[PHP-DB] Retricting Access to Menu Items

2006-03-06 Thread Jeff Broomall
Good morning everyone.

I'm building a very simple content management site that tracks tasks.

The options available are:
 1. Add Task
 2. Edit Task
 3. View Task
 4. Print Task

I need to restrict some users to only View and Print and I'm trying to find a 
way to tell the page not to load the menu options (the text) for those not 
having access to the Add and Edit functions.

IOW, they would only see View and Print.

I have three basic users:
 1. System Admin
 2. Subject Matter Expert (SME)
 3. Viewers

Obviously the System Admin and SME will have full access so it's the Viewers 
that are to have access to only View and Print.

I have a users table but haven't set it up for the distinction.  What I was 
thinking was creating a field labeled users_group and assign a numeric value 
for each user using the numbering system above.

I have my page load the menu options:

a href=home.php class=navlinkHome/abr /
a href=view_tasks.php class=navlinkView Tasks/abr /
a href=edit_task.php class=navlinkEdit Task/abr /
a href=add_task.php class=navlinkAdd Task/abr /

into here...

BODY

table width=90% border=1 cellspacing=10 cellpadding=0 align=center

 trtd colspan=2h1 id=mainheadICAO Tasks #8212; WAFS/h1/td/tr
  
tr 
td align= center valign=top nowrap=nowrap width=10%
 Menubr /
 ?php include ('./includes/menu.html'); ? --The menu above inserted here.
/td
 
td valign=top class=content 


How can I tell the system not to load the last two lines unless they are a 
System Admin or SME?

I read a chapter on Cookies/Sessions...but it wasn't that helpful for this case.

Can I setcookie('user_group', '3') and use that somehow???

Am I in the ballpark with this solution?

Thanks.

Jeff

Re: [PHP-DB] Retricting Access to Menu Items

2006-03-06 Thread Jeffrey

I've done this kind of thing with a number of web apps.

What I usually do is create a user table in MySQL with a user-name, 
password and access level, which has an integer value.


When a user logs in successfully, a session is created (see 
session_start() in php documentation), the access level is pulled from 
the user table and saved as a session variable. Then it is a simple 
matter of using bits of code like...


if ($_SESSION['access_level']  7){
echo some stuff;
}

In your example, you will also want to check the user's access level on 
each restricted page - it is not enough to hide menu options. Users 
could simply type the URL in.


I hope that's clear.

Good luck,

Jeffrey

Jeff Broomall wrote:

Good morning everyone.

I'm building a very simple content management site that tracks tasks.

The options available are:
 1. Add Task
 2. Edit Task
 3. View Task
 4. Print Task

I need to restrict some users to only View and Print and I'm trying to find a 
way to tell the page not to load the menu options (the text) for those not 
having access to the Add and Edit functions.

IOW, they would only see View and Print.

I have three basic users:
 1. System Admin
 2. Subject Matter Expert (SME)
 3. Viewers

Obviously the System Admin and SME will have full access so it's the Viewers 
that are to have access to only View and Print.

I have a users table but haven't set it up for the distinction.  What I was 
thinking was creating a field labeled users_group and assign a numeric value 
for each user using the numbering system above.

I have my page load the menu options:

a href=home.php class=navlinkHome/abr /
a href=view_tasks.php class=navlinkView Tasks/abr /
a href=edit_task.php class=navlinkEdit Task/abr /
a href=add_task.php class=navlinkAdd Task/abr /

into here...

BODY

table width=90% border=1 cellspacing=10 cellpadding=0 align=center

 trtd colspan=2h1 id=mainheadICAO Tasks #8212; WAFS/h1/td/tr
  
tr 
td align= center valign=top nowrap=nowrap width=10%

 Menubr /
 ?php include ('./includes/menu.html'); ? --The menu above inserted here.
/td
 
td valign=top class=content 



How can I tell the system not to load the last two lines unless they are a 
System Admin or SME?

I read a chapter on Cookies/Sessions...but it wasn't that helpful for this case.

Can I setcookie('user_group', '3') and use that somehow???

Am I in the ballpark with this solution?

Thanks.

Jeff


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[PHP-DB] X Tiger 10.4.5 + phpBB2 + MySQL 5 + Can't Connect Issue!

2006-03-06 Thread m i l e s

Hi,

Im getting the following: Warning: mysql_connect(): Can't connect to  
local MySQL server through socket...etc.


NOTES:

1.) MySQL 5 is running fine.
2.) Im able to connect to it via Navicat.
3.) I run Lasso on the same host, I know that I can pull and post  
data to and from MySQL. So its NOT mysql. Its PHP.


I ran across a few postings from Apple and others that had SIMILAR  
problems, however those solutions are for MySQL 4 not MySQL 5. Seems  
to be bit of confusion about how to solve this problem. I tried what  
apple suggested...and that FAILED:


http://docs.info.apple.com/article.html?artnum=301457

So now Im asking you good people, who actually BUILT the damn...ahem  
wrote the software. Im at whits end and I need to get this up and  
running.


All I did was update from 10.4.4 to 10.4.5 and now PHP will not talk  
to MySQL at all. Ok...so what's the solution folks ? Anyone ?  
Buehler ? Buehler 


Thanks ahead of time

M i l e s.

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Re: [PHP-DB] X Tiger 10.4.5 + phpBB2 + MySQL 5 + Can't Connect Issue!

2006-03-06 Thread Micah Stevens

Is there really a .sock file where you're telling php there is? MySQL can put 
the socket file anywhere you want depending on it's 
configuration. /etc/my.cnf should specify this. 

-Micah 


On Monday 06 March 2006 9:08 am, m i l e s wrote:
 Hi,

 Im getting the following: Warning: mysql_connect(): Can't connect to
 local MySQL server through socket...etc.

 NOTES:

 1.) MySQL 5 is running fine.
 2.) Im able to connect to it via Navicat.
 3.) I run Lasso on the same host, I know that I can pull and post
 data to and from MySQL. So its NOT mysql. Its PHP.

 I ran across a few postings from Apple and others that had SIMILAR
 problems, however those solutions are for MySQL 4 not MySQL 5. Seems
 to be bit of confusion about how to solve this problem. I tried what
 apple suggested...and that FAILED:

 http://docs.info.apple.com/article.html?artnum=301457

 So now Im asking you good people, who actually BUILT the damn...ahem
 wrote the software. Im at whits end and I need to get this up and
 running.

 All I did was update from 10.4.4 to 10.4.5 and now PHP will not talk
 to MySQL at all. Ok...so what's the solution folks ? Anyone ?
 Buehler ? Buehler 

 Thanks ahead of time

 M i l e s.

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[PHP-DB] X Tiger 10.4.5 + phpBB2 + MySQL 5 + Can't Connect Issue! - SOLVED!

2006-03-06 Thread m i l e s

Hi,

Figured it out:

Apple wiped out the location of the socket.

they wanted '/var/mysql/mysql.sock'

it should be: '/tmp/mysql.sock'

Problem solved.

M i l e s.

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[PHP-DB] Problems with PDO mysql

2006-03-06 Thread Escuder Nicolas
Hello,

 

I have made this little example for explain my problem : 

?

$pdo = new PDO('mysql:host=192.168.69.20;dbname=Released', 'Web',
'Master99');

$data = '/enabled=boolean:true;/description=string:-bévènements';

$pdo-query(Update Test set data =' . $data . ';);

?

 

If i execute this scripts from php cli the result will be : 

/enabled=boolean:true;/description=string:évènements

 

If i execute this scripts from apache server the result will be : 

/enabled=boolean?;/description=string:évènements

 

It’s seem that is the é and è character that make this strange change
because if i remove this character the result will be :

/enabled=boolean:true;/description=string:vnements

 

I don’t found were the charset is converted or where is the problems ….

 

For information php cli and apache module are build with the same options.

 

I need help please.

 

I hope i ask my question in the good mailing list.

 

See You Soon.

 

Nicolas.

 



[PHP-DB] PHP4 with MYSQL5

2006-03-06 Thread João Cândido de Souza Neto
Hello everyone.

I'm trying to use my php4 conecting to a mysql5 server, and then executing
some command lines to create a stored procedure but it's not working.

When a tried to execute delimiter |, my php gets de follow error:

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 'delimiter |' at
line 1

If I connect to server as local and try to execute the above command, it
works fine.

Could anyone help me about this? I'll be pleased by any tips.

Thanks.

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[PHP-DB] Problem with a login page.

2006-03-06 Thread Alex Major
Hi there everyone, just wondered if you could help me with this login scipt.
I've tried to customise a script that I found on a website, but it dosn't
seem to be working.

Here is the scipt that isn't working.
?php

// database connect script.
require 'includes/connection.php';
require 'includes/check_login.php';

if($logged_in == 1) {
die('You are already logged in, '.$_SESSION['username'].'.');
}
?
html
head
titleLogin/title
/head
body
?php

if (isset($_POST['submit'])) { // if form has been submitted


/* check they filled in what they were supposed to and authenticate */
if(!$_POST['username'] | !$_POST['password']) {
die('You did not fill in a required field.');
}

// authenticate.

if (!get_magic_quotes_gpc()) {
$_POST['username'] = addslashes($_POST['username']);
}

$check = mysql_query(SELECT username, password FROM users WHERE
username = '.$_POST['username'].');

// check passwords match

$_POST['password'] = stripslashes($_POST['password']);
$info['password'] = stripslashes($info['password']);
$_POST['password'] = $_POST['password'];

if ($_POST['password'] != $info['password']) {
die('Incorrect password, please try again.');
}

// if we get here username and password are correct,
//register session variables and set last login time.

$date = date('m d, Y');

$update_login = mysql_query(UPDATE users SET last_login = '$date' WHERE
username = '.$_POST['username'].');

$_POST['username'] = stripslashes($_POST['username']);
$_SESSION['username'] = $_POST['username'];
$_SESSION['password'] = $_POST['password'];
$mysql_disconnect();
?

h1Logged in/h1
pWelcome back ?php echo $_SESSION['username']; ?, you are logged in./p

?php

} else {// if form hasn't been submitted

?
h1Login/h1
form action=?php echo $_SERVER['PHP_SELF']? method=post
table align=center border=1 cellspacing=0 cellpadding=3
trtdUsername:/tdtd
input type=text name=username maxlength=40
/td/tr
trtdPassword:/tdtd
input type=password name=password maxlength=50
/td/tr
trtd colspan=2 align=right
input type=submit name=submit value=Login
/td/tr
/table
/form
?php
}
?
/body
/html

I will also include the .php file incase the text above dosn't read well.
Basically, the page always returns 'Incorrect Password, please try again',
and I can't work out why. I've tried with two different accounts, and
neither are working. (I am 100% sure that I am entering usernames/passwords
correctly).

Thanks for any help with this, I can't for the life of me figure this out.

Regards, 
Alex. 

?php

// database connect script.
require 'includes/connection.php';
require 'includes/check_login.php';

if($logged_in == 1) {
die('You are already logged in, '.$_SESSION['username'].'.');
}
?
html
head
titleLogin/title
/head
body
?php

if (isset($_POST['submit'])) { // if form has been submitted


/* check they filled in what they were supposed to and authenticate */
if(!$_POST['username'] | !$_POST['password']) {
die('You did not fill in a required field.');
}

// authenticate.

if (!get_magic_quotes_gpc()) {
$_POST['username'] = addslashes($_POST['username']);
}

$check = mysql_query(SELECT username, password FROM users WHERE 
username = '.$_POST['username'].');

// check passwords match

$_POST['password'] = stripslashes($_POST['password']);
$info['password'] = stripslashes($info['password']);
$_POST['password'] = $_POST['password'];

if ($_POST['password'] != $info['password']) {
die('Incorrect password, please try again.');
}

// if we get here username and password are correct, 
//register session variables and set last login time.

$date = date('m d, Y');

$update_login = mysql_query(UPDATE users SET last_login = '$date' 
WHERE username = '.$_POST['username'].');

$_POST['username'] = stripslashes($_POST['username']);
$_SESSION['username'] = $_POST['username'];
$_SESSION['password'] = $_POST['password'];
$mysql_disconnect();
?

h1Logged in/h1
pWelcome back ?php echo $_SESSION['username']; ?, you are logged in./p

?php

} else {// if form hasn't been submitted

?
h1Login/h1
form action=?php echo $_SERVER['PHP_SELF']? method=post
table align=center border=1 cellspacing=0 cellpadding=3
trtdUsername:/tdtd
input type=text name=username maxlength=40
/td/tr
trtdPassword:/tdtd
input type=password name=password maxlength=50
/td/tr
trtd colspan=2 align=right
input type=submit name=submit value=Login
/td/tr
/table
/form
?php
}
?
/body
/html
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Re: [PHP-DB] Problem with a login page.

2006-03-06 Thread dpgirago

snip...

// check passwords match

$_POST['password'] = stripslashes($_POST['password']);
$info['password'] = stripslashes($info['password']);
$_POST['password'] = $_POST['password'];

if ($_POST['password'] != $info['password']) {
die('Incorrect password, please try again.');
}
snip...

Alex,

Sticking for the moment with the error message you get, I can't see
anyplace where an array named $info with a key of password is declared
or assigned this value. Are you sure that this variable exists?

David

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[PHP-DB] Re: PHP4 with MYSQL5

2006-03-06 Thread João Cândido de Souza Neto
I receive aa answer by e-mail in order this telling me that i must upgrade
my php4 to php5.

It means that a php4 server don't works fine with mysql5 server?

João Cândido de Souza Neto wrote:

 Hello everyone.
 
 I'm trying to use my php4 conecting to a mysql5 server, and then executing
 some command lines to create a stored procedure but it's not working.
 
 When a tried to execute delimiter |, my php gets de follow error:
 
 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 'delimiter |'
 at line 1
 
 If I connect to server as local and try to execute the above command, it
 works fine.
 
 Could anyone help me about this? I'll be pleased by any tips.
 
 Thanks.

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[PHP-DB] RE: [PHP] Re: PHP4 with MYSQL5

2006-03-06 Thread jblanchard
[snip]
I receive aa answer by e-mail in order this telling me that i must upgrade
my php4 to php5.

It means that a php4 server don't works fine with mysql5 server?
[/snip]

http://www.php.net/mysqli is designed to work with versions of MySQL 4.1.n and 
above. It requires PHP5

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RE: [PHP-DB] Re: PHP4 with MYSQL5

2006-03-06 Thread Bastien Koert
it does, you need the updated mysql library that comes with the php 5 files 
(libmysql.dll for windows)


bastien



From: João Cândido de Souza Neto [EMAIL PROTECTED]
To: php-db@lists.php.net,php-general@lists.php.net
Subject: [PHP-DB] Re: PHP4 with MYSQL5
Date: Mon, 06 Mar 2006 18:38:56 -0300

I receive aa answer by e-mail in order this telling me that i must upgrade
my php4 to php5.

It means that a php4 server don't works fine with mysql5 server?

João Cândido de Souza Neto wrote:

 Hello everyone.

 I'm trying to use my php4 conecting to a mysql5 server, and then 
executing

 some command lines to create a stored procedure but it's not working.

 When a tried to execute delimiter |, my php gets de follow error:

 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 'delimiter |'
 at line 1

 If I connect to server as local and try to execute the above command, it
 works fine.

 Could anyone help me about this? I'll be pleased by any tips.

 Thanks.

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RE: [PHP-DB] Retricting Access to Menu Items

2006-03-06 Thread Bastien Koert

if you pull the user data into a session, then its a simple if them check


if ($_SESSION['user']['user_access']1)  //1 = basic user
{
 echo 'a href=edit_task.php class=navlinkEdit Task/abr /';
 echo 'a href=add_task.php class=navlinkAdd Task/abr /';

}


bastien



From: Jeff Broomall [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] Retricting Access to Menu Items
Date: Mon, 6 Mar 2006 07:27:03 -0500

Good morning everyone.

I'm building a very simple content management site that tracks tasks.

The options available are:
 1. Add Task
 2. Edit Task
 3. View Task
 4. Print Task

I need to restrict some users to only View and Print and I'm trying to find 
a way to tell the page not to load the menu options (the text) for those 
not having access to the Add and Edit functions.


IOW, they would only see View and Print.

I have three basic users:
 1. System Admin
 2. Subject Matter Expert (SME)
 3. Viewers

Obviously the System Admin and SME will have full access so it's the 
Viewers that are to have access to only View and Print.


I have a users table but haven't set it up for the distinction.  What I was 
thinking was creating a field labeled users_group and assign a numeric 
value for each user using the numbering system above.


I have my page load the menu options:

a href=home.php class=navlinkHome/abr /
a href=view_tasks.php class=navlinkView Tasks/abr /
a href=edit_task.php class=navlinkEdit Task/abr /
a href=add_task.php class=navlinkAdd Task/abr /

into here...

BODY

table width=90% border=1 cellspacing=10 cellpadding=0 
align=center


 trtd colspan=2h1 id=mainheadICAO Tasks — WAFS/h1/td/tr

tr
td align= center valign=top nowrap=nowrap width=10%
 Menubr /
 ?php include ('./includes/menu.html'); ? --The menu above inserted 
here.

/td

td valign=top class=content


How can I tell the system not to load the last two lines unless they are a 
System Admin or SME?


I read a chapter on Cookies/Sessions...but it wasn't that helpful for this 
case.


Can I setcookie('user_group', '3') and use that somehow???

Am I in the ballpark with this solution?

Thanks.

Jeff




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[PHP-DB] Very Tricky PHP Array.

2006-03-06 Thread Michael Acs

Hi there,

I have one arrray: $field_vals
There are three values in the array, however two are multi-dimension array.

i.e:
$field_vals[base] - Array (
[0] - 1
[1] - 2
)
$field_vals[make] - Array (
[0] - 5
[1] - 7
[2] - 8
)
$field_vals[qty] - 1

I have a MySQL with a table (part) for the data to be stored in: part_id 
(key), base_id, make_id, qty

I need to build a series of MySQL queries to INSERT the data into the table.

ie.
INSERT INTO part(cols...) VALUES(1,5,1);
INSERT INTO part(cols...) VALUES(1,7,1);
INSERT INTO part(cols...) VALUES(1,8,1);
INSERT INTO part(cols...) VALUES(2,5,1);
INSERT INTO part(cols...) VALUES(2,7,1);

From the example array above, (2x3x1) 6 INSERT queries would be 
required. I would want this to be processed in the site form processor 
function. Basically, all my HTML forms have:
$field[table_column] - which stores the field value or values in the 
case above.
$name[table_column] - which stores the field name, this is always a 
single value since it is a hidden field.


The [table_column] corresponds to the exact column in the database. I 
also three additional values:

$table - Mandatory, stores the table name.
$key - Mandatory, stores the primary key column for the table.
$link - Optional, however used if a entry is required to be added into 
another table that is related to the key in the primary table.


On submission, I use the $table value to list out all the columns that 
exist in the database table. Then I use a for() loop to go through each 
table column, which allows me to process both $field[table_column] and 
$field[table_column].


For me to be able to do a foreach() to achieve the list of queries I 
require to be executed in the database, I will most likely have to 
change the way I use the for loop, or will I? I am not exactly sure. My 
form processor code is quite long, however, I will place it at the 
bottom for reference.


What I could do is keep my first for() loop which outputs the submitted 
data on a confirmation page and create a separate section of the 
function altogether, which generates the queries I require. At the 
moment I generate a part of the MySQL queries inside the for loop, 
however, doesn't work quite as well when you have an array, rather then 
a single variable.


First of all, is it possible to do, if you do not want to static code to 
generate the queries?


This is how I would do it, if I used static code:

---CODE---

$db = mysql_connect();
$qty = $field_vals['qty'];
foreach ($field_vals['base'] as $base) {
   foreach ($field_vals['make'] as $make) {
  $inserts .= ($base, $make, $qty),;
   }
}
$sql = INSERT INTO part(base, make, qty) VALUES  . substr($inserts, 0, 
-1);

if (mysql_query($sql, $db)) echo done.;
else echo error.;

-/-CODE-/-

I am not asking for people to completely rewrite my code, just some tips 
or hints on how to achieve this would be greatly appreciated.


Here is my current form processor function for reference, if you need 
any additional information, please let me know.


---CODE---

?php
function formprocess($table,$key,$fname,$fval,$stage,$lnktblcol=0) {
// setup db connection
$db = new sql;
// Switch through each stage
switch ($stage) {

case Submit:
// set inital error variable
$error = false;
// lookup main table columns
$columns = $db-show_columns($table);
$count = count($columns);
// include validator configuration file
include('inc/validator.config.php');
// begin loop through each column
for ($i=0;$i$count;$i++) {
// set variables
$column = $columns[$i];
$option = ;

// check that column exists in HMTL form
if ($fname[$column] || $fval[$column]) {
// begin row and set column value
$content .= trtd width='100' 
valign='top'.$fname[$column].:/tdtd width='175'b;

$value = $fval[$column];
// validate column value
if (!is_array($value)) {
$fldmsg = validate($value,$db_col[$column]);
} else {
$fldmsg = OK;
}
// if not table key
if ($column != $key) {
// set column field and table2 for drop down values
$selcolumn = ereg_replace($table._,,$column);
$table2 = ereg_replace(_id,,$selcolumn);
// if drop down value
if (ereg(_id,$column)  !ereg(base_id,$column)) {
// grab table columns for drop down value
$columns2 = $db-show_columns($table2);
$lnkcolumn = 
ereg_replace($table2._,,$columns2[2]);

// grab link column data
if (ereg(_id,$lnkcolumn)) {
$table3 = ereg_replace(_id,,$lnkcolumn);
if (!is_array($value)) 

Re: [PHP-DB] Retricting Access to Menu Items

2006-03-06 Thread MIGUEL ANTONIO GUIRAO AGUILAR


Now that you mention it!!

I tried to use sessions but without success, i did:

session_start();
session_register(var);

but after log in as a different user, it keeps taking the data of the previous 
user. If I close the browser window and reload the page and log in, then it 
takes de current user

Maybe I'm not killing the previous session! session_unset();
Do I need to propagate the Session ID on every page that use session_start()??
Can I use session_id();?

Best Regards,
Miguel Guirao


- Mensaje original -
De: Jeffrey [EMAIL PROTECTED]
Fecha: Lunes, Marzo 6, 2006 7:37 ombr
Asunto: Re: [PHP-DB] Retricting Access to Menu Items

 I've done this kind of thing with a number of web apps.
 
 What I usually do is create a user table in MySQL with a user-name, 
 password and access level, which has an integer value.
 
 When a user logs in successfully, a session is created (see 
 session_start() in php documentation), the access level is pulled 
 from 
 the user table and saved as a session variable. Then it is a simple 
 matter of using bits of code like...
 
 if ($_SESSION['access_level']  7){
   echo some stuff;
 }
 
 In your example, you will also want to check the user's access 
 level on 
 each restricted page - it is not enough to hide menu options. Users 
 could simply type the URL in.
 
 I hope that's clear.
 
 Good luck,
 
 Jeffrey
 
 Jeff Broomall wrote:
  Good morning everyone.
  
  I'm building a very simple content management site that tracks 
 tasks. 
  The options available are:
   1. Add Task
   2. Edit Task
   3. View Task
   4. Print Task
  
  I need to restrict some users to only View and Print and I'm 
 trying to find a way to tell the page not to load the menu options 
 (the text) for those not having access to the Add and Edit functions.
  
  IOW, they would only see View and Print.
  
  I have three basic users:
   1. System Admin
   2. Subject Matter Expert (SME)
   3. Viewers
  
  Obviously the System Admin and SME will have full access so it's 
 the Viewers that are to have access to only View and Print.
  
  I have a users table but haven't set it up for the distinction.  
 What I was thinking was creating a field labeled users_group and 
 assign a numeric value for each user using the numbering system above.
  
  I have my page load the menu options:
  
  Homebr /
  View Tasksbr /
  Edit Taskbr /
  Add Taskbr /
  
  into here...
  
  BODY
  
  table width=90% border=1 cellspacing=10 cellpadding=0 
 align=center 
   trtd colspan=2h1 id=mainheadICAO Tasks #8212; 
 WAFS/h1/td/tr   
  tr 
  td align= center valign=top nowrap=nowrap width=10%
   Menubr /
   ?php include ('./includes/menu.html'); ? --The menu above 
 inserted here.
  /td
   
  td valign=top class=content 
  
  
  How can I tell the system not to load the last two lines unless 
 they are a System Admin or SME?
  
  I read a chapter on Cookies/Sessions...but it wasn't that helpful 
 for this case.
  
  Can I setcookie('user_group', '3') and use that somehow???
  
  Am I in the ballpark with this solution?
  
  Thanks.
  
  Jeff
 
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[PHP-DB] Help needed creating a social network

2006-03-06 Thread Daevid Vincent
Anyone have some pointers at a HowTo on creating a social network?

Basically I need to show people in your immediate network, and also friends
of your friends, etc... Like the whole 'six degrees of separation' thing.
Ala: myspace, friendster, etc. ad nauseum.

I prefer mySQL and PHP, but I could port from most any code. I guess I'm
mostly interested in the theory of this an how do I set up the tables
properly and what is the magic incantation (JOIN) to get this chain of
people.

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Re: [PHP-DB] Help needed creating a social network

2006-03-06 Thread Micah Stevens

CREATE TABLE `users` (userID int(11) not null auto_increment, primary 
key(userID), name tinytext not null, email tinytext not null);

CREATE TABLE `links` (userID int(11), key (userID), friendID int(11), 
key(friendID));

?php
$friends = mysql_query(select users.name, users.userID from users left join 
links on links.friendID = users.userID where links.userID = 
{$_GET['userID']});

echo h1You have friends!!/h1;
while ($f = mysql_fetch_assoc($friends)) {
echo a 
href='socialnetwork.php?userID={$f['userID']}'{$f['name']}/abr\n;
}

publish();
profit($greatly);
do (!$use) {
coldfusion('Cause it sucks');
}
?

(har har) 







On Monday 06 March 2006 9:47 pm, Daevid Vincent wrote:
 Anyone have some pointers at a HowTo on creating a social network?

 Basically I need to show people in your immediate network, and also friends
 of your friends, etc... Like the whole 'six degrees of separation' thing.
 Ala: myspace, friendster, etc. ad nauseum.

 I prefer mySQL and PHP, but I could port from most any code. I guess I'm
 mostly interested in the theory of this an how do I set up the tables
 properly and what is the magic incantation (JOIN) to get this chain of
 people.

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To unsubscribe, visit: http://www.php.net/unsub.php