如何获得 Azure DevOps 管道中构建的代码审阅者？

Question

给定构建 ID，如何在 azure DevOps 管道中获取代码审阅者姓名？假设，构建脱离了主分支——开发人员在拉取请求中审查了代码后合并了他们的功能分支。这是政策，没有人直接将他们的更改提交给 master。这意味着，每个构建背后都有一个代码审查员。我如何获得它？

谢谢！

Answer 1

您可以使用下面Rest api获取PR审稿人。

1，首先用buildId在build rest api下面调用。在响应中，您将从构建的 sourceVersion 和存储库 ID 中获取提交 ID。

GET https://dev.azure.com/{organization}/{project}/_apis/build/builds/{buildId}?api-version=5.1

2、获取commit id和repository id后。您可以调用 commit rest api 从响应中的评论中获取关联的 PR id。

GET https://dev.azure.com/{organization}/{project}/_apis/git/repositories/{repositoryId}/commits/{commitId}?api-version=5.1

3、然后调用pull request reviewer rest api获取Reviewers

GET https://dev.azure.com/{organization}/{project}/_apis/git/repositories/{repositoryId}/pullRequests/{pullRequestId}/reviewers?api-version=5.1

以下是 powershell 中的示例脚本。请参阅 this link 以获取个人访问令牌

$buildId= " "

$burl =" https://dev.azure.com/OrgName/ProjName/_apis/build/builds/$($buildId)?api-version=5.1"

$PAT="personel access token"

$base64AuthInfo= [System.Convert]::ToBase64String([System.Text.Encoding]::ASCII.GetBytes(":$($PAT)"))

$buildInfo = Invoke-RestMethod -Uri $curl -Headers @{Authorization = ("Basic {0}" -f $base64AuthInfo1)} -Method get -ContentType "application/json"
#get CommitId and repoId
$commitId = $buildInfo.sourceVersion
$repoId=$buildInfo.repository.id

#commit rest api
$curl = "https://dev.azure.com/OrgName/ProjName/_apis/git/repositories/$($repoId)/commits/$($commitId)?api-version=5.1"

$commitInfo = Invoke-RestMethod -Uri $curl -Headers @{Authorization = ("Basic {0}" -f $base64AuthInfo1)} -Method get -ContentType "application/json"
#get PR id
$prId = $commitInfo.comment.split(" ")[2].TrimEnd(":")

$prurl = "https://dev.azure.com/OrgName/ProjName/_apis/git/repositories/$($repoId)/pullRequests/$($prId)/reviewers?api-version=5.1"

Invoke-RestMethod -Uri $prurl -Headers @{Authorization = ("Basic {0}" -f $base64AuthInfo1)} -Method get -ContentType "application/json"

如果您可以在 UI 页面中使用给定的 buildId 从管道运行历史记录中找到构建。这会容易得多。您可以直接从标题中获取 PR id。见下图

您也可以点击上面截图中显示的commit id，查看commit的详细信息，在这里您将获得相关的PR。

Answer 2

这是我终于完成的工作。使用上面 Levi 的代码片段并修复一行以获取在各种情况下工作的拉取请求 ID。感谢 Levi's 的帮助！希望对大家有帮助。


$PAT="personel access token"
$base64EncodedPAT = [System.Convert]::ToBase64String([System.Text.Encoding]::ASCII.GetBytes(":$($PAT)"))
$basicAuth = @{Authorization = "Basic $base64EncodedPAT" }
$buildId= "..."

function GetCodeReviewers() {
    #Get build info
    $buildUrl = "https://dev.azure.com/OrgName/ProjName/_apis/build/builds/$($buildId)?api-version=5.1"
    $buildInfo = Invoke-RestMethod -Method Get -Uri $buildUrl -Headers $basicAuth

    # Get Commit Info
    $commitUrl = "https://dev.azure.com/OrgName/ProjName/_apis/git/repositories/$($buildInfo.repository.id)/commits/$($buildInfo.sourceVersion)?api-version=5.1"
    $commitInfo = Invoke-RestMethod -Uri $commitUrl  -Method Get -Headers $basicAuth

    #Get Code Reviewers
    $comment = $commitInfo.comment
    #$pullRequestId = $comment.split(" ")[2].TrimEnd(":") # it turns out, the 3rd item may not always be the PullRequestID so the next line may not work for all scenarios
    #note that, a comment could come as follows:
    # case 1: Merge PR 1234: some other text here including story or bug numbers
    # case 2: Merge pull request 1234 some additional text goes here including story or bug numbers
    # The following will pick the first number - which I assume will always be the PullRequestID
    $pullRequestId = $null
    $pullRequestId = $comment.Replace(':', '').Split(" ").Trim() | Where-Object {[int]::TryParse($_, $pullRequestId)} | Select-Object -First 1
    $pullRequestUrl = "https://dev.azure.com/OrgName/ProjName/_apis/git/repositories/$($buildInfo.repository.id)/pullRequests/$($pullRequestId)/reviewers?api-version=5.1"
    $reviewers = Invoke-RestMethod -Uri $pullRequestUrl -Method Get -Headers $basicAuth

    return $reviewers.value
}

如何获得 Azure DevOps 管道中构建的代码审阅者？

How do I get the code reviewer(s) of a build in Azure DevOps pipeline?

build

pull-request

build-pipeline

azure-devops-rest-api

tfs-code-review