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[Php] unexpected $end <- HILFE!
Discussion on [Php] unexpected $end <- HILFE! within the Web Development forum part of the Coders Den category.
02/07/2011, 18:10
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#1
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elite*gold: 10 
Join Date: Nov 2009
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[Php] unexpected $end <- HILFE!
Hallo, Elitepvpers.
ich habe versucht ein Registrierungs Page umzuändern aber ich habe diesen Fehler bekommen als ich fertig war (Dies ist leider nicht die einzigste seite wo das steht >.>)
Parse error: syntax error, unexpected $end in C:\xampp\htdocs\test\pages\register.php on line 158
Hier der Skript (Php-Datei):
PHP Code:
<div class="titre">
<img style="position:absolute;margin-top:-5px;margin-left:-22px;" src="img_style/icone_1.png"> S'inscrire sur <? echo $serveur ?>
</div>
<img style="position:absolute; margin-top: 30px; margin-left: -5px;" src="img_style/img_3.png">
<img style="position:absolute; margin-top: 220px; margin-left: 320px;" src="img_style/img_4.png">
<?php
mysql_connect( $dbIP, $dbUSER, $dbPSWD );
mysql_select_db( $dbNAME, $dbLOG );
mysql_set_charset ( "utf8" , $dbLOG );
$connect=mysql_connect( $dbIP, $dbUSER, $dbPSWD );
mysql_select_db( $dbNAME, $dbLOG );
mysql_set_charset ( "utf8" , $dbLOG );
if(isset($_POST['ok']))
{
$cadeau = $kdo;
$login = mysql_real_escape_string(htmlspecialchars(trim($_POST['name']) ) );
$pass_conf = mysql_real_escape_string(htmlspecialchars(trim($_POST['pass2']) ) );
$pass = mysql_real_escape_string(htmlspecialchars(trim($_POST['pass']) ) );
$mails = mysql_real_escape_string(htmlspecialchars(trim($_POST['mail']) ) );
$reponse = mysql_real_escape_string(htmlspecialchars(trim($_POST['secretanswer']) ) );
$question = mysql_real_escape_string(htmlspecialchars(trim($_POST['secretquestion']) ) );
$pseudo = mysql_real_escape_string(htmlspecialchars(trim($_POST['pseudo']) ) );
$freund = mysql_real_escape_string(htmlspecialchars(trim($_POST['freund']) ) );
if ($pass != $pass_conf){
echo "<font color=\"red\">Das Passwort darf nicht verschieden sein !</font>";
}else{
if($login=="" || $pass=="" || $mails=="" || $reponse=="" || $question=="" || $pseudo=="" || $freund=="")
{
echo '<font color="red">Fülle bitte alle Felder aus !</font><br>';
}else{
if($freund=="")
{
echo '<font color="red">Bitte gib an, wie du zu Undefined gekommen bist.</font><br>';
}else{
$sql1 = 'SELECT * FROM accounts WHERE account="'.$login.'"';
$req1 = mysql_query($sql1) or die('SQL Error !<br>'.$sql1.'<br>'.mysql_error());
$data1 = mysql_fetch_array($req1);
if($data1['account'] == $login)
{
echo '<font color="red">Der Accountname wird bereits verwendet.</font><br>';
}else{
$sql2 = 'SELECT * FROM accounts WHERE pseudo="'.$pseudo.'"';
$req2 = mysql_query($sql2) or die('SQL Error '.$sql2.''.mysql_error());
$data2 = mysql_fetch_array($req2);
if($data2['pseudo'] == $pseudo)
{
echo '<font color="red">Der Nickname wird bereits verwendet..</font><br>';
}else
{
mysql_query("INSERT INTO accounts (guid,account,pass,level,email,lastIP,lastConnectionDate,question,reponse,pseudo,banned,reload_needed,bankKamas,bank,friends,stable,freund) VALUES ('".$login."','".$login."','".$pass."','0','".$mails."','0','0','".$question."','".$reponse."','".$pseudo."','0','0','".$cadeau."','0','0','0','".$freund."')")or die(mysql_error());
echo (" ".$bienvenue." ");
}
}
}}}?>
<center><br>
<form method="post" action="" name="inscription">
Accountname :<br>
<input class="login" name="name"><br><br>
Passwort :<br>
<input class="login" name="pass" type="password"><br><br>
Passwort wiederholen :<br>
<input class="login" name="pass2" type="password"><br><br>
Email Adresse :<br>
<input class="login" name="mail"><br><br>
Geheimfrage :<br>
<input class="login" name="secretquestion"><br><br>
Geheimantwort :<br>
<input class="login" name="secretanswer"><br><br>
Nickname :<br>
<input class="login" name="pseudo"><br><br>
Wie habe ich von Undefined erfahren? :<br>
<input class="login" name="freund"><br><br>
<br>
<input style="width: 120px;" class="login" name="ok" value="Erstellen" type="submit">
</form>
</center>
Ich würde mich auf hilfe freuen 
Liebe Grüße, Defura.
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02/07/2011, 19:43
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#2
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elite*gold: 0
Join Date: Apr 2010
Posts: 1,352
Received Thanks: 788
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Quote:
Originally Posted by Defura
Hallo, Elitepvpers.
ich habe versucht ein Registrierungs Page umzuändern aber ich habe diesen Fehler bekommen als ich fertig war (Dies ist leider nicht die einzigste seite wo das steht >.>)
Parse error: syntax error, unexpected $end in C:\xampp\htdocs\test\pages\register.php on line 158
Hier der Skript (Php-Datei):
PHP Code:
<div class="titre">
<img style="position:absolute;margin-top:-5px;margin-left:-22px;" src="img_style/icone_1.png"> S'inscrire sur <? echo $serveur ?>
</div>
<img style="position:absolute; margin-top: 30px; margin-left: -5px;" src="img_style/img_3.png">
<img style="position:absolute; margin-top: 220px; margin-left: 320px;" src="img_style/img_4.png">
<?php
mysql_connect( $dbIP, $dbUSER, $dbPSWD );
mysql_select_db( $dbNAME, $dbLOG );
mysql_set_charset ( "utf8" , $dbLOG );
$connect=mysql_connect( $dbIP, $dbUSER, $dbPSWD );
mysql_select_db( $dbNAME, $dbLOG );
mysql_set_charset ( "utf8" , $dbLOG );
if(isset($_POST['ok']))
{
$cadeau = $kdo;
$login = mysql_real_escape_string(htmlspecialchars(trim($_POST['name']) ) );
$pass_conf = mysql_real_escape_string(htmlspecialchars(trim($_POST['pass2']) ) );
$pass = mysql_real_escape_string(htmlspecialchars(trim($_POST['pass']) ) );
$mails = mysql_real_escape_string(htmlspecialchars(trim($_POST['mail']) ) );
$reponse = mysql_real_escape_string(htmlspecialchars(trim($_POST['secretanswer']) ) );
$question = mysql_real_escape_string(htmlspecialchars(trim($_POST['secretquestion']) ) );
$pseudo = mysql_real_escape_string(htmlspecialchars(trim($_POST['pseudo']) ) );
$freund = mysql_real_escape_string(htmlspecialchars(trim($_POST['freund']) ) );
if ($pass != $pass_conf){
echo "<font color=\"red\">Das Passwort darf nicht verschieden sein !</font>";
}else{
if($login=="" || $pass=="" || $mails=="" || $reponse=="" || $question=="" || $pseudo=="" || $freund=="")
{
echo '<font color="red">Fülle bitte alle Felder aus !</font><br>';
}else{
if($freund=="")
{
echo '<font color="red">Bitte gib an, wie du zu Undefined gekommen bist.</font><br>';
}else{
$sql1 = 'SELECT * FROM accounts WHERE account="'.$login.'"';
$req1 = mysql_query($sql1) or die('SQL Error !<br>'.$sql1.'<br>'.mysql_error());
$data1 = mysql_fetch_array($req1);
if($data1['account'] == $login)
{
echo '<font color="red">Der Accountname wird bereits verwendet.</font><br>';
}else{
$sql2 = 'SELECT * FROM accounts WHERE pseudo="'.$pseudo.'"';
$req2 = mysql_query($sql2) or die('SQL Error '.$sql2.''.mysql_error());
$data2 = mysql_fetch_array($req2);
if($data2['pseudo'] == $pseudo)
{
echo '<font color="red">Der Nickname wird bereits verwendet..</font><br>';
}else
{
mysql_query("INSERT INTO accounts (guid,account,pass,level,email,lastIP,lastConnectionDate,question,reponse,pseudo,banned,reload_needed,bankKamas,bank,friends,stable,freund) VALUES ('".$login."','".$login."','".$pass."','0','".$mails."','0','0','".$question."','".$reponse."','".$pseudo."','0','0','".$cadeau."','0','0','0','".$freund."')")or die(mysql_error());
echo (" ".$bienvenue." ");
}
}
}}}?>
<center><br>
<form method="post" action="" name="inscription">
Accountname :<br>
<input class="login" name="name"><br><br>
Passwort :<br>
<input class="login" name="pass" type="password"><br><br>
Passwort wiederholen :<br>
<input class="login" name="pass2" type="password"><br><br>
Email Adresse :<br>
<input class="login" name="mail"><br><br>
Geheimfrage :<br>
<input class="login" name="secretquestion"><br><br>
Geheimantwort :<br>
<input class="login" name="secretanswer"><br><br>
Nickname :<br>
<input class="login" name="pseudo"><br><br>
Wie habe ich von Undefined erfahren? :<br>
<input class="login" name="freund"><br><br>
<br>
<input style="width: 120px;" class="login" name="ok" value="Erstellen" type="submit">
</form>
</center>
Ich würde mich auf hilfe freuen
Liebe Grüße, Defura. |
PHP Code:
<div class="titre">
<img style="position:absolute;margin-top:-5px;margin-left:-22px;" src="img_style/icone_1.png"> S'inscrire sur <? echo $serveur ?>
</div>
<img style="position:absolute; margin-top: 30px; margin-left: -5px;" src="img_style/img_3.png">
<img style="position:absolute; margin-top: 220px; margin-left: 320px;" src="img_style/img_4.png">
<?php
mysql_connect( $dbIP, $dbUSER, $dbPSWD );
mysql_select_db( $dbNAME, $dbLOG );
mysql_set_charset ( "utf8" , $dbLOG );
$connect=mysql_connect( $dbIP, $dbUSER, $dbPSWD );
mysql_select_db( $dbNAME, $dbLOG );
mysql_set_charset ( "utf8" , $dbLOG );
if(isset($_POST['ok']))
{
$cadeau = $kdo;
$login = mysql_real_escape_string(htmlspecialchars(trim($_POST['name']) ) );
$pass_conf = mysql_real_escape_string(htmlspecialchars(trim($_POST['pass2']) ) );
$pass = mysql_real_escape_string(htmlspecialchars(trim($_POST['pass']) ) );
$mails = mysql_real_escape_string(htmlspecialchars(trim($_POST['mail']) ) );
$reponse = mysql_real_escape_string(htmlspecialchars(trim($_POST['secretanswer']) ) );
$question = mysql_real_escape_string(htmlspecialchars(trim($_POST['secretquestion']) ) );
$pseudo = mysql_real_escape_string(htmlspecialchars(trim($_POST['pseudo']) ) );
$freund = mysql_real_escape_string(htmlspecialchars(trim($_POST['freund']) ) );
if ($pass != $pass_conf){
echo "<font color=\"red\">Das Passwort darf nicht verschieden sein !</font>";
}else{
if($login=="" || $pass=="" || $mails=="" || $reponse=="" || $question=="" || $pseudo=="" || $freund=="")
{
echo '<font color="red">Fülle bitte alle Felder aus !</font><br>';
}else{
if($freund=="")
{
echo '<font color="red">Bitte gib an, wie du zu Undefined gekommen bist.</font><br>';
}else{
$sql1 = 'SELECT * FROM accounts WHERE account="'.$login.'"';
$req1 = mysql_query($sql1) or die('SQL Error !<br>'.$sql1.'<br>'.mysql_error());
$data1 = mysql_fetch_array($req1);
if($data1['account'] == $login)
{
echo '<font color="red">Der Accountname wird bereits verwendet.</font><br>';
}else{
$sql2 = 'SELECT * FROM accounts WHERE pseudo="'.$pseudo.'"';
$req2 = mysql_query($sql2) or die('SQL Error '.$sql2.''.mysql_error());
$data2 = mysql_fetch_array($req2);
if($data2['pseudo'] == $pseudo)
{
echo '<font color="red">Der Nickname wird bereits verwendet..</font><br>';
}else
{
mysql_query("INSERT INTO accounts (guid,account,pass,level,email,lastIP,lastConnectionDate,question,reponse,pseudo,banned,reload_needed,bankKamas,bank,friends,stable,freund) VALUES ('".$login."','".$login."','".$pass."','0','".$mails."','0','0','".$question."','".$reponse."','".$pseudo."','0','0','".$cadeau."','0','0','0','".$freund."')")or die(mysql_error());
echo (" ".$bienvenue." ");
}
}
}}}
}?>
<center><br>
<form method="post" action="" name="inscription">
Accountname :<br>
<input class="login" name="name"><br><br>
Passwort :<br>
<input class="login" name="pass" type="password"><br><br>
Passwort wiederholen :<br>
<input class="login" name="pass2" type="password"><br><br>
Email Adresse :<br>
<input class="login" name="mail"><br><br>
Geheimfrage :<br>
<input class="login" name="secretquestion"><br><br>
Geheimantwort :<br>
<input class="login" name="secretanswer"><br><br>
Nickname :<br>
<input class="login" name="pseudo"><br><br>
Wie habe ich von Undefined erfahren? :<br>
<input class="login" name="freund"><br><br>
<br>
<input style="width: 120px;" class="login" name="ok" value="Erstellen" type="submit">
</form>
</center>
Du hast eine geschweifte Klammer vor dem ?> vergessen falls ich mich nicht täusche... probier aus sollte nu gehen-
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02/07/2011, 20:27
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#3
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elite*gold: 10
Join Date: Nov 2009
Posts: 1,929
Received Thanks: 471
|
Hat geklappt, danke 
Doch jetzt hab ich das gleiche problem auf der Home.php (da hab ich nichts geändert  )
PHP Code:
<?php
if (!defined('SECU'))
{
die();
}
$tNEWS = numtable("live_news");
$nPAGES = ceil(3 / 3);
if (isset($_GET['p']))
{
$p = secu($_GET['p']);
}
else
{
$p = 1;
}
$fMTS = ($p - 1) * 3;
?>
<?php if(3 == 0){
echo "<h1>Aucune news</h1><h2>Rien à signaler</h2><p>Pas de news pour le moment ...</p>";
}
else
{
$query = 'SELECT auteur, titre, date, contenu, soustitre FROM live_news ORDER BY date DESC LIMIT 0, 3;';
$sql = mysql_query($query) or die(''.$query.'<br />'.mysql_error());
while ($data = mysql_fetch_array($sql))
{
sscanf($data['date'], "%4s-%2s-%2s %2s:%2s:%2s", $an, $mois, $jour, $heure, $min, $sec);
$auteur = $data['auteur'];
?>
<div class="titre">
<img style="position:absolute;margin-top:-5px;margin-left:-22px;" src="img_style/icone_1.png"> <? echo '' , stripslashes(trim($data['titre'])) , ''; ?>
<span class="date"><? echo "Le $jour/$mois/$an à $heure:$min"; ?> par <? echo "<b>$auteur</b>"; ?></span>
</div>
<img style="float:left;" src="img_style/img_2.png">
<? echo '' , nl2br(stripslashes($data['contenu'])) , ''; ?>
<br /><br />
<?
}
mysql_free_result($sql);
}
if($tNEWS > $wNEWS) { echo '';
for ($i = 1 ; $i <= $nPAGES ; $i++)
{
echo '
';
}
echo '';}
?>
wäre cool wenn auch das gelöst wird =D
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02/07/2011, 21:29
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#4
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elite*gold: 0
Join Date: Apr 2010
Posts: 1,352
Received Thanks: 788
|
Quote:
Originally Posted by Defura
Hat geklappt, danke
Doch jetzt hab ich das gleiche problem auf der Home.php (da hab ich nichts geändert )
PHP Code:
<?php
if (!defined('SECU'))
{
die();
}
$tNEWS = numtable("live_news");
$nPAGES = ceil(3 / 3);
if (isset($_GET['p']))
{
$p = secu($_GET['p']);
}
else
{
$p = 1;
}
$fMTS = ($p - 1) * 3;
?>
<?php if(3 == 0){
echo "<h1>Aucune news</h1><h2>Rien à signaler</h2><p>Pas de news pour le moment ...</p>";
}
else
{
$query = 'SELECT auteur, titre, date, contenu, soustitre FROM live_news ORDER BY date DESC LIMIT 0, 3;';
$sql = mysql_query($query) or die(''.$query.'<br />'.mysql_error());
while ($data = mysql_fetch_array($sql))
{
sscanf($data['date'], "%4s-%2s-%2s %2s:%2s:%2s", $an, $mois, $jour, $heure, $min, $sec);
$auteur = $data['auteur'];
?>
<div class="titre">
<img style="position:absolute;margin-top:-5px;margin-left:-22px;" src="img_style/icone_1.png"> <? echo '' , stripslashes(trim($data['titre'])) , ''; ?>
<span class="date"><? echo "Le $jour/$mois/$an à $heure:$min"; ?> par <? echo "<b>$auteur</b>"; ?></span>
</div>
<img style="float:left;" src="img_style/img_2.png">
<? echo '' , nl2br(stripslashes($data['contenu'])) , ''; ?>
<br /><br />
<?
}
mysql_free_result($sql);
}
if($tNEWS > $wNEWS) { echo '';
for ($i = 1 ; $i <= $nPAGES ; $i++)
{
echo '
';
}
echo '';}
?>
wäre cool wenn auch das gelöst wird =D
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PHP Code:
<?php
if (!defined('SECU'))
{
die();
}
$tNEWS = numtable("live_news");
$nPAGES = ceil(3 / 3);
if (isset($_GET['p']))
{
$p = secu($_GET['p']);
}
else
{
$p = 1;
}
$fMTS = ($p - 1) * 3;
?>
<?php if(3 == 0){
echo "<h1>Aucune news</h1><h2>Rien à signaler</h2><p>Pas de news pour le moment ...</p>";
}
else
{
$query = 'SELECT auteur, titre, date, contenu, soustitre FROM live_news ORDER BY date DESC LIMIT 0, 3;';
$sql = mysql_query($query) or die(''.$query.'<br />'.mysql_error());
while ($data = mysql_fetch_array($sql))
{
sscanf($data['date'], "%4s-%2s-%2s %2s:%2s:%2s", $an, $mois, $jour, $heure, $min, $sec);
$auteur = $data['auteur'];
?>
<div class="titre">
<img style="position:absolute;margin-top:-5px;margin-left:-22px;" src="img_style/icone_1.png"> <? echo '' , stripslashes(trim($data['titre'])) , ''; ?>
<span class="date"><? echo "Le $jour/$mois/$an à $heure:$min"; ?> par <? echo "<b>$auteur</b>"; ?></span>
</div>
<img style="float:left;" src="img_style/img_2.png">
<? echo '' , nl2br(stripslashes($data['contenu'])) , ''; ?>
<br /><br />
<?php
}
mysql_free_result($sql);
}
if($tNEWS > $wNEWS) { echo '';
for ($i = 1 ; $i <= $nPAGES ; $i++)
{
echo '
';
}
echo '';}
?>
Bei mir zeigt er keinen Fehler mehr an.
Lieben Gruß.
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02/07/2011, 21:48
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#5
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elite*gold: 10
Join Date: Nov 2009
Posts: 1,929
Received Thanks: 471
|
Waaaa,
Hamma vielen dank
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