Help anyone?

The Coder · July 18, 2012

Hi all, I've tried to access someones profile through usersonline, however after clicking on there username from usersonline.php, it strangely opens up in a new tab, can anyone help me out on this one? I just get the feeling the issue is somewhere in here, please correct me if I am wrong, thanks. :)

$user=$r['userid'];
?>
<tr class='hover' onClick="window.location='viewuser.php?u=<?php echo $user; ?>'"><td><a href="viewuser.php?u=<?php print $user; ?>" target="_blank"><?php print $r['username']; ?> [<?php echo $r['userid']; ?>]</a></td>
<td><?php echo $la; ?> <?php print $unit; ?></td></tr>
<?php
}

Seker · July 18, 2012

Hi all, I've tried to access someones profile through usersonline, however after clicking on there username from usersonline.php, it strangely opens up in a new tab, can anyone help me out on this one? I just get the feeling the issue is somewhere in here, please correct me if I am wrong, thanks. :)
$user=$r['userid'];
?>
<tr class='hover' onClick="window.location='viewuser.php?u=<?php echo $user; ?>'"><td><a href="viewuser.php?u=<?php print $user; ?>" target="_blank"><?php print $r['username']; ?> [<?php echo $r['userid']; ?>]</a></td>
<td><?php echo $la; ?> <?php print $unit; ?></td></tr>
<?php
}

Take this out:

target="_blank"

The Coder · July 18, 2012

Thank you so much! :)

And, sorry to bother everyone with such a minor matter. :)

The Coder · July 18, 2012

Hey also, I keep getting this error:-

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/public_html/staffapps.php on line 10

This is line ten:-

><a href="index.php" >Go Home</a>";

Anyone know the issue? Many thanks! :)

Edited July 18, 2012 by The Coder
Took away "Blank" thingy and "Https" :P

Seker · July 18, 2012

Hey also, I keep getting this error:-
Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/public_html/staffapps.php on line 10

This is line ten:-
><a href="index.php" >Go Home</a>";
Anyone know the issue? Many thanks! :)

It looks like it's because you're using double quotes to wrap both your string and your echo/print.

For example, it should be either this:

print "<a href='index.php'>Go Home</a>";

Or this:

print '<a href="index.php">Go Home</a>';

But neither of these two:

print "<a href="index.php">Go Home</a>";

print '<a href='index.php'>Go Home</a>';

See the differences?

KyleMassacre · July 19, 2012

Yeah you should probably post a line or two above the line giving the so called error as most of the time its not that line with that kind of error

The Coder · July 19, 2012

Hey Seker, unfortunately I get the same error with both codes you have supplied for me. :(

Also, sorry about that Kyle here is some of the code before and after line 10. :)

require "globals.php";
if($set['stafflock'] == 'Locked')
{
echo "Staff Applications are locked at the moment, please try again later.
><a href="index.php" >Go Home</a>";
$h->endpage();
exit;
}
echo "<h3>Staff Applications</h3>";
$_POST['ID'] = abs((int) $_POST['ID']);
if($_POST['ID'])

Uridium · July 19, 2012

to solve your T_STRING error as mentioned by SEKER whos Edits were correct you can do this one of 2 ways to fix it...

require "globals.php";

if($set['stafflock'] == 'Locked'){

echo "Staff Applications are locked at the moment, please try again later.><a href=\"index.php\" >Go Home</a>";

$h->endpage();exit;}echo "<h3>Staff Applications</h3>";

$_POST['ID'] = abs((int) $_POST['ID']);

if($_POST['ID'])

or

require "globals.php";

if($set['stafflock'] == 'Locked'){

echo "Staff Applications are locked at the moment, please try again later.><a href='index.php' >Go Home</a>";

$h->endpage();exit;}echo "<h3>Staff Applications</h3>";

$_POST['ID'] = abs((int) $_POST['ID']);

if($_POST['ID'])

Remember at all times if your echo or print starts with a " or a ' when you end out your echo or print you need to use the same method as you did to start with.

example THIS WILL CREATE AN ERROR

print " how are you world ';

echo " how are you world ';

the reason for this is because you started both statements with a " and ended it with a '

However you can use more than 1 " in a statement but you need to block the next " so the script doesnt think its finished

ecample

echo "we are here <a href="index.php">HERE</a>"; <<<<< will cause an error cos you have 4 " statements and it thinks the second " is the closing out statement

so to cure this you can simply do this

echo "we are here <a href=\"index.php\">HERE</a>"; <<<< the use of the \ tells the script you havent quite finished yet so carry on...

Seker · July 19, 2012

However you can use more than 1 " in a statement but you need to block the next " so the script doesnt think its finished

ecample

echo "we are here <a href="index.php">HERE</a>"; <<<<< will cause an error cos you have 4 " statements and it thinks the second " is the closing out statement

so to cure this you can simply do this

echo "we are here <a href=\"index.php\">HERE</a>"; <<<< the use of the \ tells the script you havent quite finished yet so carry on...

I actually didn't know you can do it this way. Though, I don't know why anyone would want to. It's a lot messier, in my opinion.

Uridium · July 19, 2012

the use of the \ comes in handy sometimes when achieving mass editing especially where links are concerned

The Coder · July 19, 2012

to solve your T_STRING error as mentioned by SEKER whos Edits were correct you can do this one of 2 ways to fix it...
require "globals.php";

if($set['stafflock'] == 'Locked'){

echo "Staff Applications are locked at the moment, please try again later.><a href=\"index.php\" >Go Home</a>";

$h->endpage();exit;}echo "<h3>Staff Applications</h3>";

$_POST['ID'] = abs((int) $_POST['ID']);

if($_POST['ID'])
or
require "globals.php";

if($set['stafflock'] == 'Locked'){

echo "Staff Applications are locked at the moment, please try again later.><a href='index.php' >Go Home</a>";

$h->endpage();exit;}echo "<h3>Staff Applications</h3>";

$_POST['ID'] = abs((int) $_POST['ID']);

if($_POST['ID'])
Remember at all times if your echo or print starts with a " or a ' when you end out your echo or print you need to use the same method as you did to start with.

example THIS WILL CREATE AN ERROR

print " how are you world ';

echo " how are you world ';

the reason for this is because you started both statements with a " and ended it with a '

However you can use more than 1 " in a statement but you need to block the next " so the script doesnt think its finished

ecample

echo "we are here <a href="index.php">HERE</a>"; <<<<< will cause an error cos you have 4 " statements and it thinks the second " is the closing out statement

so to cure this you can simply do this

echo "we are here <a href=\"index.php\">HERE</a>"; <<<< the use of the \ tells the script you havent quite finished yet so carry on...

Thanks but after doing this, I get this error:-

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/gothcent/public_html/staffapps.php on line 25

With these codes:-

require "globals.php";

if($set['stafflock'] == 'Locked'){

echo "Staff Applications are locked at the moment, please try again later.><a href=\"index.php\" >Go Home</a>";

$h->endpage();exit;}echo "<h3>Staff Applications</h3>";

$_POST['ID'] = abs((int) $_POST['ID']);

if($_POST['ID'])
{
$sql = sprintf('SELECT COUNT(ID) AS `count` FROM `staffaps` WHERE ID = %u', $userid);
$sql = $db->query($sql);
$gaps = $db->fetch_row($sql);
if($gaps['count'])
{
echo "Invalid Command.
You already have submitted a application.
><a href="index.php" >Go Home</a>"; [LINE 25]
$h->endpage();
exit;
}
else if(!$gaps['count'])
{
$_POST['staff'] = abs((int) $_POST['staff']);
$_POST['about'] = mysql_real_escape_string(htmlentities($_POST['about']));
$_POST['exp'] = mysql_real_escape_string(htmlentities($_POST['exp']));
if(strlen($_POST['about']) > 75 OR strlen($_POST['exp']) > 75)
{

:( Btw I stated where line 25 is in the codes.

Edited July 19, 2012 by The Coder

Seker · July 19, 2012

Thanks but after doing this, I get this error:-

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/gothcent/public_html/staffapps.php on line 25

With these codes:-

require "globals.php";

if($set['stafflock'] == 'Locked'){

echo "Staff Applications are locked at the moment, please try again later.><a href=\"index.php\" >Go Home</a>";

$h->endpage();exit;}echo "<h3>Staff Applications</h3>";

$_POST['ID'] = abs((int) $_POST['ID']);

if($_POST['ID'])
{
$sql = sprintf('SELECT COUNT(ID) AS `count` FROM `staffaps` WHERE ID = %u', $userid);
$sql = $db->query($sql);
$gaps = $db->fetch_row($sql);
if($gaps['count'])
{
echo "Invalid Command.
You already have submitted a application.
><a href="index.php" >Go Home</a>"; [LINE 25]
$h->endpage();
exit;
}
else if(!$gaps['count'])
{
$_POST['staff'] = abs((int) $_POST['staff']);
$_POST['about'] = mysql_real_escape_string(htmlentities($_POST['about']));
$_POST['exp'] = mysql_real_escape_string(htmlentities($_POST['exp']));
if(strlen($_POST['about']) > 75 OR strlen($_POST['exp']) > 75)
{

:( Btw I stated where line 25 is in the codes.

You still haven't even fixed it.

Change this:

<a href="index.php" >Go Home</a>";

To this:

<a href='index.php' >Go Home</a>";

The Coder · July 19, 2012

Hey thanks, it works now, but another two issues, and also I'm so sorry to bother you all, but I really need mod working for the game. :)

I get an error for both of these SQL's:-

CREATE TABLE IF NOT EXISTS `staffaps` (
 `ID` int(11) NOT NULL default '0',
 `about` text NOT NULL,
 `exp` text NOT NULL,
 `position` int(11) NOT NULL default '0'
);

Error with the above SQL:-

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`ID` int(11) NOT NULL default '0', Â Â `about` text NOT NULL, Â Â `exp` text NOT' at line 2

And 2nd SQL -

INSERT INTO `settings` (
`conf_id` ,
`conf_name` ,
`conf_value`
)
VALUES (
'17', 'stafflock', 'Locked'
);

Error with 2nd SQL -

#1062 - Duplicate entry '17' for key 'PRIMARY'

Thanks to all who help me. :)

Seker · July 19, 2012

Maybe for the first SQL:

CREATE TABLE IF NOT EXISTS `staffaps` (
 `ID` int(11) NOT NULL default '0',
 `about` varchar(max) NOT NULL,
 `exp` varchar(max)  NOT NULL,
 `position` int(11) NOT NULL default '0'
);

Though, I do not know why text wouldn't work.

For the second SQL:

INSERT INTO `settings` 
(
       `conf_id` ,
       `conf_name` ,
       `conf_value`
)
VALUES 
(
       'NULL', 
       'stafflock', 
       'Locked'
);

The Coder · July 19, 2012

Didn't work mate. :(

Errors received -

SQL 1 error -

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`ID` int(11) NOT NULL default '0', Â Â `about` varchar(max) NOT NULL, Â Â `exp` ' at line 2

SQL 2 error -

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`conf_id` , Â Â Â Â Â Â Â Â `conf_name` , Â Â Â Â Â Â Â Â `conf_value` ) VALUES ' at line 3

Anyone have a clue whats wrong lol?

Thanks Seker for trying. :)

Djkanna · July 19, 2012

Didn't work mate. :(
Errors received -

SQL 1 error -

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`ID` int(11) NOT NULL default '0', Â Â `about` varchar(max) NOT NULL, Â Â `exp` ' at line 2

SQL 2 error -

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`conf_id` , Â Â Â Â Â Â Â Â `conf_name` , Â Â Â Â Â Â Â Â `conf_value` ) VALUES ' at line 3

Anyone have a clue whats wrong lol?

Thanks Seker for trying. :)

Try;

CREATE TABLE IF NOT EXISTS `staffaps` (
   `ID` INT(11) NOT NULL DEFAULT 0,
   `about` TEXT NOT NULL,
   `exp` TEXT NOT NULL,
   `position` INT(11) NOT NULL DEFAULT 0, #Probably don't need int(11) but hey.
);


INSERT INTO `settings` ( `conf_id`, `conf_name`, `conf_value`) VALUES (null, 'stafflock', 'Locked');

The Coder · July 19, 2012

Try;

CREATE TABLE IF NOT EXISTS `staffaps` (
   `ID` INT(11) NOT NULL DEFAULT 0,
   `about` TEXT NOT NULL,
   `exp` TEXT NOT NULL,
   `position` INT(11) NOT NULL DEFAULT 0, #Probably don't need int(11) but hey.
);


INSERT INTO `settings` ( `conf_id`, `conf_name`, `conf_value`) VALUES (null, 'stafflock', 'Locked');

Thanks for trying, but I get another sort of error this time:-

"SQL query:

CREATE TABLE IF NOT EXISTS `staffaps` (

`ID` INT( 11 ) NOT NULL DEFAULT 0,

`about` TEXT NOT NULL ,

`exp` TEXT NOT NULL ,

`position` INT( 11 ) NOT NULL DEFAULT 0,

#Probably don't need int(11) but hey.

);

MySQL said:

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`ID` INT(11) NOT NULL DEFAULT 0,

Â Â Â Â `about` TEXT NOT NULL,

Â Â Â Â `exp` TE' at line 2"

Once again, thanks for trying, and this seems to be a picky one eh all? :/

Seker · July 19, 2012

Thanks for trying, but I get another sort of error this time:-
"SQL query:

CREATE TABLE IF NOT EXISTS `staffaps` (

`ID` INT( 11 ) NOT NULL DEFAULT 0,

`about` TEXT NOT NULL ,

`exp` TEXT NOT NULL ,

`position` INT( 11 ) NOT NULL DEFAULT 0,

#Probably don't need int(11) but hey.

);

MySQL said:

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`ID` INT(11) NOT NULL DEFAULT 0,

Â Â Â Â `about` TEXT NOT NULL,

Â Â Â Â `exp` TE' at line 2"

Once again, thanks for trying, and this seems to be a picky one eh all? :/

After NOT NULL on the `about` and `exp` lines, you have a space before the comma. Could be messing with it, maybe.

The Coder · July 19, 2012

After NOT NULL on the `about` and `exp` lines, you have a space before the comma. Could be messing with it, maybe.

Thanks, but nothing. :(

newttster · July 19, 2012

Maybe try changing ID to staffaps ID. Perhaps there is a conflict with the field name ID??? Have you set an increment for the ID? Or a key?

Djkanna · July 19, 2012

Thanks for trying, but I get another sort of error this time:-
"SQL query:

CREATE TABLE IF NOT EXISTS `staffaps` (

`ID` INT( 11 ) NOT NULL DEFAULT 0,

`about` TEXT NOT NULL ,

`exp` TEXT NOT NULL ,

`position` INT( 11 ) NOT NULL DEFAULT 0,

#Probably don't need int(11) but hey.

);

MySQL said:

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`ID` INT(11) NOT NULL DEFAULT 0,

Â Â Â Â `about` TEXT NOT NULL,

Â Â Â Â `exp` TE' at line 2"

Once again, thanks for trying, and this seems to be a picky one eh all? :/

CREATE TABLE IF NOT EXISTS `staffaps` (
   `ID` INT(11) NOT NULL DEFAULT 0,
   `about` TEXT NOT NULL,
   `exp` TEXT NOT NULL,
   `position` INT(11) NOT NULL DEFAULT 0 #Probably don't need int(11) but hey.
);

Seems to work fine.

Help anyone?

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