Loop Problems

[Kerant]

Hi there,

I'm in the middle of developing a football predictions website, but I am having trouble forming the predictions page.

Basically I want an admin to be able to enter a numeric value ('4' for instance), and then a series of drop down boxes will appear depending on what the value was that they wanted.

For example, there are 4 football fixtures coming up. The admin enters '4' for the number of fixtures he wants to add to the site. The next page shows drop down boxes for each of the 4 fixtures, so they can select which team is playing which other team. It's hard to explain, sorry.

So far, I have tried a while loop, but had the problem of it getting stuck in a continuous loop.

The code I want to output so far is basically this....

		
<p id="favteam">Select Team 1: <br />

	<?php 
		$sql="SELECT * FROM teams ORDER BY team_name;";
		$result=mysql_query($sql) or die(mysql_error());
	?>

	<select name="TeamName1" id="TeamDropdown1" tabindex=1>

		<?php while ($getTeams=mysql_fetch_assoc($result)) { ?>

			<option value="<?php echo $getTeams['id'] ?>"><?php echo $getTeams['team_name']; ?></option>

		<?php } ?>

	</select>
	</p>

This is the basic code so far, but as you can see there is no loop involved.

Could somebody help with this one please?

I am just looking for a simple (I presume) loop to output drop down boxes based on a number (entered by the user).

Thanks in advance for taking time to understand.

[paulnaj]

Hi,

Is this the sort of thing you mean ...

<?php
$sql = "SELECT * FROM teams ORDER BY team_name;";
$result = mysql_query($sql) or die(mysql_error());
$options = '';
while($getTeams = mysql_fetch_assoc($result)){
	$options .= '<option value="'.$getTeams['id'].'">'.$getTeams['team_name'].'</option>';
} 

for($i = 1; $i <= 4; $i++){
?>
<p id="favteam">Select Team <?php echo $i; ?>: <br />
	<select name="TeamName<?php echo $i; ?>" id="TeamDropdown<?php echo $i; ?>" tabindex="<?php echo $i; ?>">
	<?php echo $options; ?>    

	</select>
</p>
<?php
}
?>

P.

[Kerant]

Absolutely spot on.

Thanks very much!

[djjjozsi]

Hy,

Save the dropdown (options ) into a string, then use a for cycle to print it using a different variable name. However, you can use an array[] name for a dropdown, then you can loop through when you process it..

[Kerant]

Just a quick queston, paulnaj.

That code you provided works like a dream, but how would I specifically get the posted variables once they have been posted by the form?

For example, this part of the code...

<select name="TeamName<?php echo $i; ?>"

Will always be different due to the variable, whereas normally it would remain static and I could get the value by using something such as...

$team1 = $_POST['TeamName1'];

Thanks for your help.

[paulnaj]

Hi again,

How about ...

<select name="<?php echo $_POST['TeamName'.$i]; ?>"

... assuming you have four $_POST variables named TeamName1, TeamName2 etc.

P.

[Kerant]

Well the user will select how many fixtures to add on a previous page, so the number could be different every time, therefore there is no guarantee it would be 4. :quiet:

[paulnaj]

Ah ... misunderstood you.

If the user has $_POSTed the number of dropdowns on a previous form, then you can use this variable to set the upper bound on the for loop. You could re-include this number in the form with the dropdowns (as a hidden field for example) and again use it to know how many teams have been chosen for the next part of your app.

P.

[Kerant]

paulnaj;10902197 wrote:

Ah ... misunderstood you.

If the user has $_POSTed the number of dropdowns on a previous form, then you can use this variable to set the upper bound on the for loop. You could re-include this number in the form with the dropdowns (as a hidden field for example) and again use it to know how many teams have been chosen for the next part of your app.

P.

Yeah thanks for that, it's what I had presumed.

I've just tried doing the code to pickup the POSTed variables but it's not working, and i just get the "undefined variable" warning in a constant loop.

Please can you tell me if I am on the right lines with this code...

$num = $_POST['num'];

for($i = 1; $i <= $num; $i++){

$team1fixture.$i = trim($_POST['Fixture'.$i.'Team1']);
$team2fixture.$i = trim($_POST['Fixture'.$i.'Team2']);
} 

Thanks a lot.

Edit - I altered the code you provided to look like this...

<select name="Fixture<?php echo $i; ?>Team1"

[djjjozsi]

You think of this?

${"team1fixture$i"}= .........

however, did you read my suggestions? 🙂 You don't need to create field names such as:

team1
team2

use an ARRAY as variable name...

<input name="variable[]" value="">
<input name="variable[]" value="">
<input name="variable[]" value="">
<input name="variable[]" value="">

This will appears in your process program, like:

$i=0;
foreach($_POST["variable"] AS $options)
{
echo "$i:  $options<br />";
$i++;
}

[Kerant]

Yeah thanks for your suggestions djjjozsi, but I'm really clueless when it comes to arrays so I would prefer to do it the way I have been trying (if possible 😕).

Is there any other suggestions you could provide me with that may fix my code? At the moment it just gets stuck in a never-ending loop of undefined variables, and I can't figure out why.

Thanks.

[djjjozsi]

$howmany=4;

for($i = 1; $i <= $howmany; $i++)
{
   if(  empty($_POST["TeamName".$i])  )
      print "Team $i is empty <br />";
   else
      print "Team $i  is " .   $_POST["TeamName".$i] ."<br />";
}

but this is the worst method you can use...

[Kerant]

Ah! Thanks very much for both your help with this. It is resolved now using this code:

$num = $_POST['num'];

for($i = 1; $i <= $num; $i++){

$number1[$i] = trim($_POST["Fixture".$i."Team1"]);
$number2[$i] = trim($_POST["Fixture".$i."Team2"]);

$sql = "INSERT INTO fixtures SET
		team_1 = '$number1[$i]',
		team_2 = '$number2[$i]'";

$result = mysql_query($sql) or die(mysql_error());

} 

Thanks again 🙂

[djjjozsi]

and Fixture is a string ? If yes, then use mysql_real_escape_string to protect the database from SQL inection.

[Kerant]

djjjozsi;10902263 wrote:

and Fixture is a string ? If yes, then use mysql_real_escape_string to protect the database from SQL inection.

Yeah it is.

Thanks for reminding me.

[Weedpacket]

Kerant wrote:

but I'm really clueless when it comes to arrays

Think of it as being able to count higher than one 🙂

[Kerant]

I'm back, with a similar problem :queasy:

I'm in the situation where I can now successfully add fixtures to the database, but I am stuck on the stage of displaying the fixtures.

I got to the point where I could pull the fixtures from the database to display them for a user, but then realised that input boxes need to be associated with the fixtures (in order for users to predict the score for each fixture). I tried creating a for loop to make each input box have a unique id, which has worked, but now the fixtures wont display properly.

The basic problem I'm having is that the loop is displaying only the first fixture from the database, but multiple times.

The code has become a bit long, but here goes...

	<?php		
		$sql = "SELECT team_1, team_2, fixture_date, date_added, gameweek, active FROM fixtures";
		$result = mysql_query($sql) or die(mysql_error());

	while ($i = mysql_fetch_array($result)) {
		$team1 = $i['team_1'];

			$getTeamName = "SELECT team_name, league_image 
							FROM teams, fixtures 
							WHERE teams.id = fixtures.team_1
							AND teams.id = $team1";
			$teamName = mysql_query($getTeamName) or die(mysql_error());

			while ($getTeam = mysql_fetch_array($teamName)) {
				$team1Name = $getTeam['team_name'];
				$image1 = $getTeam['league_image'];
			}

		$team2 = $i['team_2'];

			$getTeamName = "SELECT team_name, league_image 
							FROM teams, fixtures 
							WHERE teams.id = fixtures.team_2
							AND teams.id = $team2";
			$teamName = mysql_query($getTeamName) or die(mysql_error());

			while ($getTeam = mysql_fetch_array($teamName)) {
				$team2Name = $getTeam['team_name'];
				$image2 = $getTeam['league_image'];
			}

		$fixtureDate = $i['fixture_date'];
		$dateAdded = $i['date_added'];
		$gameweek = $i['gameweek'];
		$active = $i['active'];

		$sql = "SELECT num_fixtures FROM fixture_details";
		$result = mysql_query($sql) or die(mysql_error());
		while ($i = mysql_fetch_array($result)) {
			$num = $i['num_fixtures'];
		}

		for($i = 1; $i <= $num; $i++){
?>

		<tr>
			<th style="width: 160px"><p align="right"><?php echo $team1Name; ?></p></th>
			<td><img src="images/badge_icons/<?php echo $image1; ?>"></td>
			<td><input type="text" id="Fixture<?php echo $i; ?>Team1" name="date" style="width: 20px;" maxlength="10" /></td>
			<td><p>v</p></td>
			<td><input type="text" id="Fixture<?php echo $i; ?>Team2" name="date" style="width: 20px;" maxlength="10" /></td>
			<td><img src="images/badge_icons/<?php echo $image2; ?>"></td>
			<th style="width: 160px"><p align="left"><?php echo $team2Name; ?></p></th>		
		</tr>

<?php } } ?>

The input id is unique with this code, but only the first fixture from the database is displayed (but multiple times, depending on how many fixtures are present in the database).

Thanks very much for all the help so far, and I appreciate you taking time to look at this next problem.

[Weedpacket]

            while ($i = mysql_fetch_array($result)) {
                $num = $i['num_fixtures'];
            }

This isn't doing anything useful: it's getting the "num_fixtures" from the last record in $result - which is just a list of all the num_fixture values in fixture_details regardless of order.

for($i = 1; $i <= $num; $i++){

In this loop you're just using the same $team1Name etc. over and over again.

You're also using $i as the counter variable in all your loops - including the nested ones. They're going to keep overwriting each other.

Looping over a query using results from another query is a bad smell, too: if you've got twenty fixtures you've just made 41 queries. You could have done it in one.

$sql = "Select
	t1.team_name as home_name, t1.league_image as home_league, t2.team_name as away_name, t2.league_image as away_league,
	fixture_date, date_added, gameweek, active
from
	teams t1, teams t2, fixtures
Where
	t1.id = fixtures.team_1
and
	t2.id = fixtures.team_2";

That will get you the team name and league image of both team 1 and team 2 in each fixture, as well as the schedule information (I took the liberty of using "home" and "away" in the aliases instead of "team_1" and "team_2").

[Kerant]

Weedpacket;10902437 wrote:

This isn't doing anything useful: it's getting the "num_fixtures" from the last record in $result - which is just a list of all the num_fixture values in fixture_details regardless of order.

That is just purely getting the number of fixtures there are, to be able to form the next loop (because $num is used as the maximum number of loops to do).

Weedpacket;10902437 wrote:

In this loop you're just using the same $team1Name etc. over and over again.

Yeah, that is my main problem - but do you know how I can solve this?

Weedpacket;10902437 wrote:

You're also using $i as the counter variable in all your loops - including the nested ones. They're going to keep overwriting each other.

I tried changing them to different names for each loop, but I still have the same problem.

Weedpacket;10902437 wrote:

Looping over a query using results from another query is a bad smell, too: if you've got twenty fixtures you've just made 41 queries. You could have done it in one.

That will get you the team name and league image of both team 1 and team 2 in each fixture, as well as the schedule information (I took the liberty of using "home" and "away" in the aliases instead of "team_1" and "team_2").

That query wouldn't work as the tables are set out differently.

The tables are as follows:

teams (containing list of 20 team names & image)
fixtures (containing columns 'team_1' and 'team_2')
fixture_details (containing the number of fixtures added, everytime a new set is added)

I tried creating a single query to get the team name and image based on the team id from the fixtures table, but I could only get it to work with either team_1 or team_2, and not both.

The only way I could think of getting it to work in a single query would be to have 2 tables of teams. :queasy:

[cyberlew15]

man I know I am coming in late on this but you seem to need a lot of help on this. don't suppose you can dump your database and script up here as files then I'll have a quick look. I must say I think in future when you need this much help possibly consider using a professional to write the script for you. or even consult php.net/manual/en

Also does Preston train station still have that cool little cheap book shop? I know it's off the subject but I was just wondering.