How to ignore replace directive in go.mod

Question

I'm using "replace" statement while do my local development. So my go.mod looks like this:

require (
 gorm.io/gorm v1.21.11
 github.com/mypackages/session v1.1.0
)

replace (
 github.com/mypackages/session => ./../session
)

But I have no need in "replace" when I git commit my changes and deploy code to production, so I need to comment this line of replace code on each git commit and uncomment it then. Is there a way to ignore the "replace" statement on a production environment?

For what it's worth, https://golang.org/issue/45713 proposes an alternative to `replace` for local development, in a form that would be easier to `.gitignore` (and/or easier to override for production use by passing `-workfile=off`). — bcmills, Aug 02 '21 at 14:34
if the replace is there only because you need it in your local development, [you could use an alternate go.mod file instead](https://stackoverflow.com/questions/68764637/how-to-use-an-alternate-go-mod-file-for-local-development) — blackgreen, Jun 01 '22 at 21:23

score 5 · Answer 1 · answered Jul 23 '21 at 13:37

5

replace can't be ignored in an environment, because it's used at dependency resolution time, which comes before build, which is long before it ever gets executed in production. But to answer the root question, no, you can't "ignore" the directive. If it's there, it's there.

answered Jul 23 '21 at 13:37

Adrian

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Thanks for your answer. Are you ever faced with that problem? Maybe you know a workaround solution? – sn0rk Jul 25 '21 at 12:34

score 5 · Accepted Answer · answered Jul 23 '21 at 13:51

5

While @Adrian is correct in that there is no way to accomplish this in Go, I think this question is less about Go and more about Git. You can ignore a specific part of your file using content filters. See this SO answer for more information.

answered Jul 23 '21 at 13:51

Clark McCauley

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lazyhacker · Answer 3 · 2022-01-23T16:35:50.503

1

Have a local version of the mod file (e.g. go.local.mod) and then you can tell the go command to use it:

go build -modfile=go.local.mod main.go

edited Jan 23 '22 at 16:35

answered Jan 22 '22 at 05:55

lazyhacker

11
2

1

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score 0 · Answer 4 · answered Dec 09 '22 at 05:48

You can code a pre commit hook, which will detect whether your go.mod file contains replace or not.

Pre commit hook code -> https://gist.github.com/MohitVachhani/e331321bb6f8c3e5e26b0cb4ab792c55

Please go through this and let me know if any improvements you have!

score 0 · Answer 5 · answered Mar 01 '23 at 22:05

The recipe:

Add this to .git/config of your project

[filter "goreplace"]
  smudge = sed 's~//replace~replace~'
  clean = sed 's~replace~//replace~'

alternative add global:

git config --global filter.goreplace.smudge "sed 's~//replace~replace~'"
git config --global filter.goreplace.clean "sed 's~replace~//replace~'"

and then add (to) the .gitattributes

*.mod filter=goreplace

Notice it just works for you not for all contributors. Also it only comment out the 'replace' statement you could also remove it but this would require more sed magic ;).

How to ignore replace directive in go.mod

5 Answers5