how get Date of birth value to database

Question

0.00/5 (No votes)

See more:

how i can post this date of birth function value to my database through my registration page? Please help !! Any other suggestions, i need three drop down in my php form.

XML

<?
function date_dropdown($year_limit = 0){
        $html_output = '    <div id="date_select" >'."\n";
        $html_output .= '        <label for="date_day"></label>'."\n";

        /*days*/
        $html_output .= '           <select name="date_day" id="day_select">'."\n";
            for ($day = 1; $day <= 31; $day++) {
                $html_output .= '               <option>' . $day . '</option>'."\n";
            }
        $html_output .= '           </select>'."\n";

        /*months*/
        $html_output .= '           <select name="date_month" id="month_select" >'."\n";
        $months = array("", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December");
            for ($month = 1; $month <= 12; $month++) {
                $html_output .= '               <option value="' . $month . '">' . $months[$month] . '</option>'."\n";
            }
        $html_output .= '           </select>'."\n";

        /*years*/
        $html_output .= '           <select name="date_year" id="year_select">'."\n";
            for ($year = 1900; $year <= (date("Y") - $year_limit); $year++) {
                $html_output .= '               <option>' . $year . '</option>'."\n";
            }
        $html_output .= '           </select>'."\n";

        $html_output .= '   </div>'."\n";
    return $html_output;
}


?>


html form code:
 <td>			// or add date of birth age condition in ()
							 echo date_dropdown();
							 		?>
			</td>

Posted 30-Mar-12 7:53am

Peta2010

Updated 4-Apr-12 4:31am

ZurdoDev

v4

Add a Solution

Comments

Peta2010 30-Mar-12 14:01pm

This is date of birth function for registration page. How i can add its selected value to database ?

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

dieforwhat · Accepted Answer · 2012-04-04T12:07:00

In your action page ;
first get date of birth variables.depending on your method $_GET['date_year'] or $_POST['date_year'] etc.Then call mysql_real_escape_string to make them safe for database.($safeYear=mysql_real_escape_string($_GET['date_year']).
Then depending on your database schema call
mysql_query ("insert into youtTable values(....,'$safeDay.$safeMonth.$safeYear',....);
This is the simplest case .If your table has dateTime field than you should pass birth day after formatting obviosuly.By the way I'm assuming you are using your supplied code inside

<form>...</form>

block.

(Sorry for this barbarian answer at 2.00 a.m.)

Peta2010 · Accepted Answer · 2012-04-04T23:03:00

Solution 2

I used three drop downs in the html form and then pass all drop down values in variables and then in php i use cancatenation to make all three values together.

Its working.

$dob = $_POST['day'] . '-' . $_POST['month'] . '-' . $_POST['year'];

Posted 4-Apr-12 23:03pm