Held needed urgent!

[studgate]

I have four tables in my database, offices, members,

activities & groups.

What I want to do is if the user has a group/activity/office

in the database, to show it in his profile.

I know it should be an if/else statement but I need some

help display the data.

 

Example: Username is jean, and if jean has entered any groups,

offices,  when jean login into his account,

the group name and the office name will show so jean can see and

update them.

Thanks Guys!

 

[mpharo]

you just need to query for that information for each table and then display it...im not sure what else your asking for? maybe post a little bit of code so we can see what your dillema is??

[Barand]

Your db structure, showing table relationships would be useful too

[studgate]

just the query will be helpful. basically check the username

for each table and if the username is jean, show information.

the db structure is simple, users ask for username, address, phone,

CREATE TABLE `users` (
  `id` int(10) NOT NULL auto_increment,
  `firstname` varchar(30) NOT NULL default '',
  `lastname` varchar(30) NOT NULL default '',
  `email` varchar(50) NOT NULL default '',
  `phone` varchar(15) default NULL,
  `country` varchar(30) default NULL,
  `username` varchar(30) NOT NULL default '',
  `password` varchar(30) NOT NULL default '',
  `status` int(1) NOT NULL default '1',
  `registration_timestamp` int(20) NOT NULL default '0',
  `email_verify_code` varchar(12) default NULL,
  UNIQUE KEY `id` (`id`),
  UNIQUE KEY `username` (`username`),
  UNIQUE KEY `email` (`email`)
);';

the other tables are similar. the username field for each table

is filled when the user submit any of the form.

Thanks again guys.

[studgate]

the query is the only thing that I need some help on.

 

[Barand]

As the other tables may or may not have data related to the user, use LEFT JOIN when joining on username..

[mpharo]

<?php

$select=mysql_query("SELECT * FROM offices WHERE username='$username' ") or die(mysql_error());

while($sql=mysql_fetch_array($select)) {

echo "Username: $sql[username]"
. "Offices: $sql[offices]";

}

?>

 

just do the same thing for each table and keep giving each variable a unique name. You can also do it with a more complicated query using joins...