How to insert data into db from form with dynamic radio button..?

Question

I am trying to insert data from form with dynamically generated radio button for quiz page.I cant access the data posted by form with name as array.Please help me to do that.If you find any error it must be big help.Thankyou

        <?php
        $rs=mysql_query("select * from question where testid=$tid order by quesid ",$cn) or die(mysql_error()); 
        $n=0;
        while($row= mysql_fetch_row($rs)){?>
        <form name=myfm method=post action=Quiz.php>
        <table width=100%> <tr> <td width=30><td></td></td></tr> <table border=0>
        <?php $n=$n+1; ?>
        <tr><td>Question  <?php echo $n." "; echo $row[2]; ?></td></tr>
        <tr><td class=style8><input type="radio" name="ques['<?php echo $n; ?>'][]" value=1><?php echo $row[3]; ?></td></tr>
        <tr><td class=style8> <input type="radio" name="ques['<?php echo $n; ?>'][]" value=2><?php echo $row[4];?></td></tr>
        <tr><td class=style8><input type="radio" name="ques['<?php echo $n; ?>'][]"  value=3><?php echo $row[5];?></td></tr>
        <tr><td class=style8><input type="radio" name="ques['<?php echo $n; ?>'][]"  value=4><?php echo $row[6];?></td></tr>

    <?php 
        }
        echo "<tr><td><input type=submit name=submit id='result' value='Get Result'></form>";
        ?>
        </table></table>
        </form>

data inserting page

        <?php 
        $rs=mysql_query("select * from question where testid=$tid",$cn) or die(mysql_error());
        if($submit=='Get Result')
        { $n=0;
        while($row= mysql_fetch_row($rs)){
        for($i=0;$i<count($_POST['ques']);$i++)
        {$n=$n+1;
        $ans=$_POST['ques'][$n][$i];
        mysql_query("insert into useranswer(sessid, testid, ques, ans1,ans2,ans3,ans4,correctans,yourans) values ('".session_id()."',       $tid,'$row[2]','$row[3]','$row[4]','$row[5]', '$row[6]','$row[7]','$ans')") or die(mysql_error());
        }
        ?>

use ini_set('error_reporting', E_ALL); error_reporting(E_ALL); at top of the file. — devpro
– devpro, Commented Feb 10, 2016 at 6:14
also shows "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 1" — Siby Xavier
– Siby Xavier, Commented Feb 10, 2016 at 6:27

Codingalien · Accepted Answer · 2016-02-10 06:44:42Z

1

You have a variable $tid. Then it should be like this.

$rs=mysql_query("select * from question where testid=".$tid,$cn)

And don't forget to add mysql_real_escape_string in your queries.

mysql_real_escape_string — Escapes special characters in a string for use in an SQL statement.But this extension was deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. Instead, the MySQLi or PDO_MySQL extension should be used.

UPDATE :::

mysql_real_escape_string

 // We have not connected to MySQL

 $lastname  = "O'Reilly";
 $_lastname = mysql_real_escape_string($lastname);

 $query = "SELECT * FROM actors WHERE last_name = '$_lastname'";

UPDATE 2::

$tid = mysql_real_escape_string($tid);
$ans = mysql_real_escape_string($ans);

mysql_query("insert into useranswer(sessid, testid, ques, ans1,ans2,ans3,ans4,correctans,yourans) values ('".session_id()."',       '". $tid ."','$row[2]','$row[3]','$row[4]','$row[5]', '$row[6]','$row[7]','". $ans .'")") or die(mysql_error());

edited Feb 10, 2016 at 6:44

answered Feb 10, 2016 at 6:25

Codingalien

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5 Comments

Codingalien Over a year ago

$lastname = "O'Reilly"; $_lastname = mysql_real_escape_string($lastname); $query = "SELECT * FROM actors WHERE last_name = '$_lastname'";

Siby Xavier Over a year ago

nope.error shows"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 1"

Codingalien Over a year ago

please check you insert query

devpro Over a year ago

i think its start from $n=1; not from 0 @SibyXavier

Codingalien Over a year ago

just print the $ans, and see what is the output.

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