Function Help

[JVassie]

Hi everyone, first post here so please excuse if i miss something.

I am trying to build a function that queries from a MySQL database. The code is as follows:

function findGender($username){
   global $conn;
   $q = "SELECT gender FROM users WHERE username=$username";
   $result = mysql_query($q,$conn);
   return $result;
}

I want to use this function to return the gender of a person in the database, so i can display a list of questions based upon what gender they are.

The code for where i call the function is as follows:

<?php
session_start();
include("includelist.php");
$logged_in=checkLogin();
$username=$_SESSION['username'];
$gender=findGender($username);
echo ("Gender is" . "$gender");

Thats from the top of the .php page concerned. includelist.php is the file that contains all my functions. the checkLogin() function obviously checks whether the user is logged in and then we move to the gender. I want the function to return the result of the MySQL query and store it in the $gender variable which i will use to test against, it will either be Male or Female (obviously :p )

The echo statement is merely as a test atm to check that it works.

Currently, when the page is parsed by the server and shown in the browser, it says at the top of the page 'Gender is' followed by blank space, and then the rest of the page underneath.

Is there anything anyone can see wrong firstly with the syntax (im guessing not because i dont get any errors showing) or whether im missing any code for this sort of statement.

If there is any other code you need to look at, please let me know and i shall post it.

Thanks a lot
James

[maggerd]

There is nothing wrong with the query.

The contents of the variable $result is a resource and you must extract the data prior to use.

Use the following code after preforming your query.
$gender = mysql_fetch_assoc($result);

You may need to use these two lines if the above does not work.
$row = mysql_fetch_assoc($result);
$gender =$row['gender'];

[NogDog]

maggerd wrote:

There is nothing wrong with the query....

Actually, there is. Assuming username is non-numeric, its value must be quoted as a string literal:

$q = "SELECT gender FROM users WHERE username='$username'";

[JVassie]

Thanks a lot NogDog, i have managed to get it working now.

However im having a problem now with displaying the right set of questions, and this really baffles me, the code is:

$username=$_SESSION['username'];
$gender=findGender($username);
echo("$gender");
if($gender= 'Male'){
  include("questformm.php");
}
elseif($gender= 'Female')
{
  include("questformf.php");
}
else
{
  die("You have not filled in your Gender in your Personal Details");
}

The echo("$gender") works fine, and outputs Male or Female dependent on the user. However the if statement doesnt seem to work, and regardless of a male or female user, it displays the male set of questions.

Is there anything you can see from that that would usggest why it doesn't work?

Thanks
James

[leatherback]

This:
if($gender= 'Male'){

Tells the script to set $gender to 'Male' and if succesfull do whatever follows. You meant to do this:
if($gender == 'Male'){

The same with the elseif

[JVassie]

Thanks so much, leatherback, that solved it.