Removing all characters before last alpha character

Question

I have a mixed string such as:

Job number  45752 Subtotal price $937.50 
Job number  7852 Subtotal amount $637.50 
Job number  42 Subtotal test $427.50 
Job number  47592 Subtotal sample $976.50 

How do I detect the last alphabet character like the 1st sample 'e' on price, and get its position and then remove all other characters in front?

I know strpos can be used to find the last character

strpos(string,find,start)

But how to make it to track any alphabet instead of a fixed one? I am guessing regex might help but just no idea how to put it in. Please help.

Answer 1

With regex preg_replace() function, using flags i caseless and m multi-line mode:
To replace from ^ line start to the last alpha with optional spaces, put a greedy dot before [a-z]

$str = preg_replace('/^.*[a-z]\h*/im', "", $str);

\h* matches any amount of horizontal space. See test at regex101, eval.in, regex quickstart

Answer 2

I think this can helps:

/.* ([^a-z]+)$/igm

[Regex Demo]
or

/([^a-z]+)$/igm

[Regex Demo]