PHP - Select * Users Question

Hello, guys! Just a small first ever thread, where I ask a question where I hope anyone is able to help me out

I am working on a project with a friend of mine for a company. I have a problem with the function on a form.
I wonder; How do i change the option value of a form? Let's say if I choose a category like "Cake", I want the second drop down box to display different brands of cake. But if I choose bicycle for example, I want the option values to be different brands of bicycles. I hope you guys understand what I was trying to say here.
Anyways, here is a sample image where I want this function to occur:

Sample:


Cheers and thanks in advance,
Marius

Hi.  I have two select fields one called category and the other sub category.  Both are in the format.
<select name="category" value="

<?php $qGetCat = "SELECT * FROM club_category" ;

$rGetCat = mysql_query($qGetCat);

while ($Cat = mysql_fetch_assoc($rGetCat))

{

?>" />
                                <option value="<?php echo $Cat['categorys'];?>"><?php echo $Cat['categorys']; ?></option>
                                <?php } ?>
                                </select>
<select name="sub_category" value="

<?php $qGetSubCat = "SELECT * FROM sub_categorys" ;

$rGetSubCat = mysql_query($qGetSubCat);

while ($SubCat = mysql_fetch_assoc($rGetSubCat))

{

?>" />
                                <option value="<?php echo $SubCat['sub_categorys'];?>"><?php echo $SubCat['sub_categorys']; ?></option>
                                <?php } ?>
                                </select>


Now I want to make it so that when a particular category is chosen for example self defence.  The sub category when clicked shows all the self defence stuff.  If watersports was chosen sub category would show things like swimming.  What do I need to look up to do this? At the minute there is no link the way I have coded these to fields.

In the database its self there is a;

category table with categoryID and category_name

sub category table with sub_cat_name and categoryID

I think this will work but I have never done this so its making my head hurt just thinking about it.  Does anyone have any advice, tuts or scripts to do this? 

Thank you

I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.   my form page test.php

<?PHP
require_once("./include/membersite_config.php");

if(!$fgmembersite->CheckLogin())
{
    $fgmembersite->RedirectToURL("login.php");
    exit;
}

?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?>  "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?>  "><br><br>

<button>Submit</button>

my editform.php

<?php

$con = mysqli_connect("localhost","root","user","pass");

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");



header("Location: test.php");

?>

I just discovered that I have a major security flaw with my website.
Anyone who logs in to the website can easily access other users information as well as delete and edit other users information just by changing the ID variable in the address bar.
I have user ID Session started on these pages but still people can do anything they like with other users information just by editing the address bar.
For example if your logged in in the address bar of www.mywebsite.com/delete_mystuff.php?id=5 and change the "5" say to a "9" then you will have access to user#9 information.

Every important page that I have has this code:
Code: [Select]
 session_start();

if (!isset($_SESSION['user_id'])) {
// Start defining the URL.
         $url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['PHP_SELF']);
// Check for a trailing slash.
         if ((substr($url, -1) == '/') OR (substr($url, -1) == '\\') ) {
         $url = substr ($url, 0, -1); // Chop off the slash.
}
// Add the page.
$url .= '/index.php';
ob_end_clean(); // Delete the buffer.
header("Location: $url");
exit(); // Quit the script.
} else {
                //Else If Logged In Run The Script

if((isset($_GET['id'])) && (is_numeric($_GET['id']))) {
         $id = (int) $_GET['id'];
} elseif ((isset($_POST['id'])) && (is_numeric($_POST['id']))) {
         $id = (int) $_POST['id'];
} else {
         echo ' No valid ID found, passed in url or form element';
        
         exit();
}



What am I doing wrong?
Please help if you know how to correct this.

Many thanks in advance.

Can anybody clarify the usefulness of "AS" and elaborate on how it operates.

If I select a 'bananas' AS 'yeelow fruit' will it change the TITLE in the table?  Or the table that is being viewed?
If I dump a list of 'FRUITS' into and HTML table, will using AS rename each column for me, or is that handled by MY coing of the HTML table?

This is what i have

Code: [Select]
$traderss1=mysql_query("SELECT * FROM `poke_owned` WHERE `box`='1' AND `trainer`='" . $trading->username . "' ORDER BY `name` ASC");
How can i do this:

Code: [Select]
$traderss1=mysql_query("SELECT * FROM `poke_owned` WHERE `box`='1' AND `trainer`='" AND `traded`='0' . $trading->username . "' ORDER BY `name` ASC");
I know its not as simple as just adding  AND `traded`='0' cause i tried that and i just got this

Code: [Select]
Parse error: syntax error, unexpected '=' in /home/tpkrpgn1/public_html/trade5.php on line 57
MySQL table (in case you need to know it?)
Code: [Select]
(`id`, `trainer`, `gender`, `name`, `party1`, `party2`, `party3`, `party4`, `party5`, `party6`, `level`, `exp`, `hp`, `totalhp`, `att`, `def`, `box`, `days`, `battle_wins`, `battle_losses`, `move1`, `move2`, `move3`, `move4`, `attbon`, `defbon`, `totalexp`, `sold`, `traded`, `item`, `soldtrainer`)

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=322629.0

I am working on a quiz app

image 1 shows the index.php page
image 2 shows the first question
image 3 shows the second question
image 4 shows the third question
image 5 shows the result after completing the quiz
image 6 shows the database 'quizzer' and its tables
image 7 shows the 'questions' table
image 8 shows the 'choices' table

THIS LINK CONTAIN ALL THE CODE (and images) I HAVE DONE SO FAR

https://www.mediafir...o7f5q0fe6y/quiz

1.Now my question is how to select the question RANDOMLY from 'questions' table along with 'choices' (by adding code to the existing file or create a new one).

2.If user refresh/reload the page before starting ('Start Quiz') or click 'Take Again' after finishing the quiz, the question should appear randomly.

3.Basically I want to change the order of question appearing in the browser each time I refresh.

4.My work so far is mentioned above.........Please help me with this "RANDOM" problem !!

P.S - Will it be possible, by creating a random function in PHP which will check for repeat questions in a session and check for the 'id' of the question and if it is new display it on the page.

If so what should I do and if no then how to do?

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

I have 2 queries that I want to join together to make one row  

im new , and ... 

i hate tutorials .. books .. anything that does not make u part of the deal -  .. thats why i started by creating something and learning from my mistakes at the same time .. i like this way of learning ..

soo , while im building and trying things out .. i started thinking how the server know the person with this link is really U ? .. when u start just linking pages to each other its just a matter of finding out what is the link to do what ever u want with the users personal pages ! .. i know my questions r stupid but i just hate to go and write lessons without any effort

im confused with the concept of SESSIONS and COOKIES , r they the unswer to this security problem ? how u can work with them .. ?

im not asking for codes .. just general ideas about users and how they control their profiles and stuff with full security ?

ill be very thankful if i get any answer ^^

Hi. my browser ist telling me there are errors on line 3 and 4 for my code. It says summin like Notice: HTTP_CLIENT_IP and HTTP_X_FORWARDED_FOR are unidentified:

below is the code see if you can spot anything btw i copied down the code from beginner php tutorial 66 if you type that into youtube .
Code: [Select]
<?php

$http_client_ip = $_SERVER['HTTP_CLIENT_IP'];
$http_x_forwarded_for = $_SERVER['HTTP_X_FORWARDED_FOR'];
$remote_addr = $_SERVER['REMOTE_ADDR'];

if (!empty($http_client_ip)){
   $ip_addr = $http_client_ip;
   
}else if(!empty($http_x_forwarded_for)){
   $ip_addr = $http_x_forwarded_for;
}else{
   $ip_addr = $remote_addr;
}

echo $ip_addr;
?>

Thanks

MOD EDIT: code tags added.

how can I list a user from a table and show the results in a grid with different color eg frist in blue color second on white , 3rd on blue 4th in with etc I do need to set select command and I have db name and ip on a file called dbconfig.php from wd calendar so I just need to read the info   ps: I cant post links so search for wd calendar and see the dbconfig.php in php folder

I would get the ip address of the user that is on the site. I used $ip = $_SERVER['REMOTE_ADDR']; and it doesn't show my actual ip. Is it because im using an apache server on my computer.

 Hi guys,

im just trying to work out an app in my head and on paper. im just wondering..
when a user registers they can choose an Avatar 100px by 100px jpg, when they upload one would i then grab the file and store all Avatars in a avatar image folder and rename it to something like.. avatar[user_id].jpg and keep them all in the same folder.

or

would i crate a folder called users, each user gets their own folder with files like avatar.jpg and it finds the [user_id] folder and pulls the avatar out from that, or is there a more prefered method?

cheers