更好的方式来提取电话号码并重新格式化？(Better way to extract phone number and reformat?)

更好的方式来提取电话号码并重新格式化？(Better way to extract phone number and reformat?)

各种格式的电话号码数据（我选择这些数据是因为进入的数据不可靠，而不是预期的格式）：

+1 480-874-4666 404-581-4000 (805) 682-4726 978-851-7321, Ext 2606 413- 658-1100 (513) 287-7000,Toll Free (800) 733-2077 1 (813) 274-8130 212-363-3200,Media Relations: 212-668-2251. 323/221-2164

我的Ruby代码提取所有数字，删除美国国家代码的任何前导1，然后使用前10位数字创建所需格式的“新”电话号码：

nums = phone_number_string.scan(/[0-9]+/) if nums.size > 0 all_nums = nums.join all_nums = all_nums[0..0] == "1" ? all_nums[1..-1] : all_nums if all_nums.size >= 10 ten_nums = all_nums[0..9] final_phone = "#{ten_nums[0..2]}-#{ten_nums[3..5]}-#{ten_nums[6..9]}" else final_phone = "" end puts "#{final_phone}" else puts "No number to fix." end

结果非常好 ！

480-874-4666 404-581-4000 805-682-4726 978-851-7321 413-658-1100 513-287-7000 813-274-8130 212-363-3200 323-221-2164

但是，我认为还有更好的办法。 你可以重构这个更高效，更清晰或更有用吗？

Phone number data in various formats (I've chosen these because the data coming in is unreliable and not in expected formats):

+1 480-874-4666 404-581-4000 (805) 682-4726 978-851-7321, Ext 2606 413- 658-1100 (513) 287-7000,Toll Free (800) 733-2077 1 (813) 274-8130 212-363-3200,Media Relations: 212-668-2251. 323/221-2164

My Ruby code to extract all the digits, remove any leading 1's for the USA country code, then use the first 10 digits to create the "new" phone number in a format desired:

nums = phone_number_string.scan(/[0-9]+/) if nums.size > 0 all_nums = nums.join all_nums = all_nums[0..0] == "1" ? all_nums[1..-1] : all_nums if all_nums.size >= 10 ten_nums = all_nums[0..9] final_phone = "#{ten_nums[0..2]}-#{ten_nums[3..5]}-#{ten_nums[6..9]}" else final_phone = "" end puts "#{final_phone}" else puts "No number to fix." end

The results are very good!

480-874-4666 404-581-4000 805-682-4726 978-851-7321 413-658-1100 513-287-7000 813-274-8130 212-363-3200 323-221-2164

But, I think there's a better way. Can you refactor this to be more efficient, more legible, or more useful?

最满意答案

这是一个简单的方法，只使用正则表达式和替换：

def extract_phone_number(input) if input.gsub(/\D/, "").match(/^1?(\d{3})(\d{3})(\d{4})/) [$1, $2, $3].join("-") end end

(\d{3})(\d{3})(\d{4})所有非数字（ \D ） (\d{3})(\d{3})(\d{4}) ）和格式。

这是测试：

test_data = { "+1 480-874-4666" => "480-874-4666", "404-581-4000" => "404-581-4000", "(805) 682-4726" => "805-682-4726", "978-851-7321, Ext 2606" => "978-851-7321", "413- 658-1100" => "413-658-1100", "(513) 287-7000,Toll Free (800) 733-2077" => "513-287-7000", "1 (813) 274-8130" => "813-274-8130", "212-363-3200,Media Relations: 212-668-2251." => "212-363-3200", "323/221-2164" => "323-221-2164", "" => nil, "foobar" => nil, "1234567" => nil, } test_data.each do |input, expected_output| extracted = extract_phone_number(input) print "FAIL (expected #{expected_output}): " unless extracted == expected_output puts extracted end

Here's a much simpler approach using only regexes and substitution:

def extract_phone_number(input) if input.gsub(/\D/, "").match(/^1?(\d{3})(\d{3})(\d{4})/) [$1, $2, $3].join("-") end end

This strips all non-digits (\D), skips an optional leading one (^1?), then extracts the first remaining 10 digits in chunks ((\d{3})(\d{3})(\d{4})) and formats.

Here's the test:

test_data = { "+1 480-874-4666" => "480-874-4666", "404-581-4000" => "404-581-4000", "(805) 682-4726" => "805-682-4726", "978-851-7321, Ext 2606" => "978-851-7321", "413- 658-1100" => "413-658-1100", "(513) 287-7000,Toll Free (800) 733-2077" => "513-287-7000", "1 (813) 274-8130" => "813-274-8130", "212-363-3200,Media Relations: 212-668-2251." => "212-363-3200", "323/221-2164" => "323-221-2164", "" => nil, "foobar" => nil, "1234567" => nil, } test_data.each do |input, expected_output| extracted = extract_phone_number(input) print "FAIL (expected #{expected_output}): " unless extracted == expected_output puts extracted end