Reg ex for selecting and replacing a numeric value in a line that contains a specific pattern of string

Question

I have the following lines (using nedit)

R1_1 vsp vsp/1 23

R2_1 vsp/2 vsp/3 30

R3_2 vsp/3 vsp/4 50

R3_2 v2 v1 50

C1 vsp/1 vsp 60

I want to just replace the numbers with number*0 for the lines that start with R and contain the string vsp. I want the following in the above example

R1_1 vsp vsp/1 23*0

R2_1 vsp/2 vsp/3 30*0

R3_2 vsp/3 vsp/4 50*0

R3_2 v2 v1 50

C1 vsp/1 vsp 60

What's the regular expression for that on nedit?

Tim Biegeleisen · Accepted Answer · 2020-07-08T03:26:09.407

0

You may try the following find and replace, in regex mode:

Find:    ^(R.*\bvsp\b.*?)(\d+)$
Replace: $1$2*0

Demo

On Nedit regex, you might try using the following replacement instead:

\1\2*0

edited Jul 08 '20 at 03:26

answered Jul 08 '20 at 03:11

Tim Biegeleisen

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Like I mentioned, it should start with 'R' as well hence fifth line can't have a replacement – Venkatraman Natarajan Jul 08 '20 at 03:16
@VenkatramanNatarajan Thanks for pointing this out, I have updated my answer. – Tim Biegeleisen Jul 08 '20 at 03:18
Tried it, doesn't seem to select – Venkatraman Natarajan Jul 08 '20 at 03:19
It is working in the demo. At this point, you really should reveal which regex language/tool you are using here. – Tim Biegeleisen Jul 08 '20 at 03:21
I am using nedit. Yes it seem to work on demo. On demo, I see it removes the first character – Venkatraman Natarajan Jul 08 '20 at 03:24
Thanks, I see the problem is in finding. It's not able to locate. – Venkatraman Natarajan Jul 08 '20 at 03:30
Figured it out. ^(R.*vsp.*?)(\d+)$ this works on nedit and \1\2*0 works on nedit – Venkatraman Natarajan Jul 08 '20 at 05:25
Please _edit_ my answer then and add necessary details. I know regex, but I don't know your tool. – Tim Biegeleisen Jul 08 '20 at 05:26

Reg ex for selecting and replacing a numeric value in a line that contains a specific pattern of string

1 Answers1

Demo