Certain field wont insert into the MySQL

I have date of birth on my register form. The day, month, and year are seperate fields on MySQL. So I have used the select input for visitors to choose their D.O.B but it only enters the month and year into the users table, the day stays at 0.

The fields for the days are all numbers and are INT(11)

Here is the HTML for the D.O.B (I wont paste all the years onto here as it will take up too much room)

Date of Birth:
<select name="month" class="input">
	<option value="0">Day</option>
	<option value="1">1</option>
	<option value="2">2</option>
	<option value="3">3</option>
	<option value="4">4</option>
	<option value="5">5</option>
	<option value="6">6</option>
	<option value="7">7</option>
	<option value="8">8</option>
	<option value="9">9</option>
	<option value="10">10</option>
	<option value="11">11</option>
	<option value="12">12</option>
	<option value="13">13</option>
	<option value="14">14</option>
	<option value="15">15</option>
	<option value="16">16</option>
	<option value="17">17</option>
	<option value="18">18</option>
	<option value="19">19</option>
	<option value="20">20</option>
	<option value="21">21</option>
	<option value="22">22</option>
	<option value="23">23</option>
	<option value="24">24</option>
	<option value="25">25</option>
	<option value="26">26</option>
	<option value="27">27</option>
	<option value="28">28</option>
	<option value="29">29</option>
	<option value="30">30</option>
	<option value="31">31</option>
</select>

<select name="month" class="input">
	<option value="0">Month</option>
	<option value="1">January</option>
	<option value="2">Febuary</option>
	<option value="3">March</option>
	<option value="4">April</option>
	<option value="5">May</option>
	<option value="6">June</option>
	<option value="7">July</option>
	<option value="8">August</option>
	<option value="9">September</option>
	<option value="10">October</option>
	<option value="11">November</option>
	<option value="12">December</option>
</select>

<select name="year" class="input">
	<option value="0">Year</option>
	<option value="1998">1998</option>
	<option value="1997">1997</option>
	<option value="1996">1996</option>
	<option value="1995">1995</option>
	<option value="1994">1994</option>
	<option value="1993">1993</option>
	<option value="1992">1992</option>
	<option value="1991">1991</option>
	<option value="1990">1990</option>
</select>

HTML:

Here is the php bits that are used for D.O.B (Once again i've not added the things on this forum that don't need to be shown)

$day=$_POST['day'];
$month=$_POST['month'];
$year=$_POST['year'];

if($day == "0"){
die("You must select your Date of Birth");
}if($month == "0"){
die("You must select your Date of Birth");
}if($year == "0"){
die("You must select your Date of Birth");
}

mysql_query("INSERT INTO `users` ( `id` , `username` , `password` , `day` , `month` , `year` )
VALUES (
'', '$reg_username', '$reg_password', '$day', '$month', '$year'
)");

PHP:

 

In your HTML:

You have 2 month select tags, theirfore change the first:

<select name="month" class="input">
    <option value="0">Day</option>

HTML:

to:

<select name="day" class="input">
    <option value="0">Day</option>

HTML:

Also replace your PHP with:

//prevent sql parsing issues aswell as sql injection...
$day   = mysql_real_escape_string($_POST['day']);
$month = mysql_real_escape_string($_POST['month']);
$year  = mysql_real_escape_string($_POST['year']);

//simplified...
if ($day == 0 || $month == 0 || $year == 0) {
    die('You must select your Date of Birth');
}

mysql_query("INSERT INTO users ('username', 'password', 'day', 'month', 'year')
VALUES ('{$reg_username}', '{$reg_password}', '{$day}', '{$month}', '{$year}')") or trigger_error(mysql_error(), E_USER_ERROR);

PHP:

Let me know how that goes, also report back quoting any errors that may be apparent (if any)

 

Just insert with raw query to conform is ur query is right or not
INSERT INTO `users` ( `id` , `username` , `password` , `day` , `month` , `year` ) VALUES ('', 'test', 'test', '1', 'May', '2010')

 

you have duplicate month, and forgot to add day on your form...

 

good catch

 

Instead of escaping the dates, I'd rather (int)'em.

$year = (int) $_POST['year'];
$month = (int) $_POST['month'];
$day = (int) $_POST['day'];

PHP:

And to push date verification a little further, I suggest:

if (!checkdate($month, $day, $year))
    die('Invalid date');

PHP:

On a last note, I'm not sure what you're planning to do with this info, but is there a specific reason you're storing the birthday like this? Wouldn't it be better to keep it all in one field?

 

Thanks, this worked.

It was 'day' before, but when i copied the <Select> from month (because even before that it was not workin) i forgot to change the name. but its working now so thanks and thanks everyone else for your replies

 

No problem, but I don't deserve the credit so props to gapz101, as I didn't catch that, until gapz101 mentioned it.

You should also consider taking alook at nico_swd's post on validation

 

Sorry I didn't see this before.
Thanks I will try that later on.

I was going to put them all in one field but I didn't know how to do it so it would update to one field, and also I didn't know how to do it so it would display as a date and display their age

 

Might want to consider storing the year, month, and day as one column entry with a DATE or TIMESTAMP type. This would prevent the need for three seperate columns and would simply all the processing.