在 Azure DevOps 中输出 HTML 文件构建 YML

问题描述

我有一个创建 html 文件的 powershell 脚本，而且我是 Azure DevOps 的新手。这个 html 文件必须与构建工件一起编译和生成。我不确定我是怎么做的。有人可以指出我在何处输出发行说明的正确方向吗？我是否需要根据 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"

解决方法

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

azure-devops powershell powershell powershell-3.0