Question

所以我有这个程序使用javascript / php应该是一种时间卡系统，签入，将签入日期发送到数据库，签出，它比较日期然后它计算你的数量＆＃ 39;制作完成后，将其发送回数据库。

现在的问题是，javascript / ajax触发的php脚本似乎每个会话只运行一次，因为它运行代码，然后重新运行一个快照（？）代码，它不再与数据库一起更新，除非我进入新的会话。

以下是我使用timeSystem.php的代码:(目前只使用测试按钮，其他按钮/代码仍然在那里）

<?php
session_start();
?>
<html>
<head>


<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js"></script>
<script type="text/javascript">

    function timeCard(id) {
        if(id == "Test"){
            $(function(){
               $.get("insert.php", function(data){
                   alert(data);
               });
            });
        }
        else if(id == "Test2"){
            $(function(){
               $.get("singin.php", function(data){
                   alert(data);
               });
            });
        }
    }

</script>
<?php 
if(!$_SESSION['driverid']){
    header("location:login.php");
}
else {
?>

</head>

<body>

<h1 style="float: left" class="center">Unity Transport Authority SMS (Shift Management System)</h1> <h3>V0.2</h3>
<div style="clear: both"></div>

<table> 
    <tr>
        <td><button type="button" id="Test" onclick="timeCard(id)">Test</button></td>
        <td><button type="button" id="Test2" onclick="timeCard(id)">Test2</button></td>
        <td><form action="logout.php" mothod="post"><input type="submit"></form> </td>
    </tr>
</table>

</body>
</html>
<?php
}
?>

insert.php：

<?php
session_start();
$con=mysqli_connect(/* Removing info but this is working as intended */);
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$t = time();
$Driverno = $_SESSION['id'];

$DBDriver= mysqli_query($con,"SELECT * FROM AdminForm 
    WHERE DRIVERNO=$Driverno");

while ($row = mysqli_fetch_array($DBDriver)) {
    if ($row['CURSTATE'] == 0) {
        $sql="UPDATE AdminForm SET SINGING=$t, CURSTATE =1
        WHERE DRIVERNO=$Driverno";

        if (!mysqli_query($con,$sql)) {
            die('Error: ' . mysqli_error($con));
        }
        echo "Have a nice shift!";
    } else {
        echo "You are already signed in!";
    }
}
mysqli_close($con);
?>

singin.php：

<?php
session_start();
$con=mysqli_connect(/* Again just removed because personal info */);
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$Driverno = $_SESSION['id'];
$Payrate = 1510;

/* Select the driver from the database */
$DBDriver= mysqli_query($con,"SELECT * FROM AdminForm 
    WHERE DRIVERNO=$Driverno");

while ($row = mysqli_fetch_array($DBDriver)) {
    if ($row['CURSTATE'] == 1) { /* Checks if user is signed in */
        if ($row['RANK'] == 1) {
            $t = time();
            $DBtime=$row['SINGING'];
            echo $row['SINGING'] . "<br>";
            $time = $t - $DBtime; /*  Compare the time to the database */
            echo "Total Amount of time Worked This Session (in seconds): " . $time . "<br>";
            $Totalhours = $time + $row['TOTALHRS'];
            $Totalhours = $Totalhours / 60 / 60; /*  Conversion from seconds into hourly format */
            echo "Total Hours Worked: " . $Totalhours . "hrs";
            echo "<br>";
            $Totalpayout = $Totalhours /*/ 60 / 60*/ * $Payrate;
            echo "Total Payout is now: " . $Totalpayout . "$";

            $Paid = 0;
            $sql="UPDATE AdminForm SET TOTALHRS=$Totalhours, TOTALPAYOUT=$Totalpayout, PAID=$Paid, CURSTATE=0
            WHERE DRIVERNO=$Driverno";

            if (!mysqli_query($con,$sql)) {
              die('Error: ' . mysqli_error($con));
            }

        }
        else/* else if? */ {
            // Future code for different ranks
            echo "This is bad !";
            die();
        }
    }
    else {
        echo "Please sign in before attemping to sign out!";
    }
};

mysqli_close($con);
?>

发生的一个例子是，你登录然后按下test2按钮，如果你在值班，它会检查CURSTATE。如果没有，它会说＆＃34;有一个很好的转变＆＃34;，并且更改CURSTATE，然后如果你再次按下按钮，它将不会再次运行代码并检查你是否值班，它将只是说＆＃34;有一个很好的转变＆＃34;试。

但是如果您注销并重新登录，会重置会话，它会再次运行代码，并会说“＃34;您已经登录了！＆＃34;。

我真的希望你得到我正在谈论的内容并且可以帮助我，并且感谢您提前提供任何帮助！

PHP - 仅运行一次/不更新变量的代码

0 个答案: