user login linked to preferences

[nick_kirby]

I am trying to create a database to store certain preferences.

I have "pinched" some excellent code to enable users to create a login which is stored in a MySQL database.

What I want to do is: when user logs in, they see what they have entered into the db, and can add new items. These items are variable: this is the start of a big project where multiples selections can be made that hook into parts of a database.

I don't want them to see anyone elses entries.

e.g. Bob has entered the names Charles, James and Harriet (as text fields).

Bertram has entered Ruben, Emily and Claire.

I want each entry to create a new row in the db - eventually the fields of age, and exam results, addresses will be included.

I want Bertram to login and see his entries listed through php, with an option to edit them, but before that, I need to know how to "join" one sql table to another - if Bob is id 1, then I want everything he creates to go into table where a column labelled "user" has the entry of 1, Bertram the entry 2, Charles, entry 3.

1 1 Charles A+
2 1 Emily A+
3 1 Wilbur C-
4 2 Walter B
5 2 Clarence
6 3 Alex
7 3 Maude
8 3 Christopher

Sort of.... no, just like that!

Sorry to be long winded. I'm not good at explaining things!

login script here: http://www.niknet.homeip.net/rifts/login/login.php

Many thanks,
nik

[kakki]

maybe a "group by" statement will do this

[nick_kirby]

Ummm... pass? It's here I get stuck. I've a list of users. Now I want to link them to what they create.

[bo0]

So you would have a "users" table and an "entries" table.

In your entries table you would have a column called "userid".

To view the entries by "userid" "1" you would use:

$sql = ("SELECT * FROM entries WHERE userid='$userid'");

[nick_kirby]

ok - I've created that inside the entries table - the table for user logins has three columns: username, password adn admin.

Should I be using username instead of userid - sorry, it's a dippy question I know. I just want to be sure.

Is there any way of changing the name of a MySQL db?

Is there a way to look at the contents of a table? I do have phpadmin in here somewhere. I'll give that a go.

[bo0]

nick.kirby wrote:

ok - I've created that inside the entries table - the table for user logins has three columns: username, password adn admin.

Should I be using username instead of userid - sorry, it's a dippy question I know. I just want to be sure.

Is there any way of changing the name of a MySQL db?

Is there a way to look at the contents of a table? I do have phpadmin in here somewhere. I'll give that a go.

Add a column called "userid" into your user’s table- this way you have a unique way of identifying each user.

[nick_kirby]

Right-o. I managed that after a bit of work!

m now trying to run the sql to input into the name column:

<?php
//connect to the login db, characters table
mysql_connect('localhost', 'rifts', '***') or die('Failure to connect to the database');
mysql_select_db('login');

//insert data
$query = "INSERT INTO characters(charname) values('$name')";
$result = mysql_query($sql) or die(mysql_error());
?>

Which as I see it should take the $name variable and put it into the table?

[nick_kirby]

Sorry to reply to myself, but I can't get this to go: the query apparently runs, but on looking in phpinfo there is nothing entered into the table!

Having created a primary key field called userid in my login db, how can I now, when entering information also log that value into the characters table?

I have created a column called userid - the idea being that when a name is entered both the name AND the userid of the creator are logged.

Any help gratefully received. I'm trying, but it's a steep slope!

[thorpe]

Post your code!

[nick_kirby]

Ah - I have, for the insertion into the database, however, I really do not know where to start with entering the logged in userid key into the other table.

Am I referencing it improperly? The input script is in a different folder to the post method thus:

<form action="../scripts/input_into_db.php" method=post>
<input type="submit" value="Send!">
</form>

basically pressing submit runs the above (above) sql script. Except it doesn't!

[nick_kirby]

//insert data
$query = "INSERT INTO characters(charname) VALUES
('$_POST[$name]')";
$result = mysql_query($query) or die(mysql_error());
echo "Yep!";

I've added the $_POST missive, however the code runs, reports the echo statement, but there is nothing in the column. Nothing at all, according to phpadmin.

Does it need time to refresh?

[nick_kirby]

I've fought this, and it won't go. I really, honstly don't knwo what's going wrong. I've copied from this site, other sites, read the manual - nothing.

this is the input script:

<?php
//connect to the login db, characters table
mysql_connect('localhost', 'rifts', '***') or die('Failure to connect to the database');
mysql_select_db('login');

//insert data

//insert characters stats and name.
mysql_query ("INSERT INTO characters (charname, charstr, chardex,charcons,charint,charwis,charcha) VALUES
('$POST[$name]',
'$POST[$Str]',
'$POST[$Dex]',
'$POST[$Con]',
'$POST[$Int]',
'$POST[$Wis]',
'$_POST[$Cha]')");
mysql_query($sql) or die(mysql_error());
echo "Yep!";
?>

This is the form script:

<?php
include("config.php");
if($SESSION['logged'] != 'agk8gjf38834j2')
{
header("location:" . $login_file);
}
echo ('<a href="logout.php">logout</a>');
?>

<link href="rifts.css" rel="stylesheet" type="text/css">
<link rel="stylesheet" type="text/css" href="/rifts/login/rifts.css">

<html>
<body>

<div id="leftcontent">
<form id="contactform">
<fieldset>
<dl>
<dt><label for="name">Character Name</label></dt>
<dd><input type="text" name="name" value="" /></dd><br>
<dt><label for="email">Your Email Address</label></dt>
<dd><input type="text" name="email" value="" /></dd>
</dl>
</fieldset>
</form>

<FORM id="chkboxform">
<fieldset>
<u>Sex</u><br>
<LABEL><INPUT TYPE="radio" NAME="answer" VALUE="true" CHECKED> Male</LABEL><br>
<LABEL><INPUT TYPE="radio" NAME="answer" VALUE="false"> Female</LABEL>
</fieldset>
</FORM>

<form id="contactform">
<fieldset>
Choose your book : <SELECT NAME = book STYLE = "Width:200">
<OPTION Value = 1>Rifts Manual</OPTION>
<OPTION Value = 2>Vampire Kingdoms</OPTION>
</SELECT>
</fieldset>
</form>

<FORM id="chkboxform">
<fieldset>
<u>Character Class</u><br><br>
<LABEL><INPUT TYPE="radio" NAME="class" VALUE="true" CHECKED>OCC</LABEL><br>
<LABEL><INPUT TYPE="radio" NAME="class" VALUE="false">SCC</LABEL>
</fieldset>
</FORM>

<form id="contactform">
<fieldset>
Character template :
<SELECT NAME = occ_rcc STYLE = "Width: 200">
<OPTION Value = 1>Rifts Manual</OPTION>
<OPTION Value = 2>Vampire Kingdoms</OPTION>
</SELECT>
</fieldset>
</form>
</div>

<div id="rightcontent">
<FORM id="contactform1">
<fieldset>
<u>Attributes</u><br><br>
<dt><label for="charStr">Strength</label></dt><dd><input type="text" name="Str" value="" /></dd><br><br>
<dt><label for="charDex">Dexterity</label></dt><dd><input type="text" name="Dex" value="" /></dd><br><br>
<dt><label for="charCon">Constitution</label></dt><dd><input type="text" name="Con" value="" /></dd><br><br>
<dt><label for="charInt">Intelligence</label></dt><dd><input type="text" name="Int" value="" /></dd><br><br>
<dt><label for="charWis">Wisdom</label></dt><dd><input type="text" name="Wis" value="" /></dd><br><br>
<dt><label for="charCha">Charisma</label></dt><dd><input type="text" name="Cha" value="" /></dd><br><br>
</fieldset>
</FORM>

<form method="post" action="../scripts/input_into_db.php">
<input type="submit" value="Send!">
</form>
</div>
</body>
</html>

I am getting the error "The query was empty", which is has not been.