CODEWITHSUNDEEP
regexnotepad++
Remo Hernandez
Remo Hernandez

Reputation: 75

Notepadd++ Regex replace character equal to in certain position

can someone help me with a Notepad++ Regex to replace characters in certain positions? I know there is a lot of answers related to regex, but I'm having a hard time with this, so I apologize if this is a duplicate. Below is what I need done..

Below is a text fixed file where I need to: 1) Replace characters in position 1-3 equal to "622" and replace with "633". 2) Replace characters in position 35-42 to zeroes. This is before:

622021000021833364532        00000202050007037174     JOHN SMITH              0043000264294578
6240631075131010089984136    00000162050006912435     JOHN SMITH              0043000264294622
6242631830492020236          00000025000006912435     JOHN SMITH              0043000264294622
6222631830492020257          00000015000006912435     JOHN SMITH              0043000264294581
6220610001041000142074458    00000202050500002543     JOHN SMITH              0043000264294582

This should be after:

623021000021833364532        00000000000007037174     JOHN SMITH              0043000264294578
6240631075131010089984136    00000000000006912435     JOHN SMITH              0043000264294622
6242631830492020236          00000000000006912435     JOHN SMITH              0043000264294622
6232631830492020257          00000000000006912435     JOHN SMITH              0043000264294581
6230610001041000142074458    00000000000000002543     JOHN SMITH              0043000264294582

Below is what I have so far for request 1: Find: ^(.{3}) This finds the first 3 characters, but I need to find the first 3 characters = to "622". I can't just find "622" because line 2 and 3 have "622" near the end. Those should not be replaced.

Can someone help me with both requests? Any help is much appreciated! Thank you! - Remo

Upvotes: 0

Views: 2128

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626896

  1. Find ^622 replace: 633

  2. For 35-42 it's find ^(.{34}).{7} replace ${1}0000000 (multi-line mode). Both regex are run separately, and once.

Upvotes: 2

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