How can I verify the build version?

Hi,

I am currently evaluating Couchbase solution for my work.

I have compiled Sync Gateway based on https://github.com/couchbase/sync_gateway/blob/master/docs/BUILD.md . I have used

./bootstrap.sh -c 1bafacaece43e70cd3158b23b16a6d60f30d8c1b

1bafacaece43e70cd3158b23b16a6d60f30d8c1b is the commit id for 2.1 release. I obtained it from here:

How can I verify that I have built correct version? I am thinking along the lines of build version number that I can verify maybe in one of the generated files etc.

Any help would be greatly appreciated.

-mc

Hi mc,

The first obvious question would be what is your reason for compiling Sync Gateway from source? This isn’t supported, and you should really be running the binaries we release which have been extensively vetted by QE and performance tested before being released.

You can find these on the Couchbase downloads page here:
https://www.couchbase.com/downloads

Back to the original question though, that ID is not for 2.1.

If you really do need to run build it yourself, our bootstrap script will take a branch name, and you can specify ‘release/2.1.0’ instead of hunting for a commit.

https://github.com/couchbase/sync_gateway/tree/release/2.1.0 for which the latest commit is a036bd817d35ff1c354c644804dc588fb7c41476

Hi bbrks,

Thank you so much for your quick answer.

I decided to build it myself because I wanted to be able to use all the available features without cost concerns.

My understanding is that ‘release/2.1.0’ branch contains the latest stable GA release. Is this correct?

-mc

Hi mc,

As noted on the Couchbase downloads page, the Enterprise Edition of Sync Gateway is available for free for the purposes of evaluation and testing on non-production environments.

It is available free of charge for unlimited pre-production development and testing

Generally speaking, the head of the release branch will be the commit that went GA, but it’s unlikely your build environment will be the same as ours, and you’ll end up with a binary which will be slightly different.