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[SOLVED] PHP Simple question


Aravinthan
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Hi guys,

 

I have a problem and I dont know why it aint working.

 

Ok so, I have a page named step2.php

 

Which is the form that sends datas to the page named step3.php

 

Here is the code of step2.php:

 


$x = 1;

if ($receveur == 'Express Mistral')
{
$express_goals = $receveur_goals;
$express_pun = $receveur_pun;
}
else if ($visiteur == 'Express Mistral')
{
$express_goals = $visiteur_goals;
$express_pun = $visiteur_pun;
}

echo "<form action='step3.php' method='post' >";
echo "<table>";
echo "<thead>";
echo "<tr>";
echo "<th>Période</th><th>Temps</th><th>But</th><th>Assist</th><th>Assist</th><th>Extra(BN,DN,FO)</th>";
echo "</tr>";
echo "<tfoot><tr><td colspan='6'> </td></tr></tfoot>";
while ($express_goals >= $x)
{
echo "<tr><td><input type='textbox' name='periode_$x' /></td>";
echo "<td><input type='textbox' name='time_$x' /></td>";
echo "<td><select name='goal_$x' >";
$result = mysql_query("SELECT * FROM players ",$link);
			while($row = mysql_fetch_array($result))
			{
echo "<option value='" .$row['prenom']. "'>" .$row['prenom']. "</option>";
}
echo "</td>";
echo "<td><select name='assist1_$x' >";
$result = mysql_query("SELECT * FROM players ",$link);
			while($row = mysql_fetch_array($result))
			{
echo "<option value='" .$row['prenom']. "'>" .$row['prenom']. "</option>";
}
echo "<option value=''>Aucun</option>";
echo "</td>";
echo "<td><select name='assist2_$x' >";
$result = mysql_query("SELECT * FROM players ",$link);
			while($row = mysql_fetch_array($result))
			{
echo "<option value='" .$row['prenom']. "'>" .$row['prenom']. "</option>";
}
echo "<option value=''>Aucun</option>";
echo "</td>";
echo "<td><input type='textbox' name='extra_$x' /></td></tr>";
$x++;
}
echo "</table>";
echo "<input type='hidden' name='goals' value='$express_goals' />";
echo "<input type='hidden' name='pun' value='$express_pun' />";
echo "<input type='hidden' name='gameid' value='$gameid' />";
echo "<input type='submit' value='Continuer' />";
echo "</form>";

 

 

And here is the code for step3.php:

$gameid = $_POST['gameid'];
$express_goals = $_POST['goals'];
$express_pun = $_POST['pun'];

$x = 1;
while($express_goals >= $x)
{
$periode = $_POST['periode_$x'];
$time = $_POST['time_$x'];
$goal = $_POST['goal_$x'];
$assist1 = $_POST['assist1_$x'];
$assist2 = $_POST['assist2_$x'];
$extra = $_POST['extra_$x'];
mysql_query("INSERT INTO `boxscore` (periode, temps, but, assist1, assist2, extra, gameid) VALUES ('$periode', '$time', '$goal', '$assist1', '$assist2', '$extra', '$gameid')",$link);

echo "$goal";
echo "$x";
$x++;
}

 

 

But for some reason, There is no data inserted and the $goal doesnt show anything.

But the $x shows the numbers. Furthermore, When I change this to: $_POST['goal_$x']; to $_POST['goal_1'];

 

It shows the name of the player...

 

 

Any idea why $x is not working? And is there a way of inserting it?

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But the $x shows the numbers. Furthermore, When I change this to: $_POST['goal_$x']; to $_POST['goal_1'];

 

It shows the name of the player...

 

 

Any idea why $x is not working? And is there a way of inserting it?

 

Variable substitution only works in double quotes strings. Single quoted strings are evaluated literally.

 

http://dk.php.net/manual/en/language.types.string.php

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