Sendgrid实现-没有将nil隐式转换为String

Question

我把我的环境设置成这样：

echo "export SENDGRID_API_KEY='xxxxxxxxx'" > sendgrid.env
echo "sendgrid.env" >> .gitignore
source ./sendgrid.env

安装了Sendgrid。

我试图运行的代码：

require 'sendgrid-ruby'
include SendGrid
require 'json'

def hello_world
  from = Email.new(email: 'test@example.com')
  to = Email.new(email: 'test@example.com')
  subject = 'Sending with SendGrid is Fun'
  content = Content.new(type: 'text/plain', value: 'and easy to do anywhere, even with Ruby')
  mail = Mail.new(from, subject, to, content)

  #previous version
  #sg = SendGrid::API.new(api_key: ENV['xxxxxxx'])

  #current version
  sg = SendGrid::API.new(api_key: ENV['SENDGRID_API_KEY'])
  response = sg.client.mail._('send').post(request_body: mail.to_json)
  puts response.status_code
  puts response.body
  puts response.headers
end

hello_world

但我得到了以下错误：

no implicit conversion of nil into String

在此文件中：

/ruby/gems/2.4.0/gems/sendgrid-ruby-5.2.0/lib/sendgrid/client.rb:24:in

我不知道这里出了什么问题..。

完全错误：

/Users/pimzonneveld/.rbenv/versions/2.4.3/lib/ruby/gems/2.4.0/gems/sendgrid-ruby-5.2.0/lib/sendgrid/client.rb:24:in `+': no implicit conversion of nil into String (TypeError)
    from /Users/pimzonneveld/.rbenv/versions/2.4.3/lib/ruby/gems/2.4.0/gems/sendgrid-ruby-5.2.0/lib/sendgrid/client.rb:24:in `initialize'
    from app/helpers/mail.rb:12:in `new'
    from app/helpers/mail.rb:12:in `hello_world'
    from app/helpers/mail.rb:19:in `<main>'

Answer 1

您混淆了env变量(xxxxxxxxx)的值和它的名称(SENDGRID_API_KEY)。

在代码中设置API键时，请使用

  sg = SendGrid::API.new(api_key: ENV['SENDGRID_API_KEY'])

而不是。

Answer 2

深是对的。/ruby/gems/2.4.0/gems/sendgrid-ruby-5.2.0/lib/sendgrid/client.rb:24:in指的是

"Authorization": "Bearer ' + @api_key + '",

因此，零到字符串的不隐式转换是完全合理的。

你应该换衣服

sg = SendGrid::API.new(api_key: ENV['xxxxxxx'])

适用于：

sg = SendGrid::API.new(api_key: ENV['SENDGRID_API_KEY'])