Ruby on Rails：比较两个字符串的数据库排序规则

问题描述

我有一个单词列表，想查找数据库中已经存在的单词。

I have a list of words and want to find which ones already exist in the database.

我决定使用 SELECT，而不是进行数十个SQL查询。 word 从 table 在 word IN（array_of_words）的位置和然后遍历结果。

Instead of making tens of SQL queries, I decided to use "SELECT word FROM table WHERE word IN(array_of_words)" and then loop through the result.

问题是数据库整理。

有许多不同的字符，MySQL将它们视为相同。但是，在Ruby代码中，string1不等于string2。

There are many different characters, which MySQL treats as the same. However, in Ruby code string1 would not be equal to string2.

例如：如果单词为šuo，则如果找到，数据库也可能返回 suo （而且还可以），但是，当我要检查时，如果找到šuo的东西，Ruby当然会返回false（šuo！= suo）。

For example: if the word is "šuo", database might also return "suo", if it's found (and it's ok), but, when I want to check, if something by "šuo" is found, Ruby, of course, returns false (šuo != suo).

那么，有没有办法以相同的排序规则比较Ruby中的两个字符串？

So, is there any way to compare two strings in Ruby in terms of the same collation?

推荐答案

我用过iconv类似于以下内容：

I've used iconv like this for something similar:

require 'iconv'

class String
  def to_ascii_iconv
    Iconv.new('ASCII//IGNORE//TRANSLIT', 'UTF-8').iconv(self).unpack('U*').select { |cp| cp < 127 }.pack('U*')
  end
end

puts 'suo'.to_ascii_iconv
# => suo
puts 'šuo'.to_ascii_iconv
# => suo
puts 'suo'.to_ascii_iconv == 'šuo'.to_ascii_iconv
# => true

希望有帮助！

Zubin

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