PHP, mySQL and Form Menu's

[Anon]

What I am doing:

function getTeamNames()
{
if($link_id = mysql_connect(\"192.168.1.3\", \"PhantomStatsUser\", \"PhantomS
tatsUser\"))
{
$select = mysql_select_db(\"PhantomStats\");
$sql = \'SELECT TeamName FROM Teams\';
if($result = mysql_query($sql))
{
$x = 0;
while ($query_data = mysql_fetch_array($result))
{
$x++;
echo \"<option value=\'\" + $x + \"\'>\" + $query_data[\'TeamName\'];
}
}
}
mysql_close($link_id);
}

the function is being called as such:
<form>
<select name=\"HockeyTeam\">
<?php
getTeamNames();
?>
</select>
</form>

When I view the source of what is displayed I get this:
<form>
<select name=\"HockeyTeam\">
12345678910</select></form>

Definitely not the indended result. Idealy the end result will be such that the team names will be within the forms menu for the user to select.

As always help is always appreciated. Thank you in advance

Shaun Bramley

[Anon]

echo \"<option value=\'\" + $x + \"\'>\" + $query_data[\'TeamName\'];

that line is pretty... messy.

print("<option value=\"$x\">$query_data['TeamName']</option>");

of course, you could use echo. Just remember you don't have to escape the quotes surrounding the output you want to echo, and you have to close your options 🙂

[Anon]

<?
function getTeamNames()
{
$sql = "SELECT TeamName FROM Teams";
if($result = mysql_query($sql))
{
$x = 0;
while ($query_data = mysql_fetch_array($result))
{
$x++;
echo "<option value=$x>" .$query_data["value"];
}
}
}

?>

<form>
<select name="HockeyTeam">
<?php
if ($GetTeams)
{
getTeamNames();
}
?>
</select>
</form>

Something to play with...