chiaki*misu Posted August 13, 2009 Share Posted August 13, 2009 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.. Link to comment https://forums.phpfreaks.com/topic/170026-plissss-help/ Share on other sites More sharing options...
PFMaBiSmAd Posted August 13, 2009 Share Posted August 13, 2009 Don't double post the same question. Link to comment https://forums.phpfreaks.com/topic/170026-plissss-help/#findComment-896941 Share on other sites More sharing options...
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