Preg match in bash like in php?

Question

bash

Hello, how to get this in a bash script?

There is a string
Str = lalchlch login - 79991234567, password durhfh4fj

At the output in variables
Login = 79991234567
Pass =

durhfh4fj

Answer 1

It can be like this:

Str="лялчлч логин - 79991234567, пароль durhfh4fj"
Login=`echo "$Str" | grep -Po "логин - \K[0-9]+"`
Pass=`echo "$Str" | grep -Po "пароль \K.+"`
echo "Login=$Login"
echo "Pass=$Pass"