Login problem, I really need help

iimrii

Well I have script which wiews fields in my database...

<?php
if($loged=='yes') { ?>
<HTML><?php
$db = mysql_connect("localhost", "xxxxx", "xxxxxx");
mysql_select_db("boyem_com",$db);
$result = mysql_query("SELECT * FROM amx_regusers",$db);
echo "<TABLE BORDER=2>";
echo"<TR><TD><B>Nick</B><TD><B>Pass</B><TD><B>Flags</B><TD><B>Options</B></TR>";
while ($myrow = mysql_fetch_array($result))
{
	echo "<TR><TD>".$myrow["nick"]."<TD>".$myrow["pass"]."<TD>".$myrow["flags"];
	echo "<TD><a href=\"delete.php?nick=".$myrow[nick]."\">Delete</a>";
	echo "<TD><a href=\"edit.php?nick=".$myrow[nick]."\">Edit</a>";
}
	echo "</TABLE>";
?></HTML>
<? } else { ?>
<b>Please login <a href="index.php">here.</a></b> 
<? } ?>

ok this scripts works perfectly....when i visit view2.php i see all the columns that are generated from the database...

in this page i will give access to users to view the table in the database...but some information should be showend to the users...with given access....so I installed a authentication script where users who log in will be able to view the page that i posted above...

lets say i have 2 users

[noc]abdul
[kate]momo

well when they login they can view all the information in the page i posted above...

the thing i want is that when [noc]abdul log ins i want him to see in amx_regusers table in the database the rows which has [noc] included in the nick (field in the amx_regusers table)...

is it possible to do it...if you have questions please ask i really need it...

iimrii

my the table that is generated by view2.php looks like this

Nick

[F@NT@STiC] A K B A B A
[F@NTASTiC] Turkish

[sss] MangoExpress

[noc]mahmut
[noc]emre

so when [noc]abdul logged the view2.php should generate a page like this

Nick

[noc]mahmut
[noc]emre

if [F@NT@STiC] jack logs in the genrated view2.php shoudl look like this

Nick

[F@NT@STiC] A K B A B A
[F@NTASTiC] Turkish

i need advise please help me...

thanks in advance

scoppc

Just add a "WHERE" clause inside your SQL statement.
But first you gotta extract the portion enclosed within [] symbol out.

<?php 
	$str="[noc]Something";
	$extracted=preg_replace("/\[(.*?)\](.*)/","\\1",$str);
	print $extracted;
                //output: noc
?>

you should add a backslash symbol before [ and ] in preg_replace(), This forum takes it out...
and double backslashes for the replace parameter

After successful extraction, the just match it using the following syntax:

$sql="SELECT * FROM $tablename WHERE name LIKE '[$extracted]%'";

iimrii

but if the user logged in is [F@NT@STiC] than what happens...it seems that this one is just for [noc]
i am kind of new to php sorry for the stupid question if it is one..

iimrii

<?php  

    $str="[noc]Something"; 
    $extracted=preg_replace("/[(.*?)](.*)/","\1",$str); 
    print $extracted; 
                //output: noc 
?>

so here in $str="[noc]Something"; the [noc[something shoudl be get from the login scrpt whiever username the user logged in as...

is it possible...

scoppc

You can use the script for
[F@NT@STiC] Abdul
too.
Just try and see what you got...

Well, the name should be from your login script coz I am sure that you do set some cookies/sessions so that you can authenticate the logged in user, rite?!

iimrii

yeah....who ever logged in shoudl only see related information in the rows not all of them...i shouldnt set the script for each user it shoudl does it automatically from the login user name...if

php logs in now so he wont see anything becasue there is no[code=php]anyone in table that their nick starts with [code=php]...

scoppc

The script I've shown you should be able to do the task automatically. Just combine the extract script with the SQL statement. So, user that has

 in front of her username will only see the data of other users whose username starts with [code=php] too.

Give it a try and see whether it satisfies your needs.

jernhenrik

Backreference, shouldn't that be \$1 ?? to make it work?

$extracted=preg_replace("/[(.?)](.)/","\$1",$str);

or am i missing something here?

scoppc

or am I missing something? kidding... 🙂
Well, it works fine here... just try to copy the code and try yourself. 😉

jernhenrik

arghhh but i did...thats why I was curious....caus this:

$str="[F@NT@STIC]Something";
$extracted=preg_replace("/\[(.*?)\](.*)/","\1",$str);
print $extracted;

gives me nothing..whereas this:

$str="[F@NT@STIC]Something";
$extracted=preg_replace("/\[(.*?)\](.*)/","\$1",$str);
print $extracted;

gives me output F@NT@STIC..

:😕:

avoided code tag to make sure nothing left out

iimrii

i did it like this but it soent seems that it is working..

<?php
if($loged=='yes') { ?>
<HTML><?php
$db = mysql_connect("localhost", "xxxxxxxx", "xxxxxx");
mysql_select_db("boyem_com",$db);

$str="[noc]Something"; 
extracted=preg_replace("\[(.*?)]\(.*)/","\1",$str); 
print $extracted; 
$result = mysql_query("SELECT * FROM $amx_regusers WHERE nick LIKE '[$extracted]%'",$db);
echo "<TABLE BORDER=2>";
echo"<TR><TD><B>Nick</B><TD><B>Pass</B><TD><B>Flags</B><TD><B>Options</B></TR>";
while ($myrow = mysql_fetch_array($result))
{
	echo "<TR><TD>".$myrow["nick"]."<TD>".$myrow["pass"]."<TD>".$myrow["flags"];
	echo "<TD><a href=\"delete.php?nick=".$myrow[nick]."\">Delete</a>";
	echo "<TD><a href=\"edit.php?nick=".$myrow[nick]."\">Edit</a>";
}
	echo "</TABLE>";
?></HTML>
<? } else { ?>
<b>Please login <a href="index.php">here.</a></b> 
<? } ?>

giving me this error on line 9

Parse error: parse error in /home/boyem/boyem-www/view2.php on line 9

which the line is extracted=preg_replace("[(.?)](.)/","\1",$str);

iimrii

<?php
if($loged=='yes') { ?>
<HTML><?php
$db = mysql_connect("localhost", "xxxxx", "xxxxx");
mysql_select_db("boyem_com",$db);

$str="[noc]Something"; 
extracted=preg_replace("\[(.*?)]\(.*)/","\1",$str); 
print $extracted; 
$result = mysql_query("SELECT * FROM $amx_regusers WHERE nick LIKE '[$extracted]%'",$db);
echo "<TABLE BORDER=2>";
echo"<TR><TD><B>Nick</B><TD><B>Pass</B><TD><B>Flags</B><TD><B>Options</B></TR>";
while ($myrow = mysql_fetch_array($result))
{
	echo "<TR><TD>".$myrow["nick"]."<TD>".$myrow["pass"]."<TD>".$myrow["flags"];
	echo "<TD><a href=\"delete.php?nick=".$myrow[nick]."\">Delete</a>";
	echo "<TD><a href=\"edit.php?nick=".$myrow[nick]."\">Edit</a>";
}
	echo "</TABLE>";
?></HTML>
<? } else { ?>
<b>Please login <a href="index.php">here.</a></b> 
<? } ?>

yeah when i do code instead of php it doesnt removes is..

jernhenrik

Originally posted by jernhenrik
arghhh but i did...thats why I was curious....caus this:
$extracted=preg_replace("/[(.?)](.)/","\1",$str);

I am gonna shut up before i make a larger fool outa myself...sorry

$extracted=preg_replace("/[(.?)](.)/","\1",$str);

:rolleyes:

iimrii

this doesnt work for me

<?php
if($loged=='yes') { ?>
<HTML><?php
$db = mysql_connect("localhost", "xxx", "xxxxx");
mysql_select_db("boyem_com",$db);

$str="[F@NT@STIC]Something"; 
$extracted=preg_replace("/\[(.*?)\](.*)/","\\1",$str); 
print $extracted; 
$result = mysql_query("SELECT * FROM $amx_regusers WHERE nick LIKE '[$extracted]%'",$db);
echo "<TABLE BORDER=2>";
echo"<TR><TD><B>Nick</B><TD><B>Pass</B><TD><B>Flags</B><TD><B>Options</B></TR>";
while ($myrow = mysql_fetch_array($result))
{
	echo "<TR><TD>".$myrow["nick"]."<TD>".$myrow["pass"]."<TD>".$myrow["flags"];
	echo "<TD><a href=\"delete.php?nick=".$myrow[nick]."\">Delete</a>";
	echo "<TD><a href=\"edit.php?nick=".$myrow[nick]."\">Edit</a>";
}
	echo "</TABLE>";
?></HTML>
<? } else { ?>
<b>Please login <a href="index.php">here.</a></b> 
<? } ?>

it is givineg me this error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/boyem/boyem-www/view2.php on line 14

i am lost
😕 😕

iimrii

you might be laughnig to the code above but i have no idea i am trying my best...

if it is really easy for you please help me

jernhenrik

nope we are not laughing...well scoppc might be laughing at me, but thats ok 🙂

the warning that you got is proably because nothing was found in the database...it could be because the @ is unrecognised and you have to escape it...try and testing it , substitute:
$str="[F@NT@STIC]Something";

$str="[noc]Something";

if you have someone in your database with the nick [noc].. adn see if that changes anything...

iimrii

well it is giving me [F@NT@STiC] but the error still exist there...but this is not what i need...i dont know i am confused

scoppc

Hey, I am not laughing at anybody okay! I am not an expert so why should I laught at somebody... hehhee... (this is not 'that' laugh) 🙂

Okay, just try to run the query right from your MySQL console. observed what you got. and remember, u should already have some data inside.

iimrii

i cant it is a rented host with db 🙁