# Regex replace year I### with 1###

**URL:** <https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683>\
**Category:** Support\
**Created:** [August 19, 2018, 2:43am UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683 "2018-08-19T02:43:30Z")\
**Posts on this page:** 7\
**Page:** 1

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**Author:** ![3r47](https://community.mp3tag.de/letter_avatar_proxy/v4/letter/3/e56c9b/32.png) [@3r47](https://community.mp3tag.de/u/3r47)\
**Post date:** [August 19, 2018, 2:43am UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683/1 "2018-08-19T02:43:30Z")

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I have a bunch of wax cylinder recordings and the years listed within multiple fields have I's instead of 1's. So I need a regex to match I###, I#I#, I#II replacing the I's with 1's in all tag fields.

Thanks all for your help!

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**Author:** ![Florian](https://community.mp3tag.de/user_avatar/community.mp3tag.de/florian/32/5759_2.png) [@Florian](https://community.mp3tag.de/u/Florian)\
**Post date:** [August 19, 2018, 7:11am UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683/2 "2018-08-19T07:11:39Z")

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You don't need regular expressions for that. A normal "Replace" action for `YEAR` where you replace `I` by `1` will do.

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**Author:** ![poster](https://community.mp3tag.de/user_avatar/community.mp3tag.de/poster/32/4881_2.png) [@poster](https://community.mp3tag.de/u/poster)\
**Post date:** [August 19, 2018, 7:41am UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683/3 "2018-08-19T07:41:35Z")

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He has to replace `I` with `1` in multiple fields, not only the YEAR-Tag.

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**Author:** ![ohrenkino](https://community.mp3tag.de/user_avatar/community.mp3tag.de/ohrenkino/32/4843_2.png) [@ohrenkino](https://community.mp3tag.de/u/ohrenkino)\
**Post date:** [August 19, 2018, 9:12am UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683/4 "2018-08-19T09:12:54Z")

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OK, so much for the theoretical part.  
Try  
Search string: `I(\d+)`  
Replace string: 1$1  
You have to apply that action to every field where you suspect that year format.

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**Author:** ![3r47](https://community.mp3tag.de/letter_avatar_proxy/v4/letter/3/e56c9b/32.png) [@3r47](https://community.mp3tag.de/u/3r47)\
**Post date:** [August 19, 2018, 8:49pm UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683/5 "2018-08-19T20:49:23Z")

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> [@ohrenkino](#):
>
> (\d+)

thanks that worked for all the I###, but what about I#I#?

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**Author:** ![ohrenkino](https://community.mp3tag.de/user_avatar/community.mp3tag.de/ohrenkino/32/4843_2.png) [@ohrenkino](https://community.mp3tag.de/u/ohrenkino)\
**Post date:** [August 20, 2018, 4:41am UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683/6 "2018-08-20T04:41:47Z")

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> [@3r47](#):
>
> but what about I#I#?

It should work also.  
I just tried:  
`$regexp('I12I34',I(\d+),1$1)`  
Result: "112134"

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**Author:** ![system](https://community.mp3tag.de/uploads/default/original/2X/c/ce7035d426cb755a7916793326d23b465222a407.png) [@system](https://community.mp3tag.de/u/system)\
**Post date:** [February 9, 2026, 12:27pm UTC](https://community.mp3tag.de/t/regex-replace-year-i-with-1/42683/7 "2026-02-09T12:27:36Z")

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