Can I define a variable in a PHP if condition?

you might want something like this:

if (!is_null($my_array = wp_get_category($id)) {
    echo "asdf";
else
    echo "1234";

Assuming the function returns null upon failure. You may have to adjust it a bit.

Yes, that will work, and the pattern is used quite often.

If $my_array is assigned a truthy value, then the condition will be met.

CodePad.

<?php

function wp_get_category($id) {
   return 'I am truthy!';
}

if ($my_array = wp_get_category($id)) {
    echo $my_array;
} else {
    echo "1234";
}

The inverse is also true...

If nothing is returned by the function, I want to go into the else statement.

A function that doesn't return anything will return NULL, which is falsey.

CodePad.

<?php

function wp_get_category($id) {
}

if ($my_array = wp_get_category($id)) {
    echo $my_array;
} else {
    echo "1234";
}

This is in fact a common pattern and will work. However, you may want to think twice about using it for more complex cases, or at all. Imagine if someone maintaining your code comes along and sees

if ($x = one() || $y = two() && $z = three() or four()) {

}

It might be better to declare the variables before using them in the conditional.