All zero values passed in the table - php

Question

When I execute this code, I only get zeroes inserted into my database. Why? What can I do to solve this problem?
<select class="date" name="year"> <option value="">Year</option>
<?php $i=1; while($i<=31) { ?>
<option value="<?php $i ?>"> <? echo $i; ?> </option>
<?php $i++; } ?> </select>
And this is my mysql code:
$sql="INSERT INTO information ( Year )
VALUES ('$_POST[year]')";

Answer 1

here how it will be your code
<select class="date" name="year">
<option value="">Year</option>
<?php for($i=1;$i<=31 ;$i++ ) { ?>
<option value="<?php $i ?>"> <? echo $i; ?> </option>
<?php } ?>
</select>
EDIT :
<input type ="submit" value =" submit me"/>
<?php if (isset($_POST['year'])) // check if isset the year
{
$year = $_POST['year'] ;
} else {}
$sql="INSERT INTO information ( Year ) VALUES ('".$year."')";
?>
edit 2 :
your code is right but you should escape value to prevent sql injection
do like that
insert into ....
VALUES
('mysql_real_escape_string($_POST[user])','mysql_real_escape_string($_POST[email])'.........

Answer 2

<select class="date" name="day"> <option value="">Day</option>
<?php $i=1; while($i<=31) { ?>
<option value="<?php echo $i; ?>"> <?php echo $i; ?> </option>
<?php $i++; } ?> </select>
Always use <?php and ?> so your browser recognizes PHP
also, with your first <?php $i ?>
I changed this to echo $i; and on the same line I changed <? echo $i; ?> </option> To <?php echo $i; ?> </option>
and it worked for me.
With Your Insert:
$sql="INSERT INTO information (COL_FOR_DAY)
VALUES ('{$_POST['day']}')";