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CP,	04-08-2007 12:08 PM

calling all php and mysql guru's

Ive been searching for a day now and im going to cry soon.

I am to make a project a dummy website e-commerce website. So security and all the other things are not so important, just need things to be bland and plain but most importantly to work!

Ive got a mysql db set up, and my users can log in and log out thats working.

Now here is the problem. Once that particular user logs in i want him/her to be able to view the details stored on the db about them, i.e. their firstname, surname, email etc

Ive been trying so many methods but none seem to work or some are giving me blank results. If somebody can help me with this, it would be much appreciated.

Dismounted

04-09-2007 06:16 AM

Is this using your own backend or vBulletin's? (I'm guessing your own) You will need to query the database for the user's row and output the information.

PHP Code:


		
			
$result = mysql_query("SELECT * FROM `YOURUSERTABLE` WHERE `userid` = " . $YOURUSERIDVARIABLE . " LIMIT 1");
$row = mysql_fetch_assoc($result);

Then all your information would be in $row. To access it's data, simply use $row['FIELD'] (where FIELD is the name of the required field).

CP,	04-09-2007 06:28 PM

This is the test data in my contacts table as you can see: http://www.photo-host.org/img/958319table.jpg

ive inserted this, is this correct?:

Code:

$result = mysql_query("SELECT * FROM 'contacts' WHERE 'member_id' = " . $member_id . " LIMIT 1");

$row = mysql_fetch_assoc($result);

contacts is the table and member_id is the member id which auto increments as a new user registers.

I can login in fine with this error on top:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in e:\webareas\am663\carttest\login.php on line 8

EDIT: I added the "@" in front of mysql_fetch_assoc($result); which seemed to have removed the error message but when i call the data from the db e.g. echo $row['$member_id'];
echo $row['$username']; nothing shows up on my members page?

Adrian Schneider

04-09-2007 07:22 PM

PHP Code:


		
			
$result = mysql_query("
    SELECT * 
    FROM contacts 
    WHERE member_id = $member_id 
") or die(mysql_error());
$row = mysql_fetch_assoc($result);

CP,	04-09-2007 08:10 PM

Quote:

Originally Posted by SirAdrian (Post 1223523)

$result = mysql_query("
SELECT *
FROM contacts
WHERE member_id = $member_id
") or die(mysql_erorr());
$row = mysql_fetch_assoc($result);[/code]

you have a typo i quickly fixed it, is it suppose to be in 6 lines? as i reduce it to 1 line the colour changes... but i still followed your example and it said:

"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 '' at line 3" ?? Im on my university server, do i upload a phpinfo file to find out?

Adrian Schneider

04-09-2007 08:25 PM

What is the $member_id value set to? I'm assuming it's an integer... right?

CP,	04-09-2007 08:31 PM

Quote:

Originally Posted by SirAdrian (Post 1223572)

What is the $member_id value set to? I'm assuming it's an integer... right?

look here: http://www.photo-host.org/img/958319table.jpg member id is an auto increment integer.

Code:

CREATE TABLE contacts (

  member_id int(30) NOT NULL auto_increment,

  firstname varchar(30) NOT NULL default '',

  surname VARCHAR(30) NOT NULL default '',

  contact_no INT(11),

  housename_no VARCHAR(20),

  address_line1 VARCHAR(20),

  address_line2 VARCHAR(20),

  address_line3 VARCHAR(20),

  city_county VARCHAR(15),

  postcode VARCHAR(8),

  email varchar(30) NOT NULL default '',

  username varchar(25) NOT NULL default '',

  password varchar(255) NOT NULL default '',

  PRIMARY KEY  (member_id),

  UNIQUE KEY username (username)

) TYPE=MyISAM COMMENT='Members';

Adrian Schneider

04-09-2007 09:15 PM

In your query you use $member_id ... what is the value of that?

CP,	04-09-2007 09:28 PM

Quote:

Originally Posted by SirAdrian (Post 1223602)

In your query you use $member_id ... what is the value of that?

What do you mean? the value in the sql table or in the php code?

Adrian Schneider

04-09-2007 09:31 PM

The PHP variable... if its empty (and I'm pretty sure it is) the query will die because:

SELECT * FROM TABLE WHERE id = ;

Is not valid.

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