我正在寻找新的更新功能或建议来修复我现有的功能或建议

在代码中，我从数据库中收集数据，然后根据数据的值进行更新。我可以检索值，但无法更新它们，我需要有关更新函数或其他类型的函数语法的帮助。

<?php
$con = mysqli_connect("127.0.0.1","root","","timeclock");
if (mysqli_connect_errno($con))
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
    echo "connected";
}
$x=2;
timein($x);
function timein($x){
    $con = mysqli_connect("127.0.0.1","root","","timeclock");
    if (mysqli_connect_errno($con))
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }else{
        echo "connected";
    }
    $result=mysqli_query($con,"SELECT * FROM emplo`enter code here`yeedatabase WHERE employeeID=$x ORDER BY employeeID" );
    while($row = mysqli_fetch_array($result))
    {
        $inout = $row['inOut'];
    }
    echo $inout;
    if($inout == 1){
        echo "hello";
        $con = mysqli_connect("127.0.0.1","root","","timeclock");
        if (mysqli_connect_errno($con))
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }else{
            echo "connected";
        }
        $result=mysqli_query($con,"SELECT * FROM employeedatabase WHERE employeeID=$x ORDER BY employeeID" );
        while($row = mysqli_fetch_array($result))
        {
            $inout = $row['inOut'];
            $row['inOut']=2;
        }
        echo "</table>";
        echo $inout;
        $outa=2;
        mysqli_query($con,"UPDATE employeedatabase SET inOut='$outa' WHERE employeeID='$x'"); 
    }
}
?>-## Heading ##

mysqli 查询函数不起作用，我需要一个新函数或更好的语法来为我的 mysqli 函数。