Populate a drop down with mysql data?

[sean04]

Hey!

 

So I have the following:

 

<?PHP
$Get_Sex = mysql_query("SELECT ID, Desc FROM gender") or die(mysql_error());
$Sex_Results = mysql_fetch_array($Get_Sex);

while($val = mysql_fetch_array($Sex_Results)){
echo "<option value=$val[iD]>$val[Desc]</option>";
}
?>

 

 

How come the drop down doesn't have any data in it?

 

Thanks,

Sean

[Rustywolf]

DESC is a mysql function

 

change Desc to description in mysql and querys

 

Also, add "or DIE(mysql_error())" to the end of mysql queries

its a good practice for finding errors

[The Little Guy]

Try this:

 

echo "<option value=\"{$val['ID']}\">{$val['Desc']}</option>";

[sean04]

Thanks for the replies! Still no luck..

 

This is what its looking like:

<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post">
                            
                                Email:
                                <input type="text" name="email" size="30" maxlength="25">
                                Password:
                                <input type="password" name="password" size="30" maxlength="25">
                                Confirm Password:
                                <input type="password" name="cpassword" size="30" maxlength="25">
                                Screen Name:
                                <input type="text" name="screenname" size="30" maxlength="55">
							Gender:
							<select name="gender" value="">Gender</option>
							<?PHP
							$Get_Sex = mysql_query("SELECT ID, Description FROM gender") or die(mysql_error());
							$Sex_Results = mysql_fetch_array($Get_Sex);
							while($val = mysql_fetch_array($Sex_Results)){
							echo "<option value=\"{$val['ID']}\">{$val['Description']}</option>";
							}
							?>
							</select>



                                <input type="submit" value="Register">
                        </form>

 

I know its not to neat right now sorry

 

[The Little Guy]

Your using the wrong param for your fetch array in the while loop...

<?php
$Get_Sex = mysql_query("SELECT ID, Description FROM gender") or die(mysql_error());
while($val = mysql_fetch_array($Get_Sex)){
echo "<option value=\"{$val['ID']}\">{$val['Description']}</option>";
}
?>

[sean04]

Thank you! Works great! Thought it had to do with something like that as the error mentioned it Didn't know what to do exactly though, Thanks again!