Mysql table error?

**drkelusion** · Aug 21st, 2005, 07:24 AM

Hi Everyone!
I created a register script that works fine, and creates the correct fields and stuff in the database, but when I go to login, it comes with an error...

The error says:
Parse error: parse error, unexpected $ in /home/public_html/logscript.php on line 61

The error comes up on the logscript.php file (which is linked by a html form (<form action="logscript.php" method="post">...) on a file called login.php)

Here is the code on logscript.php

Code:

<?php

include("config.php"); 

// connect to the mysql server
$link = mysql_connect($server, $db_user, $db_pass)
or die ("Could not connect to mysql because ".mysql_error());

// select the database
mysql_select_db($database)
or die ("Could not select database because ".mysql_error());

$match = "select id from $table where username = '".$_POST['username']."'
and password = '".$_POST['password']."';"; 

$qry = mysql_query($match)
or die ("Could not match data because ".mysql_error());
$num_rows = mysql_num_rows($qry); 

if ($num_rows <= 0) { 
echo "Sorry, there is no username $username with the specified password.<br>";
echo "<a href=login.php>Try again</a>";
exit; 
} else {

setcookie("loggedin", "TRUE", time()+(3600 * 24));
setcookie("mysite_username", "$username");

if ($class != One) {
echo "You are now logged in as One!<br>"; 
echo "Continue to the <a href=one/index.php>members</a> section.";

if ($class != Two) {
echo "You are now logged in as Two!<br>"; 
echo "Continue to the <a href=two/index.php>members</a> section.";

if ($class != Three) {
echo "You are now logged in as Three!<br>"; 
echo "Continue to the <a href=three/index.php>members</a> section.";

if ($class != Four) {
echo "You are now logged in as Four!<br>"; 
echo "Continue to the <a href=Four/index.php>members</a> section.";

if ($class != Five) {
echo "You are now logged in as Five!<br>"; 
echo "Continue to the <a href=Five/index.php>members</a> section.";

if ($class != Six) {
echo "You are now logged in as Six!<br>"; 
echo "Continue to the <a href=Six/index.php>members</a> section.";

if ($class != Seven) {
echo "You are now logged in as Seven!<br>"; 
echo "Continue to the <a href=Seven/index.php>members</a> section.";

if ($class != Eight) {
echo "You are now logged in as Eight!<br>"; 
echo "Continue to the <a href=Eight/index.php>members</a> section.";

?>

Also, the "class" ($class) is found in different ROWS in the user table... I created a test user, called "test" and the class is "One".

Thanks to anyone who takes the time to read my problem and to all who try to help!

**{yak}** · Aug 21st, 2005, 04:06 PM

You missed your closing '}' on most of your if statments, there should be a corresponding closing '}' for every opened '{'. Also, you need to quote your 'One', 'Two' ect...

Code:

if ($class != 'One') {
echo "You are now logged in as One!<br>"; 
// you shoud have double quotes in your html, so you could do it like this to make it easier..
echo 'Continue to the <a href="one/index.php">members</a> section.';
}

// same for the rest of them...

**drkelusion** · Aug 21st, 2005, 05:31 PM

Hi!
Thankyou for helping!
I'm now getting this error when I editted the script:

Parse error: parse error, unexpected $ in /home/public_html/logscript.php on line 68

Thanks again for trying to help!

**ninjanutz** · Aug 21st, 2005, 05:41 PM

check and make sure that u didnt accidently make something a string on that line or add a $ on accident

**ninjanutz** · Aug 21st, 2005, 05:45 PM

ur script doesnt have 68 lines lol we'll at least the example u sent doesnt

**drkelusion** · Aug 21st, 2005, 05:48 PM

Here's the new editted code:

Code:

<?php

include("config.php"); 

// connect to the mysql server
$link = mysql_connect($server, $db_user, $db_pass)
or die ("Could not connect to mysql because ".mysql_error());

// select the database
mysql_select_db($database)
or die ("Could not select database because ".mysql_error());

$match = "select id from $table where username = '".$_POST['username']."'
and password = '".$_POST['password']."';"; 

$qry = mysql_query($match)
or die ("Could not match data because ".mysql_error());
$num_rows = mysql_num_rows($qry); 

if ($num_rows <= 0) { 
echo "Sorry, there is no username $username with the specified password.<br>";
echo "<a href=login.php>Try again</a>";
exit; 
} else {

setcookie("loggedin", "TRUE", time()+(3600 * 24));
setcookie("mysite_username", "$username");

if ($class != 'One') {
echo "You are now logged in as an One!<br>"; 
echo 'Continue to the <a href="One/index.php">members</a> section.';
}

if ($class != 'Two') {
echo "You are now logged in as an Two!<br>"; 
echo 'Continue to the <a href="Two/index.php">members</a> section.';
}

if ($class != 'Three') {
echo "You are now logged in as a Three!<br>"; 
echo 'Continue to the <a href="Three/index.php">members</a> section.';
}

if ($class != 'Four') {
echo "You are now logged in as a Four!<br>"; 
echo 'Continue to the <a href="Four/index.php">members</a> section.';
}

if ($class != 'Five') {
echo "You are now logged in as a Five!<br>"; 
echo 'Continue to the <a href="Five/index.php">members</a> section.';
}

if ($class != 'Six') {
echo "You are now logged in as an Six!<br>"; 
echo 'Continue to the <a href="Six/index.php">members</a> section.';
}

if ($class != 'Seven') {
echo "You are now logged in as a Seven!<br>"; 
echo 'Continue to the <a href="Seven/index.php">members</a> section.';
}

if ($class != 'Eight') {
echo "You are now logged in as Eight!<br>"; 
echo 'Continue to the <a href="Eight/index.php">members</a> section.';
}
?>

Did I do something wrong?
By the way, just to let you know, the "$class" is there because I want the code to check which "Class" the member is in (Which the information is found in "Users" table, in rows), which is found in the 6th column of every row for every user.

There is no "$class = [something]" code in there, am I doing something completely wrong?

Thanks again for helping...

**drkelusion** · Aug 21st, 2005, 05:49 PM

Originally Posted by ninjanutz

ur script doesnt have 68 lines lol we'll at least the example u sent doesnt

Lol.. I know, that's something I'm confused about too..
It's the full File, I didnt edit anything when I copied it to here..

**{yak}** · Aug 21st, 2005, 06:41 PM

You're getting closer, this time you missed the closing '}' of your else statement:

Code:

<?php

include("config.php"); 

// connect to the mysql server
$link = mysql_connect($server, $db_user, $db_pass)
or die ("Could not connect to mysql because ".mysql_error());

// select the database
mysql_select_db($database)
or die ("Could not select database because ".mysql_error());

$match = "select id from $table where username = '".$_POST['username']."'
and password = '".$_POST['password']."';"; 

$qry = mysql_query($match)
or die ("Could not match data because ".mysql_error());
$num_rows = mysql_num_rows($qry); 

if ($num_rows <= 0) { 
echo "Sorry, there is no username $username with the specified password.<br>";
echo "<a href=login.php>Try again</a>";
exit; 
} else {

setcookie("loggedin", "TRUE", time()+(3600 * 24));
setcookie("mysite_username", "$username");

if ($class != 'One') {
echo "You are now logged in as an One!<br>"; 
echo 'Continue to the <a href="One/index.php">members</a> section.';
}

if ($class != 'Two') {
echo "You are now logged in as an Two!<br>"; 
echo 'Continue to the <a href="Two/index.php">members</a> section.';
}

if ($class != 'Three') {
echo "You are now logged in as a Three!<br>"; 
echo 'Continue to the <a href="Three/index.php">members</a> section.';
}

if ($class != 'Four') {
echo "You are now logged in as a Four!<br>"; 
echo 'Continue to the <a href="Four/index.php">members</a> section.';
}

if ($class != 'Five') {
echo "You are now logged in as a Five!<br>"; 
echo 'Continue to the <a href="Five/index.php">members</a> section.';
}

if ($class != 'Six') {
echo "You are now logged in as an Six!<br>"; 
echo 'Continue to the <a href="Six/index.php">members</a> section.';
}

if ($class != 'Seven') {
echo "You are now logged in as a Seven!<br>"; 
echo 'Continue to the <a href="Seven/index.php">members</a> section.';
}

if ($class != 'Eight') {
echo "You are now logged in as Eight!<br>"; 
echo 'Continue to the <a href="Eight/index.php">members</a> section.';
}
}
?>

I'm not really following you on the $class issue, although looking at your code,

PHP Code:


if ($class != 'One') {

echo "You are now logged in as an One!<br>"; 

echo 'Continue to the <a href="One/index.php">members</a> section.';

}

You're saying, if $class is not equal to 'One' then echo you are now logged in as 'One', which doesn't really make much sense, I'm thinking you meant it to be:

PHP Code:


if ($class == 'One') {

echo "You are now logged in as an One!<br>"; 

echo 'Continue to the <a href="One/index.php">members</a> section.';

}

BTW:
When you get the 'Parse error: parse error, unexpected $ in ... on line...' errors, it usually means you missed a curly brace somewhere.

**PlaGuE** · Aug 21st, 2005, 11:48 PM

thats wut i was gunna say... i noticed thatas soon as i read the code. looks usefull.. tho.

**drkelusion** · Aug 22nd, 2005, 02:24 AM

Omg, thankyou so much {yak} and PlaGuE, who was about to post the answer...

Also, thankyou ninjanutz for helping earlier.

Okay, so I now know that the script works, however, Other than the page coming up with:
You are now logged in as an One!
Continue to the members section.
For some reason, it doesnt come up with anything (just a blank page)
Any ideas?

Thanks again for everyone who helped!!

**PlaGuE** · Aug 22nd, 2005, 03:23 AM

wut do you mean.. blank page.. ...

from wut i read... it looks likes its just supposed to show..
You are now logged in as an One!
Continue to the members section.(<---link to page---<<)

if i miss read any part.. plz tell me... so i can get a better understanding to help you.

**drkelusion** · Aug 22nd, 2005, 03:31 AM

No, you read correctly, but for some reason it doesnt show You are now logged in as an One!
Continue to the members section. at all..
It's just blank page (nothing is there). The file is called logscript.php still, but all it is is just a blank html page:
<html><body></body></html> is the only thing shown in the Source file.

**PlaGuE** · Aug 22nd, 2005, 03:41 AM

hmm...would i be able to see the page... so that i can visualise...

it looks like teh problem is in here

PHP Code:


include("config.php");



// connect to the mysql server

$link = mysql_connect($server, $db_user, $db_pass)

or die ("Could not connect to mysql because ".mysql_error());



// select the database

mysql_select_db($database)

or die ("Could not select database because ".mysql_error());



$match = "select id from $table where username = '".$_POST['username']."'

and password = '".$_POST['password']."';";



$qry = mysql_query($match)

or die ("Could not match data because ".mysql_error());

$num_rows = mysql_num_rows($qry);



if ($num_rows <= 0) {

echo "Sorry, there is no username $username with the specified password.<br>";

echo "<a href=login.php>Try again</a>";

exit;

} else {



setcookie("loggedin", "TRUE", time()+(3600 * 24));

setcookie("mysite_username", "$username");

but i wont know... until i can see it not showing...

I tried it myself.. and it displays stuff... but its just it cant find anything... even tho theres stuff there..lol... so im at a loss

**drkelusion** · Aug 22nd, 2005, 04:18 AM

Sure, the link to the login page is at:
http://www.animefusion.net/RPG/testing/login.php

Just use:
Username: Testing
Password: password

I did, however, edit the code and replaced "($class != 'One')" to the appropriate class to fit the signup page (eg. "One" is actually "Elements")

The account above is set to Class "Six", which is "Archer" in the editted script.
Thanks again for trying to help!

**PlaGuE** · Aug 22nd, 2005, 06:12 AM

try making $num_rows == 0... if that dont work.. take out the exit... lol.. and if that dont work.. change both..lol...

I must be missing sumtin... cuz it should work.

**drkelusion** · Aug 23rd, 2005, 01:11 AM

... I tried removing "exit", and editting "<=", and then even tried changing/removing both, but it still seems to come up with a blank page...

Code:

$qry = mysql_query($match)
or die ("Could not match data because ".mysql_error());
$num_rows = mysql_num_rows($qry); 

if ($num_rows == 0) { 
echo "Sorry, there is no username $username with the specified password.<br>";
echo "<a href=login.php>Try again</a>";
} else {

setcookie("loggedin", "TRUE", time()+(3600 * 24));
setcookie("mysite_username", "$username");

I removed "exit;" and changed "<=" to "==" and neither worked at all

Thankyou for helping though.. Maybe you know what else I may have done wrong?

**PlaGuE** · Aug 23rd, 2005, 02:16 AM

to be honest... i have no clue why its not working...
hopefully one of the others know... hmm... very wierd... cuz from the code.. it should work...

----------------------------
oohh... i just thought of sumtin... is the $class variable being passed or globalized in the script... cuz maybe, this is just a thought, its not being picked up.

**drkelusion** · Aug 23rd, 2005, 02:58 AM

Oh, okay..
Hopefully someone knows..
Thanks again for trying to help

**{yak}** · Aug 23rd, 2005, 08:48 AM

I believe your problem lies here:

Code:

$qry = mysql_query($match)

Missed a semi colon:

Code:

$qry = mysql_query($match);

Good luck

**drkelusion** · Aug 23rd, 2005, 05:30 PM

I seem to get this error now

Parse error: parse error, unexpected T_LOGICAL_OR in /home/hanime07/public_html/RPG/testing/logscript.php on line 17

Code:

<?php

include("config.php"); 

// connect to the mysql server
$link = mysql_connect($server, $db_user, $db_pass)
or die ("Could not connect to mysql because ".mysql_error());

// select the database
mysql_select_db($database)
or die ("Could not select database because ".mysql_error());

$match = "select id from $table where username = '".$_POST['username']."'
and password = '".$_POST['password']."';"; 

$qry = mysql_query($match);
or die ("Could not match data because ".mysql_error());
$num_rows = mysql_num_rows($qry); 

if ($num_rows <= 0) { 
echo "Sorry, there is no username $username with the specified password.<br>";
echo "<a href=login.php>Try again</a>";
exit;
} else

Something to do with or die ("Could not match data because ".mysql_error()); I assume?

Thanks yet again for helping!

**drkelusion** · Aug 27th, 2005, 09:06 PM

Hi Again!
As posted above, I thought the error would be in "or die ("Could not match data because ".mysql_error());" so I deleted that line, and it seems that now it just goes to a blank page, like it did earlier..

Any ideas would be greatly appreciated!
Thankyou again

**visualAd** · Aug 28th, 2005, 11:03 AM

You are making a variable assignment before using a logical or, you need to enclose the variable assignment in brackets:

PHP Code:


// connect to the mysql server

($link = mysql_connect($server, $db_user, $db_pass))

or die ("Could not connect to mysql because ".mysql_error());

It is however, better coding practice to use an IF statement instead of or die(). It makes the code easier to follow:

PHP Code:


// connect to the mysql server

if (! ($link = mysql_connect($server, $db_user, $db_pass))) {

    die ("Could not connect to mysql because ".mysql_error());

}

**drkelusion** · Aug 29th, 2005, 01:34 AM

Hi! I changed what you included, but seem to be getitng nearly the exact same problem;
Parse error: parse error, unexpected T_LOGICAL_OR in /home/hanime07/public_html/RPG/testing/logscript.php on line 18

Thankyou for helping btw!!

**visualAd** · Aug 29th, 2005, 01:41 AM

The code above it correct and won't produce a parse error. Try copying and pasting the code above. Which version of PHP are you using?

**visualAd** · Aug 29th, 2005, 01:44 AM

I've justl looked at your code again. It isn't that line which is causing the error, it is line 18, which is here:

Code:

$qry = mysql_query($match);
or die ("Could not match data because ".mysql_error());
$num_rows = mysql_num_rows($qry);

When you get an error like that, PHP will give you information about the line number and the file in which it occured, this can be very useful

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