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从 LogicApp 使用“向 DevOps 发送 HTTP 请求”时获取身份验证错误

[英]Getting authentication Error when using "Send a HTTP Request to DevOps" from LogicApp

I am Trying to automate the approval of Pull request from logic app using "Send a HTTP Request to DevOps" action.我正在尝试使用“向 DevOps 发送 HTTP 请求”操作自动批准来自逻辑应用程序的拉取请求。 Getting below authentication error.获取以下身份验证错误。

Tried below options but nothing worked.尝试了以下选项,但没有任何效果。

  1. Created new PAT with Full Access and passed to header.创建具有完全访问权限的新 PAT 并传递给 header。
  2. Enabled "Third-party application access via OAuth" in org policies在组织策略中启用“通过 OAuth 的第三方应用程序访问”
  3. Tried Basic, Bearer authentication尝试过基本,不记名身份验证

在此处输入图像描述

Error: { "status": 401, "message": "TF400813: The user 'aaaaaaaa-aaaa-aaaa-aaaa-aaaaaaaaaaaa' is not authorized to access this resource.\r\nAzure DevOps ActivityId: abcd1234-ac34-ac34-abcd1234\r\nDetails: {\"$id\":\"1\",\"innerException\":null,\"message\":\"TF400813: The user 'aaaaaaaa-aaaa-aaaa-aaaa-aaaaaaaaaaaa' is not authorized to access this resource.\",\"typeName\":\"Microsoft.TeamFoundation.Framework.Server.UnauthorizedRequestException, Microsoft.TeamFoundation.Framework.Server\",\"typeKey\":\"UnauthorizedRequestException\",\"errorCode\":0,\"eventId\":3000}\r\nclientRequestId: abcd1234-ac34-ac34-abcd1234", "error": { "message": "TF400813: The user 'aaaaaaaa-aaaa-aaaa-aaaa-aaaaaaaaaaaa' is not authorized to access this resource.\r\nAzure DevOps ActivityId: abcd1234-ac34-ac34-abcd1234\r\nDetails: {\"$id\":\"1\",\"innerException\":null,\"message\":\"TF400813: The user 'aaaaaaaa-aaaa-aaaa-aaaa-aaaaaaaaaaaa' is not authorized to access this resource.\",\"typeName\":\"Microsoft.TeamFoundation.Framework.Server.UnauthorizedRequestException, Microsoft.TeamFoundation.Framework.Server\",\"typeKey\":\"UnauthorizedRequestException\",\"errorCode\":0,\"eventId\":3000}" }, "source": "vsts-eus.azconn-eus-003.p.azurewebsites.net" }

You need to make sure you encode the PAT to base64 with no username when using basic auth in Azure DevOps.在 Azure DevOps 中使用基本身份验证时,您需要确保将 PAT 编码为 base64 而没有用户名。

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