[SOLVED] Is it possible to loop through two mysql_fetch_array funtions?

[JJohnsenDK]

Hey

Is it possible to loop through two mysql_fetch_array funtions?

In my example i want the while loop to loop through both $fet1 and $fet2, is this somehow possible?

I tried with [i]while ($fet2 = mysql_fetch_array($que2) && $fet2 = mysql_fetch_array($que2)) [/i] but this only result in that it looped through the last variable $fet2.

There most be anohter way?

[code]
<?php
include('config.php');
$sql1 = "SELECT team_id, name, home_id FROM team, game WHERE team_id = home_id";
$sql2 = "SELECT team_id, name, visitor_id FROM team, game WHERE team_id = visitor_id";

$que1 = mysql_query($sql1);
$que2 = mysql_query($sql2);

$fet1 = mysql_fetch_array($que1);

while ($fet2 = mysql_fetch_array($que2)){
echo $fet1['name']." - ".$fet2['name']."<br>";
}

?>
[/code]

[Psycho]

[quote author=JJohnsenDK link=topic=120386.msg493681#msg493681 date=1167492963]
I tried with [i]while ($fet2 = mysql_fetch_array($que2) && $fet2 = mysql_fetch_array($que2)) [/i] but this only result in that it looped through the last variable $fet2.[/quote]
Well those two statemetns are EXACTLY the same. Didn't you mean to include [b]$fet1 = mysql_fetch_array($que1)[/b] as one of the operands? The way you have it written I would expect that you would only get every other value from the second result set.

I see no reason that wouldn't work if it was properly written for both result sets. However, what exactly are you trying to do, I think a better approach would be to create a single query to return all the data you need.[/quote]

[wildteen88]

Post the databases schema for team and game tables here. And I'll be able to join those two queries into one for you. No need for two queries.

[JJohnsenDK]

Teams
-------
team_id
name
coach

Games
-------
game_id
home_id
visitor_id

[wildteen88]

Try the following query out:
[code]$sql = "SELECT t.team_id, t. name, g.home_id,  g.visitor_id FROM team AS t, game AS g
WHERE t.team_id = g.home_id AND t.team_id = g.visitor_id";[/code]
That should do it. You might want to have a read about SQL Joins. That is what I turned your query in to.

[JJohnsenDK]

I actually tried that before, but first how do i show the team name(the colum name from table team)? and second how do i show the visitor name and home name seperatly?

[code]
<?php
include('config.php');
$sql = "SELECT t.team_id, t. name, g.home_id,  g.visitor_id FROM team AS t, game AS g
WHERE t.team_id = g.home_id AND t.team_id = g.visitor_id";

$que = mysql_query($sql);

while ($fet = mysql_fetch_array($que)){
echo $fet['t.name']." - ".$fet['t.name']."<br>";
}

?>
[/code]

This doesnt work.

[wildteen88]

Your while loop is wrong.

What columns hold the team name and visitor name in your teams and games table

name is the team name and you. so you use $fet['name'] to get the team name. (note the t. and g. in the query are aliases - you don't reference them in your code. Just the actual column name).

So what coloumn holds the visitor name?

[JJohnsenDK]

The database is design in this way:

Game holds only numbers. game_id = the number of the match, home_id = the number of the home team in the team table and visitor = the number of the visitor team in the team table.

For example the first game of the season looks like this:
Game table:
game_id = 1, home_id=3 and visitor_id=1

Team table:
team_id = 1 - name = Silkeborg
team_id = 2 - name = Brøndby
team_id = 3 - name = Fc København

Then the first game of the season should look like this when it is printed to the website:
Match nr. 1  -  Fc København - Silkeborg

I hope you understand my example, else plz do write so.

[Psycho]

I think this will get you what you want:
[code]<?php
include('config.php');
$sql = "SELECT h.name as home, v.name AS visitor
          FROM game AS g
          LEFT JOIN team AS h ON g.home_id = h.team_id
          LEFT JOIN team AS v ON g.visitor_id = v.team_id";

$result = mysql_query($sql);

if (mysql_num_rows($result)==0) {
    echo "There were no results";
} else {
    $matchno = 0;
    while ($row = mysql_fetch_array($result)){
             $matchno++;
             echo "Match nr. {$matchno} - {$row['home']} - {$row['visitor']}<br>";
    }
}

>?[/code]

[JJohnsenDK]

That is what i wanted!  ;D

Im very greatfull for your, thanks alot, both of you!  :-*