Question

已更新

我试图为所有学生在最后一年制定一个基本时间表，将他们的最终项目提交给2名主管。我已成功创建了一个没有约束的计划，但现在我需要根据主管可用性创建计划。

以下是该问题的详细说明。

学生被指派一名主管，主管将监督多名学生。主管还在白天教授课程。现在我需要为所有学生创建一个时间表给他们的主管和另一个监督其他学生的主管。（目前我专注于分配给学生的主管而不是第二个主管直到我开始工作。

我想比较主管时间的主管可用性，如果他们有空，然后将他们分配到插槽，然后将该时间的可用性更新为false，以避免使用PHP sql查询进行双重预订。

所以...

到目前为止我做了什么：

// get all of the Slots. 
    $sql = "SELECT Slot_ID  FROM slot WHERE  StartTime < '18:00:00'"; 
    $result = mysqli_query($con, $sql);
    $DayTimeSlots = array(); 
    if(mysqli_num_rows($result) > 0){
        while($row = mysqli_fetch_assoc($result)){

            $DayTimeSlots [] = $row;

        }
    }

// Store slots into simple array 

    function extractSlotId($DayTimeSlots){
    return $DayTimeSlots['Slot_ID'];

    }

    $slots = array_map("extractSlotId",$DayTimeSlots);

    // SHOW SLOTS
    foreach ($slots as $slotID) {
    echo " SlotID:  = $slotID <br>"; 

    }

    // Get All students 


    $sql = "SELECT Student_ID FROM student WHERE PartTime =0"; 
    $result = mysqli_query($con, $sql);
    $FullTimeStudents = array(); 
    if(mysqli_num_rows($result) > 0){
        while($row = mysqli_fetch_assoc($result)){

            $FullTimeStudents [] = $row;

        }
    }


  // Store into a simple array 


    // Extract student id and Supervisor_ID 

    function extractStudentId($FullTimeStudents){
    return $FullTimeStudents['Student_ID'];

    }

    $students = array_map("extractStudentId",$FullTimeStudents);


    // Combine the Slot and Students array 

    $min = min(count($FullTimeStudents), count($DayTimeSlots));
    $result = array_combine(array_slice($students , 0, $min), array_slice($slots, 0, $min));

     foreach($result as $key=>$value){ // $Key = Student_ID, $value = Slot_ID
        echo " $key : $value "; 

        // get supervisor ID 

    $sql = "select Supervisor_ID FROM student where Student_ID = $key";
    $query = mysqli_query($con, $sql);
    $row = mysqli_fetch_array($query); 
    $SuperID = $row['Supervisor_ID'];
    echo  "SuperID : $SuperID "; 

    // get slotID 

    $sql = "select Date, StartTime FROM Slot where Slot_ID = $value";
    $query = mysqli_query($con, $sql);
    $row = mysqli_fetch_array($query); 
    $Date = $row['Date'];
    $StartTime = $row['StartTime'];
    echo  "Slot Date : $Date Start Time : $StartTime "; 

        // get Date id 

    $sql = "select Date_ID FROM dates where Date = '$Date'";
    $query = mysqli_query($con, $sql);
    $row = mysqli_fetch_array($query); 
    $DateID = $row['Date_ID'];
    echo  "Date ID : $DateID "; 

    // Check if the supervisor is available 

    $sql = "select `$StartTime` FROM supervisor_availability where Date_ID = $DateID AND Supervisor_ID = $SuperID";
    $query = mysqli_query($con, $sql);
    $row = mysqli_fetch_array($query); 
    $Available = $row["$StartTime"];
    echo  "Is the Lecture Available?  : $Available  "; 
    $Time = "`$StartTime`"; 

    if($Available == 1){

    $sql = "INSERT INTO student_slot (Student_ID, Slot_ID) VALUES ($key, $value)";
    $result = mysqli_query($con, $sql);
    $sql = "UPDATE `supervisor_availability` SET $Time = '0' WHERE `supervisor_availability`.`Supervisor_ID` = $SuperID AND `supervisor_availability`.`Date_ID` = $DateID" ;
    $result = mysqli_query($con, $sql); 


    } else if ($Available == 0) {

        // Not sure what to do here. 
        // Can i make the Slot it's currently checking go to the end of the 
        // array for a different student and then check the next slot and come 
        // back to it.  
    }
     }

我使用echo进行调试。

如果Supervisor可用，该算法正常工作，它会正确分配，然后更新该时段的主管可用性

如果它们不可用，只需要帮助处理它。

任何帮助或建议将不胜感激。

Supervisor_Availability Slot Table Student Table Date Table DB_Structure

使用PHPmyadmin和SQL检查调度可用性的方法和算法

0 个答案: