'Regex - add new field between semicolon separators - notepad++

I have data like this (some of the fields have values, some of them don't)

J;555;3;1990-02-19;1234;S;2020-10-17;;;;;value;3;ANOTHERVALUE

I need to add a new field right before the 5th semicolon, like this (the F value):

J;555;3;1990-02-19;1234;F;S;2020-10-17;;;;;value;3;ANOTHERVALUE

So far I have this:

^(([A-Z]);([0-9A-Z]+);([?:0-9]);([0-9\-]+);([A-Z0-9]+));(.*)$

And as replacement:

$1;$2;F;$7

As far as I know the ?: before is to mark optional, but in my case I need to represent empty fields between every semicolon

How can I solve this?

regexnotepad++


Solution 1:[1]

const regex = /^([A-Z]);([0-9]*);([0-9]*);([0-9-]*);([0-9]*);(.*)$/


let str = 'J;555;3;1990-02-19;1234;S;2020-10-17;;;;;value;3;ANOTHERVALUE'
str = str.replace(regex, '$1;$2;$3;$4;$5;F;$6')
let expected = 'J;555;3;1990-02-19;1234;F;S;2020-10-17;;;;;value;3;ANOTHERVALUE'
console.log(str === expected)

str = 'J;;3;1990-02-19;;S;2020-10-17;;;;;value;3;ANOTHERVALUE'
str = str.replace(regex, '$1;$2;$3;$4;$5;F;$6')
expected = 'J;;3;1990-02-19;;F;S;2020-10-17;;;;;value;3;ANOTHERVALUE'
console.log(str === expected)

You can find an explanation on Regex101

Basically, I made use of * which matches from zero to unlimited times to account for instances with missing values.

^([A-Z]);([0-9]*);([0-9]*);([0-9-]*);([0-9]*);(.*)$

Description

  • ([A-Z]) Match any A-Z character
  • ; Matches a semi-colon literally
  • ([0-9]*) Match any number between zero and unlimited times
  • ; Matches a semi-colon literally
  • ([0-9]*) Match any number between zero and unlimited times
  • ; Matches a semi-colon literally
  • ([0-9]*) Match any number between zero and unlimited times
  • ; Matches a semi-colon literally
  • ([0-9]*) Match any number between zero and unlimited times
  • ; Matches a semi-colon literally
  • (.*) Matches any character between zero and unlimited times

Solution 2:[2]

You may use this code in notepad++,

Find:^((.*?;){5})
Replace with:$1F;

enter image description here

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 Haji Rahmatullah

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