将curl命令转换为httparty

Question

我试图在Mailchimp V3列表中添加与HTTParty的合并字段,但无法将curl转换为HTTParty格式.
卷曲请求格式工作正常:

curl --request POST \
     --url 'https://usxx.api.mailchimp.com/3.0/lists/17efad7sd4/merge-fields' \
     --user '12:d1c1d99dr5000c63f0f73f64b88e852e-xx' \
     --header 'content-type: application/json' \
     --data '{"name":"FAVORITEJOKE", "type":"text"}' \
     --include

缺少错误API密钥的Httparty格式

response = HTTParty.post("https://us12.api.mailchimp.com/3.0/lists/17efad7sde/merge-fields", 
                :body => { 
                  :user => '12:d1c1d99dr5000c63f0f73f64b88e852e-xx',
                  :data =>  '{"name":"FAVORITEJOKE", "type":"text"}',
                  :include => ''
                }.to_json,
                :headers => { 'Content-Type' => 'application/json' } )

我也尝试没有包含选项但不工作

Answer 1

您的代码中存在多个错误.

• curl user是基本的auth用户,但是您在请求的有效负载中传递它
• data是有效负载,而是将其作为有效负载中的节点传递,然后对其进行双重序列化
• include在那里没有意义,它不是有效载荷项

这应该是正确的版本.请花点时间阅读HTTParty和curl文档和理解的差异.

HTTParty.post(
  "https://us12.api.mailchimp.com/3.0/lists/17efad7sde/merge-fields", 
  basic_auth: { username: "12", password: "d1c1d99dr5000c63f0f73f64b88e852e-xx" },
  headers: { 'Content-Type' => 'application/json' },
  body: {
    name: "FAVORITEJOKE",
    type: "text",
  }.to_json
)