Hi @luis.espeleta ,

the regular expression r'\d+$' is used to match one or more digits (\d+) at the end of the string ($)

Thank you, I tried it and got this:
(str_Factura: XDGC992679)
Why does it show 0?

Thank you, I tried it and got this:
(str_Factura: XDGC992679)
Why does it show 0?

 

It is because the specified substring is not found. In your case you should use extract / substring. I believe you are using Find.

Hi @luis.espeleta ,

I got the result with below logic : 

The key thing is I passed the RegEx in a variable and then send it to replace as it contains $ and AA checks for variable declaration.

The code works in fashion :

• List contains all the string where last numbers are to be extracted.
• Initially I replaced the last digits from the variable using RegEx : \d+$
• After that from the original string I removed the replaced text and got the desired Output.

Let me know if any query comes.

Hi @luis.espeleta ,

I got the result with below logic : 

The key thing is I passed the RegEx in a variable and then send it to replace as it contains $ and AA checks for variable declaration.

The code works in fashion :

• List contains all the string where last numbers are to be extracted.
• Initially I replaced the last digits from the variable using RegEx : \d+$
• After that from the original string I removed the replaced text and got the desired Output.

Let me know if any query comes.

Sample String has the input string and Regex has : \d+$ and is stored in Test1

Sample String has the input string, Test1 has the string other than last digits and the result is stored in OutPut

Example : Using String Replace 

First Screenshot : 

Input :

• SampleString : 12xd23fg8941332
• Regex = \d+$

Output : 

• Test1 = 12xd23fg

Second Screenshot

• SampleString : 12xd23fg8941332
• Test1 = 12xd23fg

Output : 

• OutPut = 8941332