Question

我是Ruby和Rails的新手。我没理解为什么会发生以下情况。我使用SendGrid发送电子邮件。我已经定义了一个类和一个方法：

class EmailService
  include SendGrid

  def send_email
    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)

    response = sg.client.mail._('send').post(request_body: mail.to_json)
  end

end

这完美无缺。但是，我认为最好只初始化客户端一次，而不是每次调用该方法。所以我把它作为实例变量提取出来。

class EmailService
  include SendGrid

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

  def send_email
    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)

    response = @send_grid.client.mail._('send').post(request_body: mail.to_json)
  end

end

现在我得到#<NoMethodError: undefined method 'client' for nil:NilClass>。通过招揽我看到@send_grid是零。

我用EmailService.new.send_email调用方法。据我了解，@send_grid是一个实例变量，应该用类初始化。

为什么会这样？

Answer 1

将它放在构造函数中。在您的代码段中执行了分配表达式但在send_email方法中没有的另一个范围

class EmailService
  include SendGrid

  def initialize
    @send_grid = SendGrid::API.new(api_key: ENV['SENDGRID_API_KEY'])
  end

  def send_email
    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)

    response = @send_grid.client.mail._('send').post(request_body: mail.to_json)
  end
end

初始化为实例变量时，对象为nil

1 个答案: