红宝石匹配字符串为字符串

Question

的阵列

可能重复：
In Ruby, how do I find out if a string is not in an array?

现在我有这样的代码：

partial_update = true if session[:intended_params][:controller] == 'foo' 
partial_update = true if session[:intended_params][:controller] == 'bar' 
partial_update = true if session[:intended_params][:controller] == 'ding' 
partial_update = true if session[:intended_params][:controller] == 'dong' 
partial_update = true if session[:intended_params][:controller] == 'up' 
partial_update = true if session[:intended_params][:controller] == 'down' 
partial_update = true if session[:intended_params][:controller] == 'bing' 
# ETC. 

我想改变这

array = %w[foo bar ding dong up down bing] 
partial_update = true if session[:intended_params][:controller] == array.any_of_the_possibilites 

有没有一个很好的方法来做到这一点？

Answer 1

使用红宝石Array (Enumerable) include? method：

array = %w[foo bar ding dong up down bing] 
partial_update = true if array.include?(session[:intended_params][:controller]) 

Answer 2

case session[:intended_params][:controller] 
when *%w[foo bar ding dong up down bing] 
    partial_update = true 
end 

可能被滥用case声明虽然