using variables in $_post[] - php

Question

I have been using variable in the name part of the input tag. Now while access answers as selected by users using $_post,It gives error as undefined index.Tell me how to get answers of all questions as selected .
echo "<form method=\"post\" action=\"\">";
$query=mysql_query("select q_detail,q_id from question where category=\"$value2\"",$connection);
if(!$query)
{
echo mysql_error().'query failed';
}
$ans=1;
while($value1=mysql_fetch_array($query))
{
echo "Q-$i"." ";
echo $value1['q_detail']."<br />";
$i++;
$qno=$value1['q_id'];
$query1=mysql_query("select * from answer where ans_id=$qno");
if(!$query1)
{
echo mysql_error().'query failed';
}
while($value2=mysql_fetch_array($query1))
{
$opt=$value2['option1'];
$opt1=$value2['option2'];
$opt2=$value2['option3'];
$opt3=$value2['correct'];
echo "<input type=\"radio\" value=\"$opt\" name=\"$ans\">";
echo "<span class=\"margin\">$opt</span>";
echo "<input type=\"radio\" value=\"$opt1\" name=\"$ans\">";
echo $opt1." ";
echo "<input type=\"radio\" value=\"$opt2\" name=\"$ans\">";
echo $opt2." ";
echo "<input type=\"radio\" value=\"$opt3\" name=\"$ans\">";
echo $opt3." <br /><br />";
$ans++;
}
}
echo"<input type=\"submit\" name=\"submit\" value=\"submit\">";
echo "</form>";

Answer 1

You can make the name of input as array, e.g.:
<input type="radio" value="Blah" name="answers[]" />
and in php code you can access this using the following code:
foreach($_POST['answers[]'] as $answer)
{
echo $answer;
}

Answer 2

Enclose the non-numeric value with single quotes.
$query=mysql_query("select q_detail,q_id
from question where category='$value2'");

Answer 3

Try this:
$ans=1;
if(isset($_POST))
{
echo $_POST['radio_' . $ans];
}
echo "<form method=\"post\" action=\"\">";
$query=mysql_query("select q_detail,q_id from question where category=\"$value2\"",$connection);
if(!$query)
{
echo mysql_error().'query failed';
}
while($value1=mysql_fetch_array($query))
{
echo "Q-$i"." ";
echo $value1['q_detail']."<br />";
$i++;
$qno=$value1['q_id'];
$query1=mysql_query("select * from answer where ans_id=$qno");
if(!$query1)
{
echo mysql_error().'query failed';
}
while($value2=mysql_fetch_array($query1))
{
$opt=$value2['option1'];
$opt1=$value2['option2'];
$opt2=$value2['option3'];
$opt3=$value2['correct'];
echo "<input type=\"radio\" value=\"$opt\" name=\"radio_$ans\">";
echo "<span class=\"margin\">$opt</span>";
echo "<input type=\"radio\" value=\"$opt1\" name=\"radio_$ans\">";
echo $opt1." ";
echo "<input type=\"radio\" value=\"$opt2\" name=\"radio_$ans\">";
echo $opt2." ";
echo "<input type=\"radio\" value=\"$opt3\" name=\"radio_$ans\">";
echo $opt3." <br /><br />";
$ans++;
}
}
echo"<input type=\"submit\" name=\"submit\" value=\"submit\">";
echo "</form>";
I've added the if(isset($_POST)) and before the name I've added some text, because a name as a integer, can be used as a integer index in PHP. Array's can be called like
$ar = array("Name" => "VALUE");
echo $ar[0]; // Outputs VALUE
echo $ar["Name"]; // Outputs VALUE
Will both output "VALUE". So what you're trying to do. Is getting the value of index 1 instead of key 1.

Answer 4

You have to use $_POST['actual radiobutton name'], i. e. whatever the content of $ans is at the time of the echoing.

Answer 5

You can also debug your form after submitting by using this code:
echo "<pre>";
print_r($_POST);
echo "</pre>";
This will show you the contents of the $_POST data so you can see if values are being passed correctly.