plissss help?

[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>

           

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

</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"); 

 

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

 

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

 

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

?>

 

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..

 

[PFMaBiSmAd]

Don't double post the same question.