registration2

[bleured27]

help please  next error

Parse error: syntax error, unexpected T_ECHO in /Applications/XAMPP/xamppfiles/htdocs/register2.php on line 47

<?php

$host="localhost"; 
$username=""; 
$password=""; 
$db_name="test"; 
$tbl_name="userinfo"; 

mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
$mypassword2=$_POST['mypassword2'];
$email=$_POST['email'];
$email2=$_POST['email2'];
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$mypassword2 = stripslashes($mypassword2);
$email = stripslashes($email);
$email2 = stripslashes($email2);

$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$mypassword2 = mysql_real_escape_string($mypassword2);
$email = mysql_real_escape_string($email);
$email2 = mysql_real_escape_string($email2);

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' ";
$result=mysql_query($sql);


$count=mysql_num_rows($result);



if($count==1){
echo"name already used"
;
}
else{
if ($mypassword != $mypassword1){
echo"you filld in 2 diffrend passwords";
}
else{
if ($email != $email1{
echo"you filld in 2 diffrend emails";
}
else{

ini_set('display_errors', 1); // 0 = uit, 1 = aan
error_reporting(E_ALL);

function checkemail($email)
{

    if(!stristr($email, '@'))
    {
        return false;
    }
    $email_split = explode("@", $email);

    if(count($email_split) != 2)
    {
        return false;
    }
    $email_user = $email_split[0];
    $email_host = $email_split[1];

    if(!getmxrr($email_host, $var))
    {
        return false;
    }

    if(!preg_match("/^[0-9a-z]([-_.~]?[0-9a-z])*$/i", $email_user))
    {
        return false;
    }
    return trsue;
}


$email = (isset($_POST['email'])) ? $_POST['email'] : '';


if(!checkemail($email))
{
    echo 'not failid email<br />';
}
else
{
$sql="SELECT * FROM $tbl_name WHERE email='$email'";
$result2=mysql_query($sql2);


$count2=mysql_num_rows($result2);



if($count2==1){
echo"email already used"
}
else
{

mysql_query("INSERT INTO $tbl_name VALUES ( '$myusername', $mypassword, '$email','0' )"); 

echo"registred succesfull,you can login now";  
}  
}
}
}
}
?>

 

MOD EDIT:

 . . . 

tags added.

[Pikachu2000]

When posting code, enclose it within the forum's

 . . . 

BBCode tags.

[bleured27]

OKEY can you find error?

[Pikachu2000]

The problem is on the line before the line that is reported in the error message. There's something missing from it.

[bleured27]

can you tell me what i am totaly blind now

[bleured27]

or do you mean this nothing missing here

 

Parse error: syntax error, unexpected T_ECHO in /Applications/XAMPP/xamppfiles/htdocs/register2.php on line 47

im on a mac is that something that help you my mac SKILLS are 0

[Nyuszer]

line 46

 

if ($email != $email1) {

[Pikachu2000]

if( $email != $email1  {

What goes here? -----^

[bleured27]

tnx that helped now i got this

 

Parse error: syntax error, unexpected '}', expecting ',' or ';' in /Applications/XAMPP/xamppfiles/htdocs/register2.php on line 102

<code>

<?php

 

$host="localhost";

$username="";

$password="";

$db_name="test";

$tbl_name="userinfo";

 

mysql_connect("$host", "$username", "$password")or die("cannot connect");

mysql_select_db("$db_name")or die("cannot select DB");

 

$myusername=$_POST['myusername'];

$mypassword=$_POST['mypassword'];

$mypassword2=$_POST['mypassword2'];

$email=$_POST['email'];

$email2=$_POST['email2'];

$myusername = stripslashes($myusername);

$mypassword = stripslashes($mypassword);

$mypassword2 = stripslashes($mypassword2);

$email = stripslashes($email);

$email2 = stripslashes($email2);

 

$myusername = mysql_real_escape_string($myusername);

$mypassword = mysql_real_escape_string($mypassword);

$mypassword2 = mysql_real_escape_string($mypassword2);

$email = mysql_real_escape_string($email);

$email2 = mysql_real_escape_string($email2);

 

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' ";

$result=mysql_query($sql);

 

 

$count=mysql_num_rows($result);

 

 

 

if($count==1){

echo"name already used"

;

}

else{

if ($mypassword != $mypassword1){

echo"you filld in 2 diffrend passwords";

}

else{

if ($email != $email1){

echo"you filld in 2 diffrend emails";

}

else{

 

ini_set('display_errors', 1); // 0 = uit, 1 = aan

error_reporting(E_ALL);

 

function checkemail($email)

{

 

    if(!stristr($email, '@'))

    {

        return false;

    }

    $email_split = explode("@", $email);

 

    if(count($email_split) != 2)

    {

        return false;

    }

    $email_user = $email_split[0];

    $email_host = $email_split[1];

 

    if(!getmxrr($email_host, $var))

    {

        return false;

    }

 

    if(!preg_match("/^[0-9a-z]([-_.~]?[0-9a-z])*$/i", $email_user))

    {

        return false;

    }

    return trsue;

}

 

 

$email = (isset($_POST['email'])) ? $_POST['email'] : '';

 

 

if(!checkemail($email))

{

    echo 'not failid email<br />';

}

else

{

$sql="SELECT * FROM $tbl_name WHERE email='$email'";

$result2=mysql_query($sql2);

 

 

$count2=mysql_num_rows($result2);

 

 

 

if($count2==1){

echo"email already used"

}

else

{

 

mysql_query("INSERT INTO $tbl_name VALUES ( '$myusername', $mypassword, '$email','0' )");

 

echo"registred succesfull,you can login now"; 

} 

}

}

}

}

?>

</code>

[bleured27]

oops i mean

<?php

$host="localhost"; 
$username=""; 
$password=""; 
$db_name="test"; 
$tbl_name="userinfo"; 

mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
$mypassword2=$_POST['mypassword2'];
$email=$_POST['email'];
$email2=$_POST['email2'];
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$mypassword2 = stripslashes($mypassword2);
$email = stripslashes($email);
$email2 = stripslashes($email2);

$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$mypassword2 = mysql_real_escape_string($mypassword2);
$email = mysql_real_escape_string($email);
$email2 = mysql_real_escape_string($email2);

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' ";
$result=mysql_query($sql);


$count=mysql_num_rows($result);



if($count==1){
echo"name already used"
;
}
else{
if ($mypassword != $mypassword1){
echo"you filld in 2 diffrend passwords";
}
else{
if ($email != $email1){
echo"you filld in 2 diffrend emails";
}
else{

ini_set('display_errors', 1); // 0 = uit, 1 = aan
error_reporting(E_ALL);

function checkemail($email)
{

    if(!stristr($email, '@'))
    {
        return false;
    }
    $email_split = explode("@", $email);

    if(count($email_split) != 2)
    {
        return false;
    }
    $email_user = $email_split[0];
    $email_host = $email_split[1];

    if(!getmxrr($email_host, $var))
    {
        return false;
    }

    if(!preg_match("/^[0-9a-z]([-_.~]?[0-9a-z])*$/i", $email_user))
    {
        return false;
    }
    return trsue;
}


$email = (isset($_POST['email'])) ? $_POST['email'] : '';


if(!checkemail($email))
{
    echo 'not failid email<br />';
}
else
{
$sql="SELECT * FROM $tbl_name WHERE email='$email'";
$result2=mysql_query($sql2);


$count2=mysql_num_rows($result2);



if($count2==1){
echo"email already used"
}
else
{

mysql_query("INSERT INTO $tbl_name VALUES ( '$myusername', $mypassword, '$email','0' )"); 

echo"registred succesfull,you can login now";  
}  
}
}
}
}
?>

can some one eddit that?

[Nyuszer]

line 101:

 

echo"email already used";

 

if you get parse error, always check the line mentioned in the error and the line before that