Question

假设我在管道中具有以下参数：

parameters:
- name: environment
  displayName: 'Environment'
  default: 'SERVER_0562'
  type: string
  values: 
  - 'SERVER_0562'
  - 'SERVER_5149'
  - 'SERVER_1892'

但是我不希望在运行管道的任何人从下拉列表中选择Environment时都看到这些名称。然后，我希望看到以下名称：

- QA Server
- DEV Server
- PROD Server

是否有任何“简便”的方法可以做到这一点，或者唯一的方法就是处理条件条件？

${{ if eq(parameters.environment, 'DEV Server') }}:

Answer 1

请检查以下内容：


parameters:
- name: image
  displayName: Pool Image
  type: string
  default: Ubuntu
  values:
  - Windows
  - Ubuntu
  - MacOS
  
trigger: none

jobs:
- job: build
  displayName: build
  pool: 
    ${{ if eq(parameters.image, 'Ubuntu') }}:
      vmImage: 'ubuntu-latest'
    ${{ if eq(parameters.image, 'Windows') }}:
      vmImage: 'windows-latest'
    ${{ if eq(parameters.image, 'MacOS') }}:
      vmImage: 'macOS-latest'
  steps:
  - script: echo building $(Build.BuildNumber) with ${{ parameters.image }}

更改问题后，我认为这可能对您有帮助


parameters:
- name: environment
  displayName: 'Environment'
  default: 'QA Server'
  type: string
  values: 
  - 'QA Server'
  - 'DEV Server'
  - 'PROD Server'
  
trigger: none

variables:
- name: QA
  value: 'SERVER_0562'
- name: DEV
  value: 'SERVER_5149'
- name: PROD
  value: 'SERVER_1892'

jobs:
- job: build
  displayName: build
  pool: 
    vmImage: 'windows-latest'
  steps:
  - script: echo building $(Build.BuildNumber) with ${{ parameters.environment }}
  - powershell: |
      
      If('${{ parameters.environment }}' -eq 'QA Server') {
        Write-Host '##vso[task.setvariable variable=folderName;isOutput=true]$(QA)'
      } ElseIf('${{ parameters.environment }}' -eq 'DEV Server') {
        Write-Host '##vso[task.setvariable variable=folderName;isOutput=true]$(DEV)'
      } ElseIf('${{ parameters.environment }}' -eq 'PROD Server') {
        Write-Host '##vso[task.setvariable variable=folderName;isOutput=true]$(PROD)'
      } Else {
        Write-Host 'I am here'
      }
    name: setFolderName
  - powershell: |
      Write-Host '$(setFolderName.folderName)'

Answer 2

受@Krzysztof Madej的贡献启发，我最终通过conditionally assigning variables解决了自己的问题：

         let popupShake = {
            outFields: ["*"],
           "title": function (feature){ <------------- breaks popup, but content below works
                let spl = document.createElement("span");
                spl.className = "name_plc";
                dlv.innerHTML = "<span class='name_plc'></span>";

                return spl;
           },
           "content": function (feature){
             let name_plc;
             let dlv = document.createElement("div");
             dlv.className = "popd";
             dlv.innerHTML = `<b><span class='name_plc'></span></b><br>ID: <span class="ida">${feature.graphic.attributes.id}</span><br> URL: <a href="${feature.graphic.attributes.url}" target="_blank">View</a> <br> Updated: ${feature.graphic.attributes.updated} <br>Grid_value: ${feature.graphic.attributes.grid_value}<br> Event Time: ${feature.graphic.attributes.eventTime}`;
              
             setTimeout(() => {
                getName();
             }, 20);
             ................

我认为这是一个更容易实现的方法，因为我不依赖于parameters: - name: environment displayName: 'Environment' default: 'QA Server' type: string values: - 'QA Server' - 'DEV Server' - 'PROD Server' trigger: none variables: ${{ if eq(parameters.environment, 'QA Server') }}: bacon: SERVER_0562 ${{ if eq(parameters.environment, 'DEV Server') }}: bacon: SERVER_5149 ${{ if eq(parameters.environment, 'PROD Server') }}: bacon: SERVER_1892 jobs: - job: build displayName: build steps: - script: echo building $(Build.BuildNumber) with ${{ variables.bacon }}脚本。

在Azure DevOps中是否可以对管道参数使用别名？

2 个答案: