Git命令从问题编号中获取请求主体

Question

我想通过git命令，特别是gitPython库，从问题编号中获取pull请求正文、主题和URL。我该怎么做？

Answer 1

GitPython用于git相关对象，而Pull Request与GitHub相关，因此无法用于获取GitHub数据。

您可以使用GitHub的v4 GraphQL API通过以下查询获取拉取请求的详细信息

query {
repository(name: "gitPython",owner:"gitpython-developers"){
pullRequest(number:974){
body
title
url
}
}
}

上述查询的curl请求：

curl -L -X POST 'https://api.github.com/graphql' 
-H 'Authorization: bearer <token>' 
-H 'Content-Type: text/plain' 
--data-raw '{"query":"{n repository(name: "gitPython",owner:"gitpython-developers"){n pullRequest(number:974){n bodyn titlen urln }n }n }"'

对上述请求的响应：

{
"data": {
"repository": {
"pullRequest": {
"body": "Removed A from Dockerfile that I added accidentally. THIS WILL BREAK THE BUILD",
"title": "Remove A from Dockerfile",
"url": "https://github.com/gitpython-developers/GitPython/pull/974"
}
}
}
}

注意：您需要生成令牌以访问GraphQL API，您可以按照此处给出的步骤生成

或者，您甚至可以使用GitHub的v3neneneba API来获取拉取请求的详细信息，该请求的主体、标题和url字段将作为响应的一部分

GET https://api.github.com/repos/{owner}/{repoName}/pulls/{pullRequestNumber}
GET https://api.github.com/repos/gitpython-developers/GitPython/pulls/974