在 Azure DevOps 构建 YML 中输出 HTML 文件

Question

我有一个创建 html 文件的 powershell 脚本，我是 Azure DevOps 的新手。此 html 文件必须使用构建工件进行编译和生成。我不确定我该怎么做。关于发行说明的输出位置，有人可以指出正确的方向吗？是否需要根据stagingdirectory、agent添加文件路径？

- task: PowerShell@2
  displayName: "Generate release notes"
  inputs:
    targetType: 'inline'
    script: |
     // some stuff that's irrelevant
    [string]$fileName = ".\ReleaseNotes.html"
    $releaseNotesHtml = "<!DOCTYPE html><html><head><title>Release Notes for" + 
        $applicationName + "</title></head>" +
        "<body>"

      foreach($item in $workItemResponse.value)
      {
        $releaseNotesHtml += 
          "<div>" +
            "<h4>" + $item.fields.'System.WorkItemType' + " " + 
                  $item.fields.'System.Title' + "</h4>" +
            "<b>Description</b><br/>" +
            "<p>" + $item.fields.'System.Description' + "</p>" +
          "</div>"
      }

      $releaseNotesHtml += "</body></html>"
      #=== Write the release notes to the output file === 
      Write-Host "Starting to write release notes to: "$fileName
      $releaseNotesHtml | out-file $fileName
      Write-Host "Finished writing release notes"

Answer 1

将文件复制到 $(Build.ArtifactStagingDirectory) 中，然后将其作为工件发布。参见 https://docs.microsoft.com/en-us/azure/devops/pipelines/artifacts/build-artifacts?view=azure-devops&tabs=yaml

在 Azure DevOps 构建 YML 中输出 HTML 文件

Outputting HTML file in Azure DevOps build YML

powershell

powershell-3.0

azure-devops