What Regex would capture everything from ' mark to the end of a line?

Question

I have a text file that denotes remarks with a single \'.

Some lines have two quotes but I need to get everything from the first instance of a \

Answer 1

https://regex101.com/r/Jjc2xR/1

/(\w*\(Hex\): w*)(.*?)(?= |$)/gm

I'm sure this one works, it will capture de hexa serial in the badly structured text multilined bellow

     Space Reservation: disabled
         Serial Number: wCVt1]IlvQWv
   Serial Number (Hex): 77435674315d496c76515776
               Comment: new comment

I'm a eternal newbie in regex but I'll try explain this one

(\w*(Hex): w*) : Find text in line where string contains "Hex: "

(.*?) This is the second captured text and means everything after

(?= |$) create a limit that is the space between = and the |

So with the second group, you will have the value