php - 我正在寻找新的更新功能或建议来修复我现有的功能

Question

在代码中，我从数据库中收集数据，然后根据数据的值更新它。我可以检索这些值，但我无法更新它们 我需要有关更新函数的语法或不同类型函数的帮助。

<?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 查询函数不工作 我需要一个新函数或更好的 mysqli 函数语法。

Answer 1

“不工作”不是很有帮助，在未来的错误中，放入和没有发生的事情是有帮助的。

但看起来这条线需要从

SELECT * FROM emplo`enter code here`yeedatabase WHERE employeeID=$x ORDER BY employeeID" );

至

SELECT * FROM employeedatabase WHERE employeeID=$x ORDER BY employeeID" );