如何通过点击按钮来执行php文件？

我有一个HTML页面，类似于“请假申请表”，其中有一个名为“接受”的按钮

我已经在下面编写了PHP代码，当我点击accept按钮时需要执行这些代码

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("leave", $con);
$leaveType=$_POST['leavename'];
$totalDays=$_POST['days'];
$EId=$_POST[empid];

$sql="SELECT 
          LFA, SickLeave, TransferLeave, HolidayLeave, AnnualLeave, UnpaidLeave 
      FROM leaverecords 
      WHERE empid='$EId'";

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

$result=mysql_query($sql);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
    $lfa = $row['LFA'];
    $sleave = $row['SickLeave'];
    $tleave = $row['TransferLeave'];
    $hleave = $row['HolidayLeave'];
    $aleave = $row['AnnualLeave'];
    $unpaidleave = $row['UnpaidLeave'];
}

if($leaveType=="LFA")
{
    if( $lfa>0 && $lfa>=$totalDays)
    {
        $newlfa = $lfa - $totalDays;
        $sql="UPDATE leaverecords SET LFA = '$newlfa' WHERE empid='$EId'";
        if (!mysql_query($sql,$con))
        {
            die('Error: ' . mysql_error());
        }  
    }
    else
        echo "Leave not available!";

}    

elseif($leaveType=="Sick Leave")
{        
    $newsleave = $sleave + $totalDays;
    $sql="UPDATE leaverecords SET SickLeave = '$newsleave' WHERE empid='$EId'";
    if (!mysql_query($sql,$con))
    {
        die('Error: ' . mysql_error());
    }        
}    

elseif( $leaveType=="Transfer Leave")
{       
    if($tleave!=5)
    { 
        $newtleave = $tleave + $totalDays;
        $sql="UPDATE leaverecords SET TransferLeave = '$newtleave' WHERE empid='$EId'";
        if (!mysql_query($sql,$con))
        {
            die('Error: ' . mysql_error());
        }
    }
    else
        echo "Leave not available!";         
}  

elseif($leaveType=="Holiday Leave")
{        
    $newhleave = $hleave + $totalDays;
    $sql="UPDATE leaverecords SET HolidayLeave = '$newhleave' WHERE empid='$EId'";
    if (!mysql_query($sql,$con))
    {
        die('Error: ' . mysql_error());
    }        
}

elseif($leaveType=="Annual Leave")
{        
    if($aleave>0 && $aleave>=$totalDays)
    {
        $newaleave = $aleave - $totalDays;
        $sql="UPDATE leaverecords SET AnnualLeave = '$newaleave' WHERE empid='$EId'";
        if (!mysql_query($sql,$con))
        {
            die('Error: ' . mysql_error());
        }

    }
    else
        echo "Leave not available!";         
}

elseif($leaveType=="Unpaid Leave")
{       
    if($unpaidleave<90)
    { 
        $newunpaidleave = $unpaidleave + $totalDays;
        $sql="UPDATE leaverecords SET UnpaidLeave = '$newunpaidleave' WHERE empid='$EId'";
        if (!mysql_query($sql,$con))
        {
            die('Error: ' . mysql_error());
        }
    }
    else
        echo "Leave not available!";         
}                
?>

HTML在哪里？你在哪里创建表单？没有PHP教程警告人们吗？hv是使用dreamweaver8创建的……在我的html文件中，所有这些都在下面。如果$Eid变量是一个，请确保将其清理为整数，否则你将很容易进行sql注入。这仍然是你的做法。“不工作”是对症状的模糊描述，无法提供诊断。该死！！：（在给出一些测试数据时犯了一个愚蠢的错误：|
<form method="post" action="url_to_that_script.php">
    <input value="Accept" type="submit">
    <!-- other fields -->
</form>