如何从json文件中排除数据以转换为csv文件powershell [英] How to exclude data from json file to convert into csv file powershell
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问题描述
我想将json文件转换为csv-
I want to convert my json file into 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",
"email": "john.doe@gmail.com",
"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",
"email": "john.doe@gmail.com",
"date": "2020-06-09T17:03:33Z"
},
"committer": {
"name": "john doe",
"email": "john.doe@gmail.com",
"date": "2020-06-09T17:03:33Z"
},
"comment": "Updated FGDH",
"changeCounts": {
"Add": 0,
"Edit": 1,
"Delete": 0
},
"url": "https://dev.azure.com/",
"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.
I don't want all the fields in my CSV file. I want commitid,commiter.name, commiter.date and comment only.
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?
with this I get all data , How can I get selected data ?
另一个json文件-
{
"value": [
{
"id": "5c264dd2-bbcf-4537-8a63-19a0a4d2dc",
"name": "Develop",
"url": "https://dev.azure.com/",
"project": {
"id": "0042dc5c-fd13-4e3c-bfd7-7feb52e287",
"name": "test",
"url": "https://dev.azure.com/",
"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",
"url": "https://dev.azure.com/",
"project": {
"id": "0e2df786-94a0-42c8-b068-c2656b",
"name": "automation",
"url": "https://dev.azure.com/",
"state": "wellFormed",
"revision": 19,
"visibility": "private",
"lastUpdateTime": "2020-05-02T03:32:50.937Z"
},
"size": 0,
"remoteUrl": "https://dev.azure.com/",
"sshUrl": "git@ssh.dev.azure.com:v3",
"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
from this file I only want to select projects with size - 0 , CSV should look like this
name size
a 0
b 0
推荐答案
注意:为简便起见,省略了以下命令到 Export-Csv
.
您需要将 Select-Object
与计算出的属性:
(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
进行过滤在感兴趣的对象中:
In the second case, add Where-Object
to filter in the objects of interest:
(Get-Content other.json -Raw | ConvertFrom-Json).value |
Where-Object size -eq 0 |
Select-Object name, size
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