Curl Ansible convert CULL请求与--data urlencode

我想用Ansible通过RESTAPI将一个HTML文件上传到GitLab

我的卷发请求运行良好：

 curl -H "Content-Type: application/x-www-form-urlencoded" --request POST  --header 'PRIVATE-TOKEN: my_tocken' --data-urlencode content@/tmp/report.html 'https://my_server/api/v4/projects/3/repository/files/my_customer%2Freportname%2Ehtml?branch=master&commit_message=create%20a%20new%20report' -k

如何使用uri模块翻译它

uri:
  url: "https://my_server/api/v4/projects/3/repository/files/my_customer%2Freportname%2Ehtml?branch=master&commit_message=create%20a%20new%20report"
  validate_certs: no
  method: POST
  headers:
     Content-Type: application/x-www-form-urlencoded
     PRIVATE-TOKEN: "my_tocken"
  status_code: 200
  body: "data-urlencode=content@/tmp/report.html"

我得到：

 "json": {
    "error": "content is missing"
    },

如果/tmp/report.html位于Ansible controller计算机上，则：

uri:
  url: "https://my_server/api/v4/projects/3/repository/files/my_customer%2Freportname%2Ehtml?branch=master&commit_message=create%20a%20new%20report"
  validate_certs: no
  method: POST
  headers:
     Content-Type: application/x-www-form-urlencoded
     PRIVATE-TOKEN: "my_tocken"
  status_code: 200
  body: content={{ lookup('file', '/tmp/report.html') | urlencode }}

---

如果它位于不同的目标上，您需要先查看数据。

谢谢Techraf，您是对的

正确的要求是：

 - name:  Gitlab | upload file
       uri:
         url: "https://my_server/api/v4/projects/3/repository/files/my_customer%2Freportname%2Ehtml?branch=master&commit_message=create%20a%20new%20report%20for%20server"
        validate_certs: no
         method: POST
         headers:
            Content-Type: application/x-www-form-urlencoded
            PRIVATE-TOKEN: "my_tocken"
      status_code: 201
      body: "content={{ lookup('file', '/tmp/report.html')|urlencode }}"
    delegate_to: localhost

---

没有必要发布单独的答案，我可以编辑我的答案。更好的是，请修复您的问题，因为您编写的-data urlencode content@/tmp/report.html工作正常。