Question

我正在Rails应用程序中集成SMS API。从用户表中获取数字时，我收到的错误为：

TypeError (no implicit conversion of nil into String):
  app/models/message.rb:16:in `+'
  app/models/message.rb:16:in `sms_message'

这是我的message.rb

def sms_message
    require "rubygems"
    require "net/https"
    require "uri"
    require "json"
    uname = "demo@gmail.com"
    hash = "54030737b105dsds7777 "
    numbers = numbers
    message = body
    sender = "TXTLCL" 
    requested_url = 'http://api.textlocal.in/send/?' + "username=" + uname + "&hash=" + hash + "&numbers=" + numbers + "&message=" + body + "&sender=" + sender
    uri = URI.parse(requested_url)
    http = Net::HTTP.start(uri.host, uri.port)
    request = Net::HTTP::Get.new(uri.request_uri)
    res = http.request(request)
    response = JSON.parse(res.body)
    puts (response)
end

def message_number
  numbers = Student.new do |u|
    u.phone1 = numbers
  end   
end

Answer 1

Rails有一个很好的方法叫做to_query。首先，像这样构建一个params哈希：

params = {username: uname, hash: hash, numbers: numbers, message: body, sender: sender}

然后，只需致电"http://api.textlocal.in/send/?#{params.to_query}"

params.to_query将处理零值。

但是，这只是编写代码的好方法。如果值为nil，您仍会得到相同的错误。最好是在进行API调用之前添加一些验证。此外，请确保您是否发送了API所期望的确切密钥。

Answer 2

您的numbers或body为nil

要避免这种情况，您可以将这些值转换为字符串或使用插值

转换为字符串

numbers = numbers.to_s
message = body.to_s # Use message instead of body while joining using +
requested_url = 
  'http://api.textlocal.in/send/?' + "username=" + uname + "&hash=" + hash + 
  "&numbers=" + numbers + "&message=" + message + "&sender=" + sender

使用插值

requested_url = 
  "http://api.textlocal.in/send/?username=#{uname}&hash=#{hash}&numbers=#{numbers}&message=#{message}&sender=#{sender}"

MessagesController中的TypeError＃create_message没有将nil隐式转换为String

2 个答案: