Help my codes

[itohideo]

I try to run this codes but just only sex is not coming. I input name, japanese , sex,but sex doesn't come to my database. I did so many times. But still it is not coming only sex. Please help again.

Thank you for helping before


<?php
$username = "itohideo";
$password = "1234";
$host = "localhost";

$dbcon = mysql_connect($host, $username, $password);
$database = mysql_select_db("itohideo");

if (isset($_POST["submit"])) {
$name = $_POST["name"];
$japanese = $_POST["japanese"];
$sex = $_POST["sex"];


$query_insert = "insert into japan (name, japanese, sex) values ('$name', '$japanese', '$sex')";
$query_result_insert = mysql_query($query_insert, $dbcon) or die (mysql_error($dbcon));

$query = "select * from japan";
$query_result = mysql_query($query, $dbcon);

$num = mysql_num_rows($query_result);
print "<table border=\"1\">";
while ($row = mysql_fetch_array($query_result)) {
print "<tr>";
print "<td><b>".$row[0]. "</b> </td><td><b>".$row[1]."</b><br></td>";
print "<td><b>".$row[2]. "</b> </td><td><b>".$row[3]."</b><br></td>";
print "<td><b>".$row[4]. "</b> </td><td> <b>".$row[5]."</b><br></td>";
print "</tr>";
}
print "</table>";
print "there are $num result in the table Japan";
exit;
}
?>

<form method="post" action="dc1.php">
<input name="name" type="text">
<input name="japanese" type="text">
<input name="sex" type="text">

<input type="submit" name="submit">



</form>

[itohideo]

My php is 4.2 and

my php says here is wrong

$sex = $_POST["sex"]; I don't know why

[Husain]

Can you please post the EXACT error message that PHP is giving you. It would help others understand the problem better. Thanks

[johnn]

What is the error message?

Anyway, you should not placed these codes at top of your script

$username = "itohideo"; 
$password = "1234"; 
$host = "localhost"; 

$dbcon = mysql_connect($host, $username, $password); 
$database = mysql_select_db("itohideo"); 

You place them where just before you're ready to insert or select