如何从json文件中排除数据以转换为csv文件powershell

如何解决如何从json文件中排除数据以转换为csv文件powershell

我想将json文件转换为csv-

{
  "count": 28,"value": [
  {
    "commitId": "65bb6a911872c314a9225815007d74a","author": {
    "name": "john doe","email": "john.doe@gmail.com","date": "2020-06-09T17:03:33Z"
     },"committer": {
    "name": "john doe","date": "2020-06-09T17:03:33Z"
    },"comment": "Merge pull request 3 from dev into master","changeCounts": {
    "Add": 6,"Edit": 0,"Delete": 0
     },"url": "https://dev.azure.com/","remoteUrl": "https://dev.azure.com/"
     },{
     "commitId": "bdcb4a1e1e671c15333eacc31aa9795fe32","author": {
     "name": "john doe","date": "2020-06-09T17:03:33Z"
      },"committer": {
     "name": "john doe","comment": "Updated FGDH","changeCounts": {
     "Add": 0,"Edit": 1,"Delete": 0
      },"remoteUrl": "https://dev.azure.com/"
    }
  ]
 }

我不希望CSV文件中的所有字段。我只需要commitid,commiter.name,commiter.date和注释。

 Get-Content "\commitinfo.json" -Raw | ConvertFrom-Json | Select -Expand value |Export-Csv 
 "\commitinfo.csv" 

以此获取所有数据,如何获取所选数据?

另一个json文件-

{
 "value": [
  {
    "id": "5c264dd2-bbcf-4537-8a63-19a0a4d2dc","name": "Develop","project": {
    "id": "0042dc5c-fd13-4e3c-bfd7-7feb52e287","name": "test","state": "wellFormed","revision": 11,"visibility": "private","lastUpdateTime": "2020-04-15T04:04:30.01Z"
    },"defaultBranch": "refs/heads/master","size": 55438,"remoteUrl": "https://dev.azure.com","sshUrl": "git@ssh.dev.azure.com:v3","webUrl": "https://dev.azure.com/"
   },{
    "id": "3219e8e2-281d-40ad-81c8-1cecf8","name": "automation","project": {
    "id": "0e2df786-94a0-42c8-b068-c2656b","revision": 19,"lastUpdateTime": "2020-05-02T03:32:50.937Z"
    },"size": 0,"remoteUrl": "https://dev.azure.com/","webUrl": "https://dev.azure.com/"
    }
  ]
 }

从该文件中,我只想选择大小为-0的项目,CSV应该看起来像这样

     name   size 
     a       0
     b       0

解决方法

注意:为简洁起见,省略了以下命令Export-Csv的使用。

您需要将Select-Objectcalculated properties结合使用:

(Get-Content commitinfo.json -Raw | ConvertFrom-Json).value |
  Select-Object commitid,@{n='committerName'; e={ $_.committer.name } },@{n='committerDate'; e={ $_.committer.date } },comment

在第二种情况下,添加Where-Object以过滤感兴趣的对象:

(Get-Content other.json -Raw | ConvertFrom-Json).value |
  Where-Object size -eq 0 |
    Select-Object name,size

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