[PHP] Nie mogę znaleźć błedu Opcje

[CryWolf]

Cześć, wiecie może czemu na pierwszym elseif wyrzuca błąd - linijka 52?
http://www.wklejto.pl/37188

[PHP] pobierz, plaintext 
<? 
require_once "mysql.php";
echo '
<TITLE>tytuł</TITLE>
<META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=utf-8"><center><br><br>';
 
  function twojeip() {
if ($_SERVER['HTTP_X_FORWARDED_FOR']) {    $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];  }
else                                  {    $ip = $_SERVER['REMOTE_ADDR'];  } 
                     }
twojeip();
 
 
if (!get_magic_quotes_gpc()) 
{      $wname = addslashes(htmlspecialchars($_POST['name']));
       $wsurname = addslashes(htmlspecialchars($_POST['surname'])); 
       $wage = addslashes(htmlspecialchars($_POST['age']));  
       $wcountry = addslashes(htmlspecialchars($_POST['country'])); 
       $wcomment = addslashes(htmlspecialchars($_POST['comment']));
       $wsex = addslashes(htmlspecialchars($_POST['sex']));
}
 
else
{      $wname = htmlspecialchars($_POST['name']);
       $wsurname = htmlspecialchars($_POST['surname']);
       $wage = htmlspecialchars($_POST['age']);
       $wcountry = htmlspecialchars($_POST['country']);
       $wcomment = htmlspecialchars($_POST['comment']);
       $wsex = htmlspecialchars($_POST['sex']);
 
}
 
 
 
 
 
 
 
 
    $result = mysql_query("SELECT * FROM vanhalen"); 
    $checked = mysql_num_rows($result);   
 
if ($_POST['accepted'] == 1) {
if ( empty($_POST['name']) OR empty($_POST['surname']) OR empty($_POST['age']) OR empty($_POST['country']) OR empty($_POST['fantime']) ) { echo '<b>Fill in all fields.<meta http-equiv="refresh" content="2;url=add.php"/></b>'; }
              }
 
 
 
 
    $result = mysql_query("SELECT id FROM vanhalen WHERE email = '$_POST[email]'"); 
    $checkedemail = mysql_num_rows($result); 
elseif ($checkedemail > 0) { echo '<b>This email is already in use! <meta http-equiv="refresh" content="2;url=add.php"/></b>'; }
 
 
 
 
 
    $result = mysql_query("SELECT id FROM vanhalen WHERE ip = '$ip'"); 
    $checkedip = mysql_num_rows($result); 
elseif ($checkedip > 0) { echo '<b>Your comment is still in the database! You cant add more than one.<meta http-equiv="refresh" content="2;url=add.php"/></b>'; }
 
 
 
 
 
 
 
 
 
 
 
 
 
elseif ($_POST['accepted'] == 1) 
{ 
mysql_query(blablabla);
echo 'Everything is ok!<meta http-equiv="refresh" content="2;url=index.php"/>';
}
 
 
 
 
else 
{
echo '<table border=10 bordercolor=black><center>
<form method=POST>
<td width=70><center>Name:</td><td><input type=text name=name SIZE="7" maxlength=15"><td><tr>
<td width=70><center>Surname:</td><td><input type=text name=surname SIZE="7" maxlength=15"><td><tr>
<td width=70><center>Age:</td><td><input type=text name=age SIZE="7" maxlength=15"><td><tr>
<td width=70><center>Country:</td><td><input type=text name=country SIZE="7" maxlength=15"><td><tr>
<td width=70><center>E-Mail:</td><td><input type=text name=email SIZE="7" maxlength=15"><td><tr>
<td width=70><center>Short comment:</td><td><input type=text name=email SIZE="7" maxlength=30"><td><tr>
<td width=70><center>Sex:</td><td><select name=sex><option value=Female>Female</option><option value=Male>Male</option></select><td><tr>
 
    
 
<input type=hidden name=accepted value=1>
<td colspan=2><center><br><input type=submit value=Confirmation><br><br></td>
</table>';  
}
 
 
 
 
 
?>
[PHP] pobierz, plaintext 

Ten post edytował CryWolf 26.06.2009, 17:55:18

[Pawel_W]

o ile wiem to pomiędzy if() { } i elseif() { } nie może być innych instrukcji..

Ten post edytował Pawel_W 26.06.2009, 18:30:47