班级明智时间表

Question

我想创建一个学生时间表，每周有5个科目，5个老师，5个班级，每天5小时和5天。我已经完成了所有事情，但现在我面临的问题是，如果没有特定的一天插入数据（假设星期一没有课程），我正在为星期二插入数据。然后星期二的数据显示在星期一。

这里有一些图片：

输出TimeTable（此处No Class字段的时间是下午2点至下午3点，日期是星期四。但它会在星期五和上午10点至11点显示）

我的TimeTable数据库表

我想显示一天是否插入了行。然后在那一天，数据应显示为“No Class＆＃39;在前端时间表中。我在这里写下我的逻辑：

   <body>
    <?php include 'sidenav.php';?>
    <article>
        <table border="2" style= "background-color: #84ed86; color: #761a9b; margin: 0 auto;" >
              <tr>
                <th><b>TIME</b></th>
                <th><b>Monday</b></th>
                <th><b>Tuesday</b></th>
                <th><b>Wednesday</b></th>
                <th><b>Thursday</b></th>
                <th><b>Friday</b></th>
              </tr>
            <tbody>
                <h1>Class Wise Time Table</h1>
                <?php
                    error_reporting(1);
                    session_start();
                    if ((isset($_SESSION['login'])) && (isset($_REQUEST['cls']))) {
                        $classname = $_REQUEST['cls'];
                        $conn = mysqli_connect("localhost", "root", "", "finaltask");
                        $sql = "select DISTINCT (time) as time from time_table where classname = "."'$classname'";
                        $rest = mysqli_query($conn, $sql);
                        while($t = mysqli_fetch_assoc($rest)){
                        $sql = "select * from time_table where classname = "."'$classname'". " AND time = '".$t['time']."'";
                        //print_r($sql);exit;
                        $res = mysqli_query($conn, $sql);
                        while($row = mysqli_fetch_assoc($res)){
                            //echo '<pre>'; print_r ($row);
                            if(($row['day']=='Monday')&&($row['time']==$t['time'])){
                                echo"<tr> <td>{$row['time']}</td>";
                                if(($row['subject']==' ')&&($row['teacher']== ' ')){
                                    echo "<td>No Class</td>";//die('');
                                }else{
                                    echo  "<td>".$row['subject']."<br>".$row['teacher']."</td>";
                                }
                            }
                            if(($row['day']=='Tuesday')&&($row['time'])){
                                if(($row['subject']=='')&&($row['teacher']=='')){
                                echo "<td>No Class</td>";
                                }else{
                                    echo  "<td>".$row['subject']."<br>".$row['teacher']."</td>";
                                }
                            }
                            if(($row['day']=='Wednesday')&&($row['time'])){
                                if(($row['subject']=='')&&($row['teacher']=='')){
                                echo "<td>No Class</td>";
                                }else{
                                    echo  "<td>".$row['subject']."<br>".$row['teacher']."</td>";
                                }
                            }
                            if(($row['day']=='Thurshday')&&($row['time'])){
                                if(($row['subject']=='')&&($row['teacher']=='')){
                                echo "<td>No Class</td>";
                                }else{
                                    echo  "<td>".$row['subject']."<br>".$row['teacher']."</td>";
                                }
                            }
                            if(($row['day']=='Friday')&&($row['time'])){
                                if(($row['subject']=='')&&($row['teacher']=='')){
                                echo "<td>No Class</td>";
                                }else{
                                    echo  "<td>".$row['subject']."<br>".$row['teacher']."</td></tr>";
                                }
                            }
                        }
                      }
                    }        
                ?>
                </tbody>
            </table>

Answer 1

这是我给你的建议。您可以定义一个新数组，并存储由工作日名称组织的while循环中的值，然后使用创建另一个数组，其中包含您想要查看的工作日名称，并使用foreach循环可以检查是否获得了任何值当前循环中的那一天，然后在没有数据的情况下打印No class一天，在有数据的那天打印time table，如下所示。

<?php
  $days_data = array();
  while($row = mysqli_fetch_assoc($res)){
        $days_data[$row['day']] == $row;
  }

if($days_data):
    $week_days = array('Monday','Tuesday','Wednesday','Thursday','Friday','Saturday','Sunday');
    foreach($week_days as $wday){
        if(!empty($days_data[$wday])){
            $row_data = $days_data[$wday]; // $row_data contains all of the data from that specific day time,name,teacher and etc...
            // there is data for this day
        } else {
            // no data for the day currently in loop
        }
    }
endif;

Answer 2

班级明智时间表

        <table border="2" style= "background-color: #84ed86; color: #761a9b; margin: 0 auto;" >
              <tr>
                <th><b>TIME</b></th>
                <th><b>Monday</b></th>
                <th><b>Tuesday</b></th>
                <th><b>Wednesday</b></th>
                <th><b>Thursday</b></th>
                <th><b>Friday</b></th>
              </tr>
            <tbody>

                <?php
                    error_reporting(1);
                    session_start();
                    if ((isset($_SESSION['login'])) && (isset($_REQUEST['cls']))) {
                        $classname = $_REQUEST['cls'];
                        $conn = mysqli_connect("localhost", "root", "", "finaltask");
                        $sql = "select DISTINCT (time) as time from time_table where classname = "."'$classname'";
                        $rest = mysqli_query($conn, $sql);
                        while($t = mysqli_fetch_assoc($rest)){
                        $sql = "select * from time_table where classname = "."'$classname'". " AND time = '".$t['time']."'";
                        //print_r($sql);exit;
                        $res = mysqli_query($conn, $sql);
                        ?>
                        <tr>
                            <td><?php echo $t['time'] ?></td>
                        <?php
                        $week_days = array('Monday','Tuesday','Wednesday','Thurshday','Friday');
                        $classes = array();
                        while($row = mysqli_fetch_assoc($res)) {
                            $classes[$row['day']] = $row;
                        }
                        foreach ($week_days as $day) {
                        ?>
                            <?php if (array_key_exists($day, $classes)) { $row = $classes[$day]; ?>
                            <td><?php echo $row['subject'] . '<br />' . $row['teacher'] ?></td>
                            <?php } else { ?>
                            <td>No Class</td>
                            <?php } ?>
                        <?php    
                        }
                        ?>
                        </tr>
                             <?php
                        }     
                ?>
                </tbody>
            </table>

如何在php中创建时间表

2 个答案:

班级明智时间表