-3

を取得してParse error: syntax error, unexpected T_IFいます。if (mail($to, $subject, $body, $header))を閉じましたsemi-colons。Ifステートメントに問題がある理由がわかりません

 <?php
    include('dbconnect.php');

    if (isset($_POST['submit']))
    {

        ...

        $to = 'sent@sent.com';
        $subject = $_POST['Subject'];
        $header = 'From:' .$_POST['Email'];
        $message = 
        'New fedback sent by  
        First Name: '.$_POST['FirstName'].' 
        Last Name: '.$_POST['LastName'].'
        about '.$_POST['Subject'].' 
        Message: '.$_POST['msg'].' 
        '

        if (mail($to, $subject, $body, $header))
        {

    }   
?>
4

2 に答える 2

2

$messageたとえば、割り当てを閉じる必要があります;

 $message = 
    'New fedback sent by  
    First Name: '.$_POST['FirstName'].' 
    Last Name: '.$_POST['LastName'].'
    about '.$_POST['Subject'].' 
    Message: '.$_POST['msg'].' 
    ';
于 2013-07-02T02:40:12.087 に答える
1

これ：

    $message = 
    'New fedback sent by  
    First Name: '.$_POST['FirstName'].' 
    Last Name: '.$_POST['LastName'].'
    about '.$_POST['Subject'].' 
    Message: '.$_POST['msg'].' 
    '

閉じセミコロンがありません;。あなたはそれを閉じる必要があります：

    $message = 
    'New fedback sent by  
    First Name: '.$_POST['FirstName'].' 
    Last Name: '.$_POST['LastName'].'
    about '.$_POST['Subject'].' 
    Message: '.$_POST['msg'];
于 2013-07-02T02:40:03.290 に答える