[RESOLVED] Inserting Dates into mysql table

[tams]

Hi,
I want to insert dates from a drop down date menu into a mysql table. The problem is that mysql won't read any of the selected dates. The code for inserting in mysql is correct. I just need some help with the PHP code for reading dates.

Thank you

[Flyier]

Hello.

Please be more specific, post some code.

What is the format of the date forms ?

• Mitko Kostov

[tams]

Thank you for replying.
The PHP code for creating the drop down menus is as follows:
<td <?php

$monthName = array(1=> "January", "February", "March","April", "May", "June", "July","August", "September", "October","November", "December");

$today = time();

$f1_today = date("Y-M-D",$today);

echo "<div style = 'text-align: center'>\n";
/ build selection list for the month /

$todayMO = date("n",$today); //get the month from $today

echo "<select name='dateMO1'>\n";

for ($n=1;$n<=12;$n++)

{

echo "<option value=$n\n";

if ($todayMO == $n)

{

 echo " selected";

}

echo "> $monthName[$n]\n";

}

echo "</select>";

/ build selection list for the day /

$todayDay= date("d",$today); //get the day from $today

echo "<select name='dateDay1'>\n";

for ($n=1;$n<=31;$n++)

{

echo " <option value=$n";

if ($todayDay == $n )

{

 echo " selected";

}

echo "> $n\n";

}

echo "</select>\n";

/ build selection list for the year /

$startYr = date("Y", $today); //get the year from $today

echo "<select name='dateYr1'>\n";

for ($n=$startYr-66;$n<=$startYr;$n++)

{

echo " <option value=$n";

if ($startYr == $n )

{

 echo " selected";

}

echo "> $n\n";

}

echo "</select>\n";

?></td></tr>

And the code for inserting it into mysql is:
$query = "INSERT INTO personalinfo (birthdate)
VALUES ('$_POST[f1_today]');

[Flyier]

Hello again.

I really don't know what you are trying to do.
You want to generate forms with today date and the user to choose a date which will be sent in MySQL table ?

Please, show me exactly how you assign value to $_POST[f1_today].

• Mitko Kostov

[tams]

Yes I want to make a form with todays dates and let the user choose the date to be sent into mysql table.

The assigning value part is the part I am having trouble with. I have absolutely no idea how to assign a value to the date selected by the user and then send it to mysql table. If you could help me with that part, I would greatly appreciate it.

Thank you

[Flyier]

Hello.

Here is a fucntion to generate select field. I know it can be done simple, but I haven't slept for 4 days 😉

function genSelect()
{

 $date = array(date(d),date(m), date(Y));

  echo "<select type=\"text\" name=\"day\" value=\"\" size=\"1\">";
  for ($i=1;$i<=31;$i++) {
    if($i==$date[0]) { echo "<option SELECTED>&nbsp;$i&nbsp;</option>"; } 
 	else { echo "<option>$i</option>"; }
 }
 echo "</select>"; 

 echo "<select type=\"text\" name=\"month\" value=\"\" size=\"1\">";
 for ($i=1;$i<=12;$i++) {
    if($i==$date[1]) { echo "<option SELECTED>&nbsp;$i&nbsp;</option>"; } 
 	else { echo "<option>$i</option>"; }
 }
 echo "</select>";

 echo "<select type=\"text\" name=\"year\" value=\"\" size=\"1\">";
 for ($i=2007;$i>=1950;$i--) {
    if($i==$date[2]) { echo "<option SELECTED>&nbsp;$i&nbsp;</option>"; } 
 	else { echo "<option>$i</option>"; }
 }
	echo "</select>";

 }

?>

This you include in HTML :

<form action="" method="POST">
<?php genSelect();
           $date = array($_POST['year'], $_POST['month'], $_POST['day']);
           $date = implode("-", $date);

?>
<input type="submit" name="submit" value="submit">
</form>

PS: I leave to you to make day and month smaller than 10 in DD and MM format 🙂

[tams]

Thank you.
I appreciate your help a lot. Good luck with your sleep🙂

[bradgrafelman]

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