Constant T_VARIBALE syntax unexpected error

2l8vsan

I've tried changing my code ensuring that the values are established and declared before the variable is actually being used. But I keep getting
Parse error: syntax error, unexpected '$query' (T_VARIABLE)

<?php
require 'database.php';

$indexCounter = 0;
$login = 0;

$welcome_msg = "Welcome, " . "$userID" . ".";
$error_msg = "Invalid. Try again";

if (isset($_POST['userID'])) {
$userID = $_POST['userID'];
}

if (isset($_POST['userPassword'])) {
$userPassword = $_POST['userPassword'];
}

$userID = $_POST['userID'];
$userPassword = $_POST['userPassword']

//Access Database
$query = "SELECT * FROM table1";
$getUserID = $db->query($query);
$getUserPassword = $db->query($query);



//Cross reference database to form
foreach ($getUserID as $gUID) {
if ($userID == $gUID['userID'] && $userPassword == $gUID['userPassword']) {
	$login = 1;
	echo $welcome_msg;
	}
	}
?>



<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns = "http://www.w3.org/1999/xhtml">

<body>
<br />
<?php if ($login == 1) { ?>
	<p><?php echo $welcome_msg; ?> </p>
	<table>
	<tr>
		<th>Crt. No.</th>
		<th>Student ID</th>
		<th>Password</th>
	</tr>
	<?php foreach ($gS2 as $gUID): ?>
		<tr>
			<?php $indexCounter = ($indexCounter + 1); ?>
			<td><?php echo $indexCounter; ?></td>
			<td><?php echo $gS1["userID"]; ?></td>
			<td><?php echo $gS2["userPassword"]; ?></td>
		</tr>
	<?php endforeach; ?>
</table>
</p>
<?php } ?>
<?php if($login == 0) { ?>
	<?php echo $error_msg; ?>
	<?php } ?>
	<br />
</body>
</html>

Right now, it is saying line22 or "$query" is causing the problem. But query is defined. What should I do to make this work?

leatherback

I do not see you initiating $db anywhere, which may cause the problem. Then, it is not saying $query is not defined. It is telling you that it did not expect $query where you used it.

What does $dh->query() take as input?

NogDog

Missing semi-colon on the preceding command.

2l8vsan

$db =
Gets its value from another page.
('include database.php')
Where is the missing semicolon? I'm sorry but I can't find it, or what you are referring to.

bradgrafelman

2l8vsan;11036025 wrote:
Where is the missing semicolon?

No one knows - it's missing. 😉

On line 19, there is no terminating semicolon:

$userPassword = $_POST['userPassword']

EDIT: Also, rather than being horribly inefficient and looping through every single row in the DB, why not use a WHERE condition on that SQL query to only SELECT the single row you care about?

2l8vsan

I know it is inefficient. It's part of a requirement, so it has to be done that way, where everything is displayed.
I now get an undefined index error for userID, all on the same line 31.

Notice: Undefined index: userID
and the notice repeats 6 times. Interestingly as I have 6 records in the table I need to talk to.
So the main issue it is pointing at is

foreach ($getUserID as $gUID) {
if ($userID == $gUID['userID'] && $userPassword == $gUID['userPassword']) {
	$login = 1;
	echo $welcome_msg;
	}
	}
?>

I do not see how the index is undefined. As userID is well established way up at Line 5.

Any ideas?

NogDog

As I don't know for sure what your database functions do, my best guess is that there is no column named 'userID' in the table being queried? (It's complaining about $gUID['userID'], not the $userID variable.)

2l8vsan

My table schema is table students (student_id, password), with primary key being student_id.

I keep getting 'undefined index: student_id'. But student_id is the field I need to query in this page.
And it is compared to the form input, where userID is compared to userID['student_id']. Hence the associative array.

<?php
require 'database.php';
session_start();
$indexCounter = 0;
$login = 0;


$userID = ($_POST['userID']);
$userPassword = ($_POST['userPassword']);

if (isset($_POST['userID'])) {
$userID = ($_POST['userID']);
}

if (isset($_POST['userPassword'])) {
$userPassword = ($_POST['userPassword']);
}


$welcome_msg = "Welcome, " . "$userID" . ".";
$error_msg = "Invalid logon or password.";

$query = "SELECT * FROM students;";
$student_logins = $db->query($query);
$student_logint = $db->query($query);

$userID = $_POST['userID'];
$userPassword = $_POST['userPassword'];

foreach ($student_logins as $student_login) {
if ($userID == $student_login['student_id'] && $userPassword == $student_login['password'] )
{
	$login = 1;
	echo $welcome_msg; }}

?>



<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns = "http://www.w3.org/1999/xhtml">
<body>
<br />
<?php if ($login == 1) { ?>
	<p><?php echo $welcome_msg; ?> </p>
	<table>
	<tr>
		<th>Number</th>
		<th>Student Logon</th>
		<th>Password</th>
	</tr>
<?php foreach ($student_logint as $student_login): ?>
	<tr>
		<?php $indexCounter = $indexCounter + 1; ?>
		<td><?php echo $indexCounter; ?></td>
		<td><?php echo $student_login['student_id']; ?></td>
		<td><?php echo $student_login['password']; ?></td>
		</tr>
		<?php endforeach; ?>
</table>
</p>
<?php } ?>
<?php if($login == 0) { ?>
	<?php echo $error_msg; ?>
	<?php } ?>
	<br />
</body>
</html>

Any ideas?

2l8vsan

I should also mention it displays an undefined index notice for password, too.
Essentially, the two fields I attempt to display via arrays (student_id, password) errors both display for both fields of the table saying they are undefined indices.

NogDog

Try injecting some print_r() or var_dump() commands to find out what the result arrays really look like, and see if that points you toward the actual problem.

I have a macro in my editor that outputs the following for me: 🙂

echo "<pre>".print_r($var, 1)."</pre>\n";

(change $var to whatever the variable is you want to inspect)

2l8vsan

Wow that worked!
I have 5 records in my table but I'll show the output of one from the above code you gave me (print_r) ..

Array 
(
       [STUDENT_ID] => user1
       [0] => user1
       [PASSWORD] => 1234
       [1] => 1234
)

I don't get it. It clearly gets the information from the array, it just doesn't want to show it via <td> cells when I try via XHTML.
It gives me an undefined index error for password and student_id, but it's clearly defined lol.

Weedpacket

Based on what you've shown it's because the fields are named [font=monospace]STUDENT_ID[/font] and [font=monospace]PASSWORD[/font], not [font=monospace]student_id[/font] and [font=monospace]password[/font].

2l8vsan

Unbelievable how it's case sensitive in php code, but not case-sensitive in myphpadmin when creating the schema. It finally works. Thank you.

Weedpacket

Column names are case-insensitive in SQL (and table names are supposed to be as well, but MySQL doesn't do that). But array keys in PHP are entirely different things, and are case-sensitive.

Derokorian

Weedpacket;11036077 wrote:
(and table names are supposed to be as well, but MySQL doesn't do that on linux).

Fixed that for you.

johanafm

unless "lower_case_table_names" option is set to 1 or 2. Or wait, is that only if set to 1? :o

Windows - case insensitive
Unix & Linux - case sensitive
Unless it's OSX using HFS+ file system - case insensitive

lower_case_table_names
0: store casing, compare sensitive
1: store in lower-case, compare insensitive
2: store casing, but compare as if lower-case

For extra fun: it's apparently possible and allowed to set a value of 0 on case insensitive file systems, but doing so while refering to tables using different casings may lead to MyISAM index corruption.
http://dev.mysql.com/doc/refman/5.7/en/identifier-case-sensitivity.html
http://dev.mysql.com/doc/refman/5.7/en/server-system-variables.html#sysvar_lower_case_table_names