通过 Github API 判断一个提交是否为合并提交

Question

是否可以通过 Github API 判断一个提交是否为合并提交？它是从哪个分支合并的？

在比较一个分支上的两个提交时，我试图查看所有更改的文件，但我想忽略它所分支的分支的合并提交所做的任何更改。任何从分支分支出来的分支都不会忽略它们的合并提交。

Answer 1

curl \
-H "Accept: application/vnd.github.v3+json" \
https://api.github.com/repos/octocat/hello-world/commits

您可以使用此 API 来获取所有提交。获取所有提交后，根据 GitHub 的 API 文档，响应将如下所示。

[


{
    "url": "https://api.github.com/repos/octocat/Hello-World/commits/6dcb09b5b57875f334f61aebed695e2e4193db5e",
    "sha": "6dcb09b5b57875f334f61aebed695e2e4193db5e",
    "node_id": "MDY6Q29tbWl0NmRjYjA5YjViNTc4NzVmMzM0ZjYxYWViZWQ2OTVlMmU0MTkzZGI1ZQ==",
    "html_url": "https://github.com/octocat/Hello-World/commit/6dcb09b5b57875f334f61aebed695e2e4193db5e",
    "comments_url": "https://api.github.com/repos/octocat/Hello-World/commits/6dcb09b5b57875f334f61aebed695e2e4193db5e/comments",
    "commit": {
      "url": "https://api.github.com/repos/octocat/Hello-World/git/commits/6dcb09b5b57875f334f61aebed695e2e4193db5e",
      "author": {
        "name": "Monalisa Octocat",
        "email": "support@github.com",
        "date": "2011-04-14T16:00:49Z"
      },
      "committer": {
        "name": "Monalisa Octocat",
        "email": "support@github.com",
        "date": "2011-04-14T16:00:49Z"
      },
      "message": "Fix all the bugs",
      "tree": {
        "url": "https://api.github.com/repos/octocat/Hello-World/tree/6dcb09b5b57875f334f61aebed695e2e4193db5e",
        "sha": "6dcb09b5b57875f334f61aebed695e2e4193db5e"
      },
      "comment_count": 0,
      "verification": {
        "verified": false,
        "reason": "unsigned",
        "signature": null,
        "payload": null
      }
    },
    "author": {
      "login": "octocat",
      "id": 1,
      "node_id": "MDQ6VXNlcjE=",
      "avatar_url": "https://github.com/images/error/octocat_happy.gif",
      "gravatar_id": "",
      "url": "https://api.github.com/users/octocat",
      "html_url": "https://github.com/octocat",
      "followers_url": "https://api.github.com/users/octocat/followers",
      "following_url": "https://api.github.com/users/octocat/following{/other_user}",
      "gists_url": "https://api.github.com/users/octocat/gists{/gist_id}",
      "starred_url": "https://api.github.com/users/octocat/starred{/owner}{/repo}",
      "subscriptions_url": "https://api.github.com/users/octocat/subscriptions",
      "organizations_url": "https://api.github.com/users/octocat/orgs",
      "repos_url": "https://api.github.com/users/octocat/repos",
      "events_url": "https://api.github.com/users/octocat/events{/privacy}",
      "received_events_url": "https://api.github.com/users/octocat/received_events",
      "type": "User",
      "site_admin": false
    },
    "committer": {
      "login": "octocat",
      "id": 1,
      "node_id": "MDQ6VXNlcjE=",
      "avatar_url": "https://github.com/images/error/octocat_happy.gif",
      "gravatar_id": "",
      "url": "https://api.github.com/users/octocat",
      "html_url": "https://github.com/octocat",
      "followers_url": "https://api.github.com/users/octocat/followers",
      "following_url": "https://api.github.com/users/octocat/following{/other_user}",
      "gists_url": "https://api.github.com/users/octocat/gists{/gist_id}",
      "starred_url": "https://api.github.com/users/octocat/starred{/owner}{/repo}",
      "subscriptions_url": "https://api.github.com/users/octocat/subscriptions",
      "organizations_url": "https://api.github.com/users/octocat/orgs",
      "repos_url": "https://api.github.com/users/octocat/repos",
      "events_url": "https://api.github.com/users/octocat/events{/privacy}",
      "received_events_url": "https://api.github.com/users/octocat/received_events",
      "type": "User",
      "site_admin": false
    },
    "parents": [
      {
        "url": "https://api.github.com/repos/octocat/Hello-World/commits/6dcb09b5b57875f334f61aebed695e2e4193db5e",
        "sha": "6dcb09b5b57875f334f61aebed695e2e4193db5e"
      }
    ]
  }
]

在上面的payload中，对于数组中的每一项，如果“parents”键中有多行，那么这意味着它是一个合并提交。上面的有效负载示例在“parents”数组中只有一行，因此它不是合并提交。

要找出它是从哪个分支合并的，我相信目前没有使用Github API的解决方案。

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