  'end'   => $row["End"]
 );
}

echo json_encode($data);

?>

根据登录的用户从表中读取数据

0 个答案: