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"Is this possible with PHP?"

I have a table in my database which contains a users ID and a number.
What I want to be able to do is for each user when they view a particular page to have a pull down menu which contains the numbers from the table in the databse that apply to them. Some users will have no numbers and some users may have quite a few 10+

eg

userID | number
1......| 345
2......| 345
1......| 121
1......| 567

So when user 1 views the page they get the menu with 345, 121 and 567 in it but if user 2 views it they only get 345.

At the moment they have to enter a number from memory and I think it would be much better if they could select it from a menu, other alternative is just to display the numbers that apply to them but I feel that the pull down menu would be a better solution.

So can it be done? If so how easy would it be?

Cheers

I have a table in my database which contains a users ID and a number.
What I want to be able to do is for each user when they view a particular page to have a pull down menu which contains the numbers from the table in the databse that apply to them. Some users will have no numbers and some users may have quite a few 10+

eg

userID | number
1......| 345
2......| 345
1......| 121
1......| 567

So when user 1 views the page they get the menu with 345, 121 and 567 in it but if user 2 views it they only get 345.

At the moment they have to enter a number from memory and I think it would be much better if they could select it from a menu, other alternative is just to display the numbers that apply to them but I feel that the pull down menu would be a better solution.

So can it be done? If so how easy would it be?

Cheers

Easy as pie, though I'm tired and about to go to bed, so I'll give you the jist, and hopefully you can do the rest.

1. Run the query SELECT number FROM table WHERE id=1; or somthing like that (haven't done SQL for a little while, so my syntax may be wrong.)

2. Output an HTML select list and form:

while ($row = mysql_fetch_row($result))
echo '$row[0]',"\n";

?>

Sorry if I completely mess up the syntax, but like I said ... zzzzZZZZ

Thanks for the reply.

I was trying something very similar. But I was using a printf in the while loop so the menu was being displayed but with no entries in it.
Tried using the echo instead just there and its displaying stuff in the menu, not what I wanted though as its displaying the variable name and not what it has been assigned too. Sure its more likely something to do with my code so I'll look over it again.

Edit: Still just displaying $number instead of the actual value from the table, what I've tried hasn't worked. Going to get some sleep and give it another go tomorrow.

thanks again

Try:
----------------------

include("dbconnect.php"); // takes care of MySQL connection and DB selection

$userid = $_POST['userid']; // change this to wherever you have the userID stored
$result = mysql_query("SELECT number FROM table WHERE userID='".$userid."';");

while ($row = mysql_fetch_assoc($result)) {
echo "" . $row['number'] . "\n";
}
?>

Excellent, thanks.

After a few little changes to the code that seemed to produce the right results.
Nice one

Much appreciated.