Question

我正在使用phpmyadmin创建一个考勤系统。我目前正在使用以下列：日期，CardUID（因为我计划将RFID与之集成），超时，超时，迟到和宵禁。我还制作了一个html文件，该文件目前处于基本形式，没有CSS和其他内容。 The html screenshot

因此该代码旨在从html获取CardUID和时间。然后，代码将检查是否已经在数据库中输入了UID（包括超时），如果已输入，则将“ Time in in-time-”列放入“ Time in Time-in”，如果未输入，则将其更新为“ Time-out”柱。它还会检查“报到”日期的日期，如果它是在星期一降落的，则迟到时间与工作日的其余时间不同。所有时间的宵禁都是一样的。

<?php
    include("Connect.php");

    $dbselect = mysqli_select_db($conn, "att");

    $DATE = date('d-m-y');
    $UID = $_GET['uid'];
    $TIME = $_GET['time'];
    $TDY = FALSE;
    $CURF = FALSE;
    $sql_u = "SELECT * FROM `stuatte` WHERE EXISTS(SELECT * FROM `stuatte` WHERE `CardUID`='$UID' AND `Time-out`=NULL)"; 
    $dayOfWeek = date("w", strtotime($DATE));


    if($sql_u == TRUE){ 
        if (date($TIME) < 1700) { 
            $CURF = true;
            $TOUT = $TIME;
            $sql = "UPDATE `stuatte` SET `Time-out`='$TOUT' WHERE `CardUID`='$UID'";
        }
        else{
            $TOUT = $TIME;
            $sql = "UPDATE `stuatte` SET `Time-out`='$TOUT' WHERE `CardUID`='$UID'";
        }
        mysqli_query($conn, $sql);
    }
    else{
        if($dayOfWeek == 1){
            if (date($TIME) > 1315) { 
                $TDY = TRUE;
                $TIN = $TIME;
                $sql ="INSERT INTO `stuatte`(`Date`, `CardUID`, `Time-in`, `Time-out`, `Tardy`, `Curfew`) VALUES ('$DATE','$UID','$TIN','$TOUT','$TDY','$CURF')";
            }
            else{
                $TIN = $TIME;
                $sql ="INSERT INTO `stuatte`(`Date`, `CardUID`, `Time-in`, `Time-out`, `Tardy`, `Curfew`) VALUES ('$DATE','$UID','$TIN','$TOUT','$TDY','$CURF')";
            }
        mysqli_query($conn, $sql);
        }
        else if($dayOfWeek == 2 || $dayOfWeek == 3 || $dayOfWeek == 4 || $dayOfWeek == 5){
            if (date($TIME) > 1330) { 
                $TDY = TRUE;
                $TIN = $TIME;
                $sql ="INSERT INTO `stuatte`(`Date`, `CardUID`, `Time-in`, `Time-out`, `Tardy`, `Curfew`) VALUES ('$DATE','$UID','$TIN','$TOUT','$TDY','$CURF')";
            }
            else{
                $TIN = $TIME;
                $sql ="INSERT INTO `stuatte`(`Date`, `CardUID`, `Time-in`, `Time-out`, `Tardy`, `Curfew`) VALUES ('$DATE','$UID','$TIN','$TOUT','$TDY','$CURF')";
            }
        mysqli_query($conn, $sql);          
        }
    }

    header("refresh:1; url=UI.php");
?>

我的问题在于代码没有显示，据说是在html表中。感谢您抽出宝贵时间阅读本文，并可能对我有所帮助。

为什么没有数据发送到我的phpmyadmin数据库？

0 个答案: