Get the branch name from tag using Azure DevOps api - azure-devops

Question

Can someone help me to get the actual branch name if I have the tag associate with it?
Azure DevOps API would be best in my case. Thanks in advance.
SO far I get the API that gives all the tags for particularly repo but this does not solve my issue.
https://dev.azure.com/{{organization}}/{{project}}/_apis/git/repositories/{{repository_id}}/refs?filter=tags/&api-version=6.0
But it does not give me any branch information associated with it.

Answer 1

There isn’t such REST API to get branch name by a tag or commit. In the Azure DevOps, Tags are created based on the commit id, and a commit or tag can belong more than one branches. Let’s illustrate by below graph:
A---B---E---F master
\ /
C---D dev
This is a branch structure with master and dev branch. And dev branch merge into master with commit F. If you want to get the branch name based on commit D, you will get two branch names. Commit D belongs to two branch: dev (first parent) and master (second parent).
We can get the branch name via condition and multiple REST APIs, also we could get the branch name via git cmd.
a. List all tags and get the objectId.
GET https://dev.azure.com/{Org name}/{Project name}/_apis/git/repositories/{repo ID}/refs?filter=tags/&api-version=6.0
b. Get the tag info and commit ID via objectId
Note: the field taggedObject.objectId is commit ID
GET https://dev.azure.com/{organization}/{project}/_apis/git/repositories/{repositoryId}/annotatedtags/{objectId}?api-version=6.0-preview.1
Workaround 1
Now we could get the commit details and get the branch name(s) contain the given commit, we can use git command:
git branch --contains <commit>
Workaround 2
We could list all commits info on a branch via this API
GET https://dev.azure.com/{Org name}/_apis/git/repositories/{repositoryId}/commits?searchCriteria.itemVersion.version={branch name}&api-version=6.0
And now, we get commit id which contain the tag and all commit ID on a branch, add if{} or eq() to check it, if it return True, we could know this branch contain the tag
Update1
My test repo contains branch main and test01, the branch main contains the tag Tag01, the branch test01 contains the tag Tag01 and tag02
Power shell script:
$ListAllTagsURL="https://dev.azure.com/{Org name}/{Project name}/_apis/git/repositories/{Repo ID}/refs?filter=tags/&api-version=6.0"
$PAT="{PAT}"
$base64AuthInfo= [System.Convert]::ToBase64String([System.Text.Encoding]::ASCII.GetBytes(":$($PAT)"))
$ListAllTags = Invoke-RestMethod -Uri $ListAllTagsURL -Headers #{Authorization = "Basic {0}" -f $base64AuthInfo} -Method get
#List all tags name
#write-host $ListAllTags.value.name
foreach($Tag in $ListAllTags.value){
if($Tag.name -eq "{tag name such as refs/tags/Tag01}"){
$objectId = $Tag.objectId
$TagName = $Tag.name
}
}
#Get tag details and commit ID
$TagURL="https://dev.azure.com/{Org name}/{Project name}/_apis/git/repositories/{Repo ID}/annotatedtags/$($objectId)?api-version=6.0-preview.1"
$TagInfo = Invoke-RestMethod -Uri $TagURL -Headers #{Authorization = "Basic {0}" -f $base64AuthInfo} -Method get
$TagCommitID = $TagInfo.taggedObject.objectId
#write-host $TagCommitID
#List all branch
$url="https://dev.azure.com/{Org name}/{Project name}/_apis/git/repositories/{Repo ID}/refs?filter=heads&api-version=6.1-preview.1"
$BranchInfo = Invoke-RestMethod -Uri $url -Headers #{Authorization = "Basic {0}" -f $base64AuthInfo} -Method get
foreach($Branch in $BranchInfo.value){
#write-host $Branch.name
$BranchName = $Branch.name.split("/",3)[-1]
#write-host $BranchName
#List all commit ID via Branch Name
$BranchDetailUrl="https://dev.azure.com/v-viliu/_apis/git/repositories/{Repo ID}/commits?searchCriteria.itemVersion.version=$($BranchName)&api-version=6.0"
$BranchDetailInfo = Invoke-RestMethod -Uri $BranchDetailUrl -Headers #{Authorization = "Basic {0}" -f $base64AuthInfo} -Method get
#write-host $BranchDetailInfo.value.commitId
foreach($CommitID in $BranchDetailInfo.value.commitId){
If($CommitID -eq $TagCommitID){
write-host $BranchName "Contain this tag" $TagName
}
}
}
Result: