将两个哈希与自定义规则进行比较

Question

我需要比较 HTTP URL 中的查询。这是一个典型的查询：

a1 = "name=blabla&id=123456"

查询可能有变量，例如：

a2 = "name=blabla&id={CustomID}"

blur_compare(a1, a2) 应该是true。以下是其他情况：

a3 = "height=170&sex=male"
a4 = "name=blabla"
a5 = "name=blabla&id=654321"
a6 = "id={CustomeID}&name={CustomeName}"

blur_compare(a3, a2) #=> false, params doesn't match.
blur_compare(a4, a2) #=> false, param's number doesn't match.
blur_compare(a5, a1) #=> false, the id doesn't match.
blur_compare(a6, a2) #=> true. order doesn't matter.

我正在考虑将字符串转移到哈希，例如

{"name" => "blabla", "id" => "123456"}

然后进行比较。但是我的代码很乱，又长又丑。很多if-else 条件。我想知道是否有更好的方法来做到这一点。

抱歉，这是代码。只是觉得有点害羞分享它，尤其是当它写得不好的时候。但我正在学习:)。

path_query.rb

class PathQuery

  def self.is_var?(str)
    regex = /^{.*}$/
    str =~ regex ? true : false
  end

  def self.blur_match_query?(query1, query2)
    # deal with the condition when query1 or query2 maybe nil
    if query1 == nil && query2 == nil
      return true
    elsif query1 == nil || query2 == nil
      return false
    else
    end

    res = true
    # return false directly if number of keys is not equal
    if query1.keys.size == query2.keys.size
      query1.each{|k,v|
        # when the value of same key is diff, then need to found out if one of them is variable.
        if v != query2[k]
          if not (PathQuery.is_var?(v) || PathQuery.is_var?(query2[k]))
            res = false
            break
          end
        end
      }
    else
      res = false
    end
    res
  end

end

test_path_query.rb

class TestPathQuery < Minitest::Test
  def test_blur_match_query?()
    a1 = {"name" => "blabla", "id" => "123456"}
    a2 = {"name" => "blabla", "id" => "{CustomID}"}
    a3 = {"height" => "170", "sex" => "male"}
    a4 = {"name" => "blabla"}
    a5 = {"name" => "blabla", "id" => "654321"}
    a6 = {"id" => "{CustomeID}", "name" => "{CustomeName}"}

    assert_equal true, PathQuery.blur_match_query?(a1, a2)
    assert_equal false, PathQuery.blur_match_query?(a1, a3)
    assert_equal false, PathQuery.blur_match_query?(a4, a2)
    assert_equal false, PathQuery.blur_match_query?(a5, a1)
    assert_equal true, PathQuery.blur_match_query?(a6, a2)
    assert_equal true, PathQuery.blur_match_query?(nil, nil)
    assert_equal false, PathQuery.blur_match_query?(nil, a2)
  end
end

Answer 1

试试这个

def self.blur_match_query?(query1, query2)
  (query1.keys-query2.keys).empty?
end