So I have this registration script:


The HTML:

<form action="register.php" method="POST">
    <label>Username:</label> <input type="text" name="username" /><br />
    <label>Password:</label> <input type="text" name="password" /><br />

    <label>Gender:</label>
    <select name="gender">
      <optgroup label="genderset">
        <option value="Male">Male</option>
        <option value="Female">Female</option>
        <option value="Hermaphrodite">Hermaphrodite</option>
        <option value="Not Sure!!!">Not Sure!!!</option>
      </optgroup>
    </select><br />
<input type="submit" value="Register" />
</form>

The PHP/SQL:

PHP Code:

<?php

$username = $_POST['username'];
$password = $_POST['password'];
$gender = $_POST['gender'];

mysql_query("INSERT INTO registration_info (username, password, gender) VALUES ('$username', '$password', '$gender')
")  

?>

The problem is, the username and password gets inserted into the
"registration_info" table just fine. But the Gender input from the
select drop down menu doesn't. Can some one tell me how to fix this,

P.S. I know this is an unsafe script as I am inserting form data directly into the database without validating it in any way first. That's because I am just learning PHP/SQL and want to get the basics down first. This script will never actually go online.
thanks.

Have you printed the values to screen after submissions?

Originally Posted by chrishirst

Have you printed the values to screen after submissions?

I'm not sure what you mean, I know the gender input hasn't been submitted because when I enter the Select All command in mysql monitor : "SELECT * from registration_info" I can see that the username and password has been inputted but the gender column is empty. It's not Null but empty.

echo the variables to the browser instead of running the SQL.

It's debugging 101 rather than guessing where it is not working.

I made a stupid mistake, the problem was I used:

$gender = $_POST['gender'];

when the name of my drop down list was actually "dropdown"

Now that I clanged it to

$gender = $_POST['dropdown'];

It inserts just fine.

Easily spotted by echoing the POSTed form keys and values to the browser.

you might want to change the name back to gender, once you get a second dropdown or more your coding will begin to be harder to follow. Fine if its for yourself, but i've realised the value of using logical names, variables in my coding as well as documenting throughout the file.

One of my sites is now 2yrs old and in its 4th or 5th version, so sometimes tracking parts can take a while. Dont think "ill just do it later" cause you never will, unless you are really disciplined. When a page is finished and working correctly go back and explain the things that arent obvious