Pass a dynamic variable through URL php

Question

I'm not sure about the title, I tried my best. I have a table displayed with information from a database using this file

display.php

<?php

    mysql_connect("localhost", "root", "root") or die(mysql_error());
    mysql_select_db("tournaments") or die(mysql_error());

    $result = mysql_query("SELECT * FROM tournies") 
    or die(mysql_error());  

    echo '<table id="bets" class="tablesorter" cellspacing="0" summary="Datapass">
<thead>
    <tr>
        <th>Tournament <br> Name</th> 
        <th>Pot</th> 
        <th>Maximum <br> Players</th> 
        <th>Minimum <br> Players</th> 
        <th>Host</th> 
        <th></th>
        <th></th>
    </tr>
</thead>
<tbody>';



    while($row = mysql_fetch_array( $result )) {

       $i=0; if( $i % 2 == 0 ) {
            $class = "";
        } else {
            $class = "";
        }

        echo "<tr" . $class . "><td>"; 
        echo $row['tour_name'];
        $tour_id = $row['tour_name'];
        echo "</td><td>"; 
        echo $row['pot']," Tokens";
        echo "</td><td class=\"BR\">"; 
        echo $row['max_players']," Players";
        echo "</td><td class=\"BR\">"; 
        echo $row['min_players']," Players";
        echo "</td><td class=\"BR\">";
        echo $row['host'];
        echo "</td><td>";
        echo "<input id=\"delete_button\" type=\"button\" value=\"Delete Row\" onClick=\"SomeDeleteRowFunction(this)\">";
        echo "</td><td>";
        echo "<form action=\"join.php?name=$name\" method=\"POST\" >";
        echo "<input id=\"join_button\" type=\"submit\" value=\"Join\">";
        echo "</td></tr>"; 
    } 

    echo "</tbody></table>";
?>

Basically I want the user to press a button from a row of the table and they go to a new page called join.php. I need the persons username and the name of the tournament from the row the clicked.

For example here's my page: my page

When they click the join button at the end of row one it should send them to

'join.php?name=thierusernamehere&tourname=dfgdds'

Any help much appreciated. Thanks.

Just add a hidden field (echo "<input name=\"name\" type=\"hidden\" value=\"".$name."\">"; for example) to the form with the username & tourname. — Kep
– Kep, Commented Sep 4, 2013 at 0:53
PSA: The mysql_* functions are deprecated in PHP 5.5. It is not recommended for writing new code as it will prevent you from upgrading in the future. Instead, use either MySQLi or PDO and be a better PHP Developer. — Jason McCreary
– Jason McCreary, Commented Sep 4, 2013 at 0:55

ManreeRist · Accepted Answer · 2013-09-04 00:58:55Z

2

echo '<td><a href="join.php?name='.$row['name'].'&tourname='.$row['tour_name'].'">Join</a></td>'

answered Sep 4, 2013 at 0:58

ManreeRist

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2 Comments

Phil Over a year ago

name doesn't appear to be part of the resultset.

user2288508 Over a year ago

@AndreeCrist Just had to fix a few naming errors and it worked like a charm, thanks.

CodinCat · Accepted Answer · 2013-09-04 01:05:11Z

There are many way to approach.

The easiest way is just echo '<a href="join.php?name=' . $row['col_name'] . '">JOIN</a>';

or you can use a form with hidden input and submit button.

BUT

Your code is really a mess, try to make your code more maintainable and readable. And do NOT use any mysql_* functions, they are deprecated.

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