Regular Expression PHP Help with if not equal to

Question

I am trying to use regular expressions to make my inventory upload script a little better.

I sell clothing and accessories. In my inventory file I have a field called ProductDepartment which is a combination of gender/item type. I have a MySQL table for gender with the following values:

Mens = 1
Womens = 2
Unisex = 3

Departments in the uploaded file look like the following, for example.

Mens Outerwear
Mens Watches
Womens Jeans
Headphones

So basically Im trying to make a statment that goes, if $ProductDept contains Mens, set $gender_id = 1, if $ProductDept contains Womens, set $gender_id = 2, if $ProductDept does not contain either, set $gender_id = 3

And I'm having trouble figuring out how to write the 3rd if statement correctly. Here's what I have.

if(preg_match('/Mens/g', $ProductDept)) {
    $gender_id = '1';
}

if(preg_match('/Womens/g', $ProductDept)) {
    $gender_id = '2';
}

if( != preg_match('/Mens|Womens/g', $ProductDept)) {
    $gender_id = '3';
}

Answer 1

Change != to !:

if (!preg_match('/Mens|Womens/g', $ProductDept)) { ... } 

You could also rewrite it using elseif as follows:

if (preg_match('/Womens/g', $ProductDept)) {
    $gender_id = '2';
} elseif (preg_match('/Mens/g', $ProductDept)) {
    $gender_id = '1';
} else {
    $gender_id = '3';
}

Answer 2

I'll do :

if(preg_match('/Womens/', $ProductDept)) {
  $gender_id = '2';
} elseif(preg_match('/Mens/', $ProductDept)) {
  $gender_id = '1';
} else {
  $gender_id = '3';
}