Parse error: syntax error, unexpected $end in...

**Atli** · Jun 23 '09, 05:59 PM

Hi.

You don't close your string properly on line #26.
That's why you get this error.

Also, to close a huge security hole in that code.
Don't put user input directly into SQL queries... or anything, for that matter.
Always validate it and sanitize it before using it.
The mysql_real_esca pe_query function is extremely helpful here.

Check out SQL Injection in the manual. Explains why this is so important.

**mideastgirl** · Jun 23 '09, 10:24 PM

unexpected $end

How would I end line 26? I have tried removing the semi-colon and using ) instead of } and I still keep getting the same result. I am not really sure what to try next.

**Atli** · Jun 24 '09, 12:32 AM

Close the string with a double-quote before the semi-colon.

The string should be enclosed in quote-marks, but you only ever open the string, forgetting to close it with the second quote-mark.

The problem is basically:
[code=php]
// You are doing this:
$sql = "INSERT INTO whatever;

// Where you should be doing
$sql = "INSERT INTO whatever";
[/code]
The variables, which you correctly enclose in brackets, are not a part of the problem, and should work fine if left alone as they are.

**mideastgirl** · Jun 24 '09, 12:39 PM

syntax error, unexpected '{' ...on line 26

Now it is telling me the brackets are wrong on line 26. The issue the whole time has been with line 26, however I did go ahead and make the changes you suggested:
[code=php]
mysql_connect($ hostname,$usern ame, $password) OR DIE ('Unable to connect to database! Please try again later.');
mysql_select_db ($dbname);

$query = 'SELECT * FROM $usertable';
$result = mysql_query($qu ery);
if($result) {
while($row = mysql_fetch_arr ay($result)){
$name = $row['$yourfield'];
echo 'Name: '.$name;

mysql_select_db ("mideasthonors 2");

$sql="INSERT INTO $usertable (Name, Address 1, Address 2, City, State, Zip Code, Website, Contact Name, Position, Phone Number, Email Address)";

VALUES

{$_POST['name']},{$_POST['address 1']},{$_POST['address 2']},{$_POST['city']},{$_POST['state']},{$_POST['zip code']},{$_POST['website']},{$_POST['contact name']},{$_POST['position']},{$_POST['phone number']},{$_POST['email address']};

if (!mysql_query($ sql,$con))

{

die('Error: ' . mysql_error());

}

echo '1 record added';

mysql_close($co n);
}
}
?>[/code]

**Atli** · Jun 24 '09, 01:38 PM

You closed the string in the wrong line.

Everything you want to have inside a string must be enclosed inside the quotes.
You have like 3 lines there that should be inside your string, but you close the string after the first line.

You are doing something like this:
[code=php]
$str = "This string spans";
multiple lines and
should not be closed
until after the last line;
[/code]
So PHP, having closed the string after the first line, tries to parse the rest of what should be inside the string as PHP code, obviously failing.

It should look like:
[code=php]
$str = "This string spans
multiple lines and
should not be closed
until after the last line";
[/code]

**mideastgirl** · Jun 24 '09, 01:49 PM

syntax error, unexpected '}' in ...on line 23

So I put the quote at the end of line 26, which solved that problem. Now at the bottom of the script where the last 2 curly brackets are, I am now receiving this error:

syntax error, unexpected '}' in /home/content/m/i/d/mideasthonors/html/admintaskstable .php on line 23

I thought these brackets were necessary to close the connection?

**dlite922** · Jun 24 '09, 03:57 PM

Originally posted by mideastgirl

So I put the quote at the end of line 26, which solved that problem. Now at the bottom of the script where the last 2 curly brackets are, I am now receiving this error:

syntax error, unexpected '}' in /home/content/m/i/d/mideasthonors/html/admintaskstable .php on line 23

I thought these brackets were necessary to close the connection?

It's not that hard. All opened brackets must have closing brackets.

Try this code:

Code:

<?php
mysql_connect($hostname,$username, $password) OR DIE ('Unable to connect to database! Please try again later.');
mysql_select_db($dbname);
 
$query = 'SELECT * FROM $usertable';
$result = mysql_query($query);

if($result) 
{
    while($row = mysql_fetch_array($result))
    {
        $name = $row['$yourfield'];
        echo 'Name: '.$name;
 
		mysql_select_db("mideasthonors2");
 
		$sql="INSERT INTO $usertable (Name, Address 1, Address 2, City, State, Zip Code, Website, Contact Name, Position, Phone Number, Email Address) VALUES {$_POST['name']},{$_POST['address 1']},{$_POST['address 2']},{$_POST['city']},{$_POST['state']},{$_POST['zip code']},{$_POST['website']},{$_POST['contact name']},{$_POST['position']},{$_POST['phone number']},{$_POST['email address']}";
 
		if (!mysql_query($sql,$con))		 
		{
			die('Error: ' . mysql_error());
		}
		 
		echo '1 record added'; 
    }
}

mysql_close($con);    // should be at the same level as your connect().

**mideastgirl** · Jun 24 '09, 04:14 PM

Apparently it is more difficult for me than others:(

I tried the code you gave me and this is the error I received:

Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in /home/content/m/i/d/mideasthonors/html/admintaskstable .php on line 28

You did not include the database connection which I added on line 2, which gave me an error that no connection was possible, or something to that degree. And now it is giving me the error above. This is the code I am using:
[code=php]
<?php
include ("admintasks-dbcon.php");
mysql_connect($ hostname,$usern ame, $password) OR DIE ('Unable to connect to database! Please try again later.');
mysql_select_db ($dbname);

$query = 'SELECT * FROM $usertable';
$result = mysql_query($qu ery);

if($result)
{
while($row = mysql_fetch_arr ay($result))
{
$name = $row['$yourfield'];
echo 'Name: '.$name;

mysql_select_db ("mideasthonors 2");

$sql="INSERT INTO $usertable (Name, Address 1, Address 2, City, State, Zip Code, Website, Contact Name, Position, Phone Number, Email Address) VALUES {$_POST['name']},{$_POST['address 1']},{$_POST['address 2']},{$_POST['city']},{$_POST['state']},{$_POST['zip code']},{$_POST['website']},{$_POST['contact name']},{$_POST['position']},{$_POST['phone number']},{$_POST['email address']}";

if (!mysql_query($ sql,$con))
{
die('Error: ' . mysql_error());
}

echo '1 record added';
}
}
mysql_close($co n);
?>[/code]

**code green** · Jun 25 '09, 07:41 AM

I find it hard to believe you do not understand.
The error message tells you exactly what is wrong.

Code:

mysql_close($con);

Where is the resource $con?
Is it in

include ("admintasks-dbcon.php");

If not then it just comes randomly into play half way through your script.
Even so it is not a critical error so it should still work.
I would have expected a more serious error earlier because the mysterious $con
is used here

Code:

 if (!mysql_query($sql,$con))

but php is good at guessing which database you mean

**mideastgirl** · Jun 25 '09, 12:58 PM

Thank you for your help. I deleted the start session at the top of this page and no error message came up, but it still will not go to the next page of admintasks.php which is where it should go to after logging in. Here is my script:

Code:

<?php
include("admin-dbcon.php");

//Check to see if the username and password is valid (if it is in the database)
$validate = mysql_query("select * from admin_login where username = '$_POST[username]' and password = '$_POST[password]'"); 

$isvalid=mysql_num_rows($validate);

//if valid login send on, if not send to error page
if ($isvalid != 0) {
	$_SESSION['username'] = $_POST[username];
	while ($row=mysql_fetch_array($validate)){
		$_SESSION['userid']=$row["ID"];
	}
 	header("Location: admintasks.php");
} 
//else {
//	header("Location: login-error.php");
//}
?>

**mideastgirl** · Jun 25 '09, 01:37 PM

Headers already sent

I am still getting an error message when I try to login from page www.mideasthonors.org/adminlogin.php this is the error I am receiving:

Warning: Cannot modify header information - headers already sent by (output started at /home/content/m/i/d/mideasthonors/html/adminloginproce ss.php:1) in /home/content/m/i/d/mideasthonors/html/adminloginproce ss.php on line 15
This is the code I am using:
[code=php]
<?php
include("admin-dbcon.php");

//Check to see if the username and password is valid (if it is in the database)
$validate = mysql_query("se lect * from admin_login where username = '$_POST[username]' and password = '$_POST[password]'");

$isvalid=mysql_ num_rows($valid ate);

//if valid login send on, if not send to error page
if ($isvalid != 0) {
$_SESSION['username'] = $_POST[username];
while ($row=mysql_fet ch_array($valid ate)){
$_SESSION['userid']=$row["ID"];
}
header("Locatio n: admintasks.php" );
}
//else {
// header("Locatio n: login-error.php");
//}
?>
[/code]

**code green** · Jun 25 '09, 02:32 PM

Cannot modify header information - headers already sent by

This generally means the browser has already output something or recieved HTML headers before receiving some header information.

I am not very experienced with header stuff but session_start must be called before any other output to the browser, including header().

What is admin-dbcon.php up to?

Just check you have no white space either side of <?php, this has got me a few times

**mideastgirl** · Jun 25 '09, 02:41 PM

No white space on any pages!

This is the code in admin-dbcon.php:

[code=php]
<?php
$dbHost = 'mideasthonors. db.4381361.host edresource.com' ;
$dbName = 'mideasthonors' ;
$dbUser = 'xxx';
$dbPass = 'xxx';

mysql_connect(" $dbHost","$dbUs er","$dbPass" ) or die("Error: Unable to connect to database! Please try again later.");

//Select the database we want to use
mysql_select_db ($dbName) or die("Error: Could not select database");
?>
[/code]

I have not opened a session in either one or a header at the top. So I will re-add session start () and see what happens

**mideastgirl** · Jun 25 '09, 02:47 PM

I feel like code needs to be in adminlogin.php and admintasks.php to get the two to connect to the actual loginprocess.ph p page. I am trying to get from adminlogin.php to admintasks.php through a login process which is in loginprocess.ph p. The code to connect to mysql is in admin-dbcon.php...in other words, I feel like all five pages should be linked someone??? Maybe I am wrong...but I am literally only a week into all the php stuff and have looked up how to do what I have so far online. I really do appreciate the help you all are giving.

Parse error: syntax error, unexpected $end in...

Parse error: syntax error, unexpected $end in...

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