如何通過點擊按鈕執行一個PHP文件？

-1

我有一個HTML頁面，就像一個「離開申請表」，其中有一個名爲「接受」的按鈕如何通過點擊按鈕執行一個PHP文件？

我寫了下面的PHP代碼，當我點擊該接受時需要執行按鈕。

<?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!";   
}     
?>

當我點擊它時，我該怎麼做按鈕來使PHP代碼運行？

來源

2011-11-23 ZoeHime

HTML在哪裏？你在哪裏創建表單？ – ManseUK

沒有PHP教程警告人們[Bobby Tables]（http://bobby-tables.com/）？ – Quentin

hv使用dreamweaver8創建....在我的html文件中，所有這些都在

– ZoeHime

<form method="post" action="url_to_that_script.php"> 
    <input value="Accept" type="submit"> 
    <!-- other fields --> 
</form>

來源

2011-11-23 16:32:00 Quentin

我已經完成了..那不是工作 – ZoeHime

這仍然是你如何做到的。「不工作」對症狀的描述過於模糊，無法提供診斷。 – Quentin

該死！ :(在提供一些測試數據時犯了一個愚蠢的錯誤：| – ZoeHime

如何通過點擊按鈕執行一個PHP文件？

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