Could you explain that in terms that I can understand? Maybe dumb it down a little bit?

And, could you please provide an example?


No error message.

Everything looks like its included in the $sql = "INSERT INTO but nothing is going into the date fields of the database.

From what I have read on this topic so far, this appears to be the solution that I would use.

If you are not sure where your error is in your sql statement, I suggest getting that issue corrected before trying to insert the date into the database. A single error in the sql statement can leave you scratching your head for hours if you're not sure how to debug the issue.

Here is how I would work the problem:
Take your sql statement and print it out on the screen (echo or print command allows you to do this) and then copy and paste the displayed sql (this should have all of your variables populated at this point) and drop it into your sql console (phpmyadmin or sql server management studio).

The sql should execute correctly (if you have trouble finding the sql window in phpmyadmin, I can walk you through that separately). If your sql is not executing correctly when you place it in the sql console, start reworking the query until you are able to get an executable query.

(Ok, it appears to have an issue with dates, so remove the date columns, and try and execute again, no there is an issue with the number of columns not matching the values sent, rip out extra value columns and try again, ... repeat the process until you can execute the sql.)

Once you have found and fixed the basic issue (data is now appearing in the database, but there still is an issue with some of the data not making it into the database). I would then search on google for a solution to that problem - google: "How to insert dates into database mysql" - read over a couple of articles related to that solution, and try and solve the problem using the solutions discussed in the articles.

These are basic debugging techniques that any developer uses to solve problems. Trust me, there have been times when I have been tripped up on this very same issue, and usually it's more related to the basic sql syntax than the actual technique of how I'm inserting the data. (Oh I forgot a comma there, STUPID! - and then pretty soon, everyone will wonder why you're yelling at your computer screen).

I hope this information is helpful. If you want any more tips, let me know (send me a PM, because I might not check back to this post), I'd be happy to share.

Uh,,,, no - I can not post a link to the form. Its for a medical provider, and nobody from the outside world can gain access to it.

I'am a hardware / networking guy, and management dropped this project in my lap. They told me they needed a database to track certain patient information, so here I am.


Here is a copy of the drop down menu and insert codes - A lot of the other codes have been removed:

$sql = "INSERT INTO patientformdata (birthdate)
PrepSQL($varBirthdate) . ", " .


<p>
Select Birthday:

<?php

$display .= '</select>
<select name="DateOfBirth_Year">
<option value="">- Year -</option>';

$thisyear = date('Y');
$now = $thisyear - 2;
$end = $thisyear - 100 ;

for ($now; $now >= $end; $now--) {
if ($_POST['DateOfBirth_Year'] == $now){
$display .= '<option value="'.$now.'" selected>'.$now.'</option>';
}
else{
$display .= '<option value="'.$now.'" >'.$now.'</option>';
}
}
$display .= '</select> ';

$display .='</select>
<select name="DateOfBirth_Month">
<option value="" >- Month -</option>';
for ($i = 1; $i <= 12; $i++) {
if ($_POST['DateOfBirth_Month'] == $i){
$display .= '<option value="'.$i.'" selected>'.GetMonthString($i).'</option>';
}
else{
$display .= '<option value="'.$i.'" >'.GetMonthString($i).'</option>';
}
}
function GetMonthString($n){
$timestamp = mktime(0, 0, 0, $n, 1, 2010);
return date("F", $timestamp);
}

$display .= '<select name="DateOfBirth_Day">
<option value="" >- Day -</option>';

for ($i = 1; $i <= 31; $i++) {
if ($_POST['DateOfBirth_Day'] == $i){
$display .= '<option value="'.$i.'" selected>'.$i.'</option>';
}
else{
$display .= '<option value="'.$i.'" >'.$i.'</option>';
}
}
$display .= '</select> ';

echo $display;

$DateOfBirth_Year = $_POST['yyyy'];
$DateOfBirth_Month = $_POST['mm'];
$DateOfBirth_Day = $_POST['dd'];
$Birthdate = "$DateOfBirth_Year, $DateOfBirth_Month, $DateOfBirth_Day"; // or whatever format you want

?>
</p>

Sorry, I did not realize that I had not posted the code I was using. I edited my post just above yours and added the code.

The actual insert should be performed after the form has been tested as being submitted. I'm guessing that's done somewhere else...