it does not submitted to database

[chiaki*misu]

 

hello everyone..

 

i am using php script, n the input data is to be filled in database-sqlyog..i've question,the database is just for storing data right? so i just create tables in sqlyog. as for the server connection, apache and mysql, i use xampp.

 

i code:

 

<form name="noledge" action="sectionc.php" method="post">

            <td><input type="radio" name="rad1" value="1"/></td>

            <td><input type="radio" name="rad1" value="2"/></td>

            <td><input type="radio" name="rad1" value="3"/></td>

            <td><input type="radio" name="rad1" value="4"/></td>

            <td><input type="radio" name="rad1" value="5"/></td>

            <td><input type="radio" name="rad1" value="6"/></td>

            <td><input type="radio" name="rad1" value="7"/></td>

            <td><input type="radio" name="rad1" value="8"/></td>

            <td><input type="radio" name="rad1" value="9"/></td>

            <td><input type="radio" name="rad1" value="10"/></td>

            </form>

 

then..in sectionc.php

 

<?php

 

session_start();

$user="root";

$host="localhost";

$password="password";

$database="borang";

 

$link= mysql_connect($host,$user,$password) or die ("unable to connect to server");

 

$db = mysql_select_db($database,$link) or die ("unable to locate database");

 

$code=mysql_set_charset($link,'utf8') or die ("unable to set connection coding");

 

$sql="INSERT INTO section_c (C1) VALUES ($_REQUEST[ $rad1[0] ]);

 

$result=mysql_query($sql) or die("unable to execute query");

?>

 

i suppose that the value of 1 will be set in column C1 in the table section_c, correct? it does not, so..what did i miss here?

 

:(Onegai shimasu..

[ohdang888]

do this:

 

$value = $_REQUEST[$rad1[0]];

$sql="INSERT INTO section_c (`C1`) VALUES ('$value') ";

 

[chiaki*misu]

thanks for the reply..

ive tried that..still dont work..or it is about the database?

[PFMaBiSmAd]

The form data will be in $_POST['rad1']

 

To see what you are actually getting from the form, use the following debugging code -

 

<?php
echo "<pre>";
echo "POST:";
print_r($_POST);
echo "</pre>";
?>

[chiaki*misu]

so what i should do is:

 

<?php

 

session_start();

include("bore.inc");

include("connection.inc");

include("debug.inc");

//i've saved the code given in an .inc file

 

$value = $_POST['rad1'];

 

$sql="INSERT INTO section_c (`C1`) VALUES ('$value') ";

 

mysql_query($sql) or die(mysql_error());

?>

 

then..where will the debug file notify me the value selected from the $_POST?

[chiaki*misu]

:'(

plisss..anyone..?