PHP - Simple Mysql Query. Help

The following query or while loop is only increasing the ArticleID variable every 3rd time the script is run, I've narrowed it down to the following code snippet. Can you spot a problem with this, I'm in my first week of PHP and MySQL and I can't see any problem with it.

Any help would be mighty appreciated by this idiot 

Code snippet:
---
$result = mysql_query("SELECT ArticleID FROM test_top ORDER BY ArticleID ASC LIMIT 1") or die(mysql_error()); 
   while($row = mysql_fetch_array($result))
   {
   $ArticleID=$row['ArticleID'];
   }
$ArticleID=intval($ArticleID);
$ArticleID++;
---

Hello, I am trying to pick up php again and just exercising my skills. So I have it so that it fills my form with the values of what I want to edit, and when I click the edit button, it doesn't edit any of the information. When I echo out $result, I get a MYSQL query string that has the same values as the table, so its not getting the new values that are edited.


<?php

@mysql_connect('localhost', 'root', '') or die("Could not connect to Mysql Server. " . mysql_error());

@mysql_select_db('tutorials') or die("Could not connect to Database. " . mysql_error());

if(isset($_GET['edit']))

{

$id = $_GET['edit'];

$query = "SELECT `username`, `password` FROM `users` WHERE `id` = '$id'";

$result = mysql_query($query);

$row = mysql_fetch_array($result);

$name = $row['username'];

$password = $row['password'];

}

if(isset($_POST['edit']))

{

$id = $_GET['edit'];

$query = "UPDATE `users` SET `username` = '$name', `password` = '$password' WHERE `id` = '$id'";

$result = mysql_query($query);

echo $query;

if(!$result)

{

echo mysql_error();

}else{

echo 'updated post';

}

}
?>
<form method="POST" action="" >

<input type="text" name="name" value="<?php echo $name; ?>" /> First name <br />

<input type="text" name="password" value="<?php echo $password; ?>" /> Last name <br />

<input type="submit" name="edit" value="edit" />
</form>


I believe it has something to do with the values of $name and $password in the form conflicting with the first if isset and the second if isset.

Thanks for any help possible   

I'm attempting to implement a simple social networking system but at the moment am confused about how to create a multiple query which will display a certain user's friends list.

The database contains four tables, the two tables that I'm using at the moment at 'usersTable' and 'friendshipsTable' are detailed below.


usersTable           | Table that stores all the user data

UserID         | Default primary key
Forename              |
Surname         |
Username           |
Password              |
Email Address      |


friendshipTable   | Table that stores information about friendships between users

FriendshipID      |  Default primary key
userID_1                |  UserID
userID_2         |  UserID
Status         |  Either Pending or Confirmed.

The user's id is parsed into the url, and then saved into a variable.

blah.com/userprofile.php?id=6

$id = $_GET['id']; 

I am familiar with creating simple queries, but can't quite work out how to set up multiple table queries.

What the query needs to do is to check the userID that is parsed with the url, and then check the friendshipsTable by checking if either the userID_1 or userID_2 field  matches the userID to grab the records from the table where there is a match.

The next step is to check to see if the friendship is 'Confirmed' or 'Pending' and if it is 'Pending' to ignore it. Once the records have then been chosen I need the query to then check the value in either userID_1 or userID_2 that doesn't match userID and then pull the user's username and name from the usersTable so it can be displayed on a webpage.

I've no idea hoe much I may or may not be overcomplicating this, an example of the code that I've got so far for this query can be found below, but that's as far as I've got at the moment.

$displayFriends = mysql_query("SELECT * FROM friendshipTable, usersTable WHERE friendshipTable.userID_1='$id' OR friendshipTable.userID_2='$id' ");

Cheers for any help.

Here is my code:
// Start MySQL Query for Records
$query = "SELECT codes_update_no_join_1b" .
"SET orig_code_1 = new_code_1,
       orig_code_2 = new_code_2" .
"WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)";
$results = mysql_query($query) or die(mysql_error());
// End MySQL Query for Records


This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this???

Code: [Select]
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 '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1

If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

  Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");
 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>
 
<div id="wrap">
 
<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>
 
<? include_once("includes/navigation.php"); ?>
 
<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php
 
if(!$_SESSION['user'])
{
 
        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);
 
        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>
       
        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>
       
        </select></td></tr>
       
        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");
               
                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }
       
}
else
{
        echo "You are currently logged in!";
}
 
?>
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