我面临着API的几个问题,
第一:
发送方法询问’id'(消息ID或线程ID)..但为什么?
我正在发送新消息,所以它不应该要求.根据Gmail Api documnetation
它是可选的.
https://developers.google.com/gmail/api/v1/reference/users/messages/send
ArgumentError: Missing required parameters: id.
第二:
即使在指定消息ID后,它也会返回此消息.
Your client has issued a malformed or illegal request.
码
require 'mime' include MIME msg = Mail.new msg.date = Time.Now msg.subject = 'This is important' msg.headers.set('Priority','urgent') msg.body = Text.new('hello,world!','plain','charset' => 'us-ascii') msg.from = {'hi@gmail.com' => 'Boss Man'} msg.to = { 'list@example.com' => nil,'john@example.com' => 'John Doe','jane@example.com' => 'Jane Doe',} @email = @google_api_client.execute( api_method: @gmail.users.messages.send(:get),body_object: { raw: Base64.urlsafe_encode64(msg.to_s) },parameters: { userId: 'me' } )
并且当然认证工作正常.
其他一些方法也很好
喜欢:
get list of messages(Users.messages.list) get single message(Users.messages.get)
但
发送消息不起作用.
解决方法
我认为
@gmail.users.messages.send(:get) is equal to @gmail.users.messages.get
因为“.send”是ruby方法
@gmail.users.messages.to_h['gmail.users.messages.send']
例:
msg = Mail.new msg.date = Time.Now msg.subject = options[:subject] msg.body = Text.new(options[:message]) msg.from = {@_user.email => @_user.full_name} msg.to = { options[:to] => options[:to_name] } @email = @google_api_client.execute( api_method: @gmail.users.messages.to_h['gmail.users.messages.send'],parameters: { userId: 'me',} )
谢谢.