--- In [email protected], "siham_bakashwain"
<[EMAIL PROTECTED]> wrote:
>
> hi all
> i have in my form three <option>
> day , month and year
> and i want combine the three and insert it into one field with type
> date how i can make this ??
> please tell me
> thanks
>
Hello,
This is a very simple example, but it should do what you need. You
will need to enter all of the month values (1-12) into the "month"
select menu, all of the day values (1-31) into the "day" select menu,
and any "year(s)" you wish to display in the "year" select menu. For
this example, I am using 11/19/2007 as the 'date string' combination
the user selected from the select menus.
Notice in this example, the form below will post back to itself using
$_SERVER['PHP_SELF'].
NOTE: You do not need to set the field in your database table where
you wish to store the date to a DATE field type. Instead, use VARCHAR
and make the field length in your database table 10 characters or
VARCHAR(10). Do not use any default values (e.g. 0000-00-00, etc.).
This should get you started...
<html>
<head>
<title>Date Formatting Tutorial</title>
</head>
<body>
<form name="date" action="<?php $_SERVER['PHP_SELF'] ?>"
method="POST">
Month: <select name="month" selected><option
value="11" />11</select><br />
Day: <select name="day" selected><option
value="19" />19</select><br />
Year: <select name="year" selected><option
value="2007" />2007</select><br />
<input type="submit" name="dateform" value="Submit">
</form>
<?php
if(isset($_POST['dateform']))
{
$date = $_POST['month']."/".$_POST['day']."/".$_POST['year']."";
echo $date."<br />";
$date2 = $_POST['month']."-".$_POST['day']."-".$_POST['year']."";
echo $date2."<br />";
$date3 = $_POST['year']."-".$_POST['month']."-".$_POST['day']."";
echo $date3;
}
?>
</body>
</html>
----------------------
Please refer to www.php.net/date for more guidance on how to format
dates. Best of luck!
- William
