获取 GitHub 存储库中文件的所有提交

Question

如果我使用下面的url,

https://api.github.com/repos/raspberrypi/linux/commits?path=drivers/gpu/drm/i915/intel_display.c

我获得了前 30 次提交。但是那个文件有大约 3000 次提交。我怎样才能得到这个文件的所有提交？

Answer 1

您可以查询与给定文件关联的所有提交：

var stats = new Dictionary<string, int>();

foreach (var path in allPaths)
{
    var request = new CommitRequest { Path = "path/to/file.cs" };
    var commitsForFile = await client.Repository.Commit.GetAll(Owner, Name, request);
    stats.Add(path, commitsForFile.Length);
}

此代码使用由GitHub官方维护和支持的Octokit。所有 Octokit 库均在 MIT 许可证下发布，这意味着它们可以在任何项目中修改和使用。

取自 - https://github.com/octokit/octokit.net/issues/1293

Answer 2

在查看 Pagination on the GitHub API documentation. But, you have to be careful about the API request rate as discussed here 之后，我最终做了类似以下代码的事情。

import requests
import json

url = 'https://api.github.com/repos/raspberrypi/linux/commits?access_token=<your github user OAuth token>&path=drivers/gpu/drm/i915/intel_display.c'
commits = []

i = 1
r = requests.get(url + '&per_page=100&page=' + str(i))
while len(r.json()) != 0:
    commits.extend(r.json())
    i += 1
    r = requests.get(url + '&per_page=100&page=' + str(i))

with open('commits.txt', 'w') as outfile:
    json.dump(commits, outfile)

获取 GitHub 存储库中文件的所有提交

Get all commits of a file in a GitHub repo

github-api