I have a string representation of a MD5 hex digest for a file, that I want to convert to base64 in order to use the Content-MD5 HTTP header when uploading it. Is there a clearer or more efficient mechanism to do than the following?
def hex_to_base64_digest(hexdigest)
[[hexdigest].pack("H*")].pack("m").strip
end
hex_digest = "65a8e27d8879283831b664bd8b7f0ad4"
expected_base64_digest = "ZajifYh5KDgxtmS9i38K1A=="
raise "Does not match" unless hex_to_base64_digest(hex_digest) === expected_base64_digest
Seems pretty clear and efficient to me. You can save the call to strip by specifying 0 count for the 'm' pack format (if count is 0, no line feed are added, see RFC 4648)
def hex_to_base64_digest(hexdigest)
[[hexdigest].pack("H*")].pack("m0")
end
Related
I have a string representation of a MD5 hex digest for a file, that I want to convert to base64 in order to use the Content-MD5 HTTP header when uploading it. Is there a clearer or more efficient mechanism to do than the following?
def hex_to_base64_digest(hexdigest)
[[hexdigest].pack("H*")].pack("m").strip
end
hex_digest = "65a8e27d8879283831b664bd8b7f0ad4"
expected_base64_digest = "ZajifYh5KDgxtmS9i38K1A=="
raise "Does not match" unless hex_to_base64_digest(hex_digest) === expected_base64_digest
Seems pretty clear and efficient to me. You can save the call to strip by specifying 0 count for the 'm' pack format (if count is 0, no line feed are added, see RFC 4648)
def hex_to_base64_digest(hexdigest)
[[hexdigest].pack("H*")].pack("m0")
end
i may recieve these two strings:
base = Base64.encode64(File.open("/home/usr/Desktop/test", "rb").read)
=> "YQo=\n"
string = File.open("/home/usr/Desktop/test", "rb").read
=> "a\n"
what i have tried so far is to check string with regular expression i-e. /([A-Za-z0-9+\/]{4})*([A-Za-z0-9+\/]{4}|[A-Za-z0-9+\/]{3}=|[A-Za-z0-9+\/]{2}==$)/ but this would be very heavy if the file is big.
I also have tried base.encoding.name and string.encoding.name but both returns the same.
I have also seen this post and got regular expression solution but any other solution ?
Any idea ? I just want to get is the string is actually text or base64 encoded text....
You can use something like this, not very performant but you are guaranteed not to get false positives:
require 'base64'
def base64?(value)
value.is_a?(String) && Base64.strict_encode64(Base64.decode64(value)) == value
end
The use of strict_encode64 versus encode64 prevents Ruby from inadvertently inserting newlines if you have a long string. See this post for details.
Documentation for a Third-Party API that I'm working with states:
"[O]ur API only accepts padded Base64 encoded strings."
What are "padded Base64 encoded strings" and how can I generate them in Ruby. The code below is my first attempt at creating JSON formatted data converted to Base64.
xa = Base64.encode64(a.to_json)
The padding they are talking about is actually part of Base64 itself. It's the "=" and "==" at the end. Base64 encodes packets of 3 bytes into 4 encoded characters. So if your input data has length n and
n % 3 = 1 => "==" at the end for padding
n % 3 = 2 => "=" at the end for padding
No need for you to change your code.
It looks like the base64 library pads by default; padding in Base64 would be the = characters on the end of the data.
You can see this by running the following in the irb console:
irb(main):002:0> require 'base64'
=> true
irb(main):003:0> Base64.encode64('a')
=> "YQ==\n"
Without the padding, you couldn't be sure whether YQ was everything or whether it was missing something.
I am trying to convert a hex value to a binary value (each bit in the hex string should have an equivalent four bit binary value). I was advised to use this:
num = "0ff" # (say for eg.)
bin = "%0#{num.size*4}b" % num.hex.to_i
This gives me the correct output 000011111111. I am confused with how this works, especially %0#{num.size*4}b. Could someone help me with this?
You can also do:
num = "0ff"
num.hex.to_s(2).rjust(num.size*4, '0')
You may have already figured out, but, num.size*4 is the number of digits that you want to pad the output up to with 0 because one hexadecimal digit is represented by four (log_2 16 = 4) binary digits.
You'll find the answer in the documentation of Kernel#sprintf (as pointed out by the docs for String#%):
http://www.ruby-doc.org/core/classes/Kernel.html#M001433
This is the most straightforward solution I found to convert from hexadecimal to binary:
['DEADBEEF'].pack('H*').unpack('B*').first # => "11011110101011011011111011101111"
And from binary to hexadecimal:
['11011110101011011011111011101111'].pack('B*').unpack1('H*') # => "deadbeef"
Here you can find more information:
Array#pack: https://ruby-doc.org/core-2.7.1/Array.html#method-i-pack
String#unpack1 (similar to unpack): https://ruby-doc.org/core-2.7.1/String.html#method-i-unpack1
This doesn't answer your original question, but I would assume that a lot of people coming here are, instead of looking to turn hexadecimal to actual "0s and 1s" binary output, to decode hexadecimal to a byte string representation (in the spirit of such utilities as hex2bin). As such, here is a good method for doing exactly that:
def hex_to_bin(hex)
# Prepend a '0' for padding if you don't have an even number of chars
hex = '0' << hex unless (hex.length % 2) == 0
hex.scan(/[A-Fa-f0-9]{2}/).inject('') { |encoded, byte| encoded << [byte].pack('H2') }
end
Getting back to hex again is much easier:
def bin_to_hex(bin)
bin.unpack('H*').first
end
Converting the string of hex digits back to binary is just as easy. Take the hex digits two at a time (since each byte can range from 00 to FF), convert the digits to a character, and join them back together.
def hex_to_bin(s) s.scan(/../).map { |x| x.hex.chr }.join end
i have ruby code to parse data in excel file using Parseexcel gem. I need to save 2 columns in that file into a Hash, here is my code:
worksheet.each { |row|
if row != nil
key = row.at(1).to_s.strip
value = row.at(0).to_s.strip
if !parts.has_key?(key) and key.length > 0
parts[key] = value
end
end
}
however it still save duplicate keys into the hash: "020098-10". I checked the excel file at the specified row and found the difference are " 020098-10" and "020098-10". the first one has a leading space while the second doesn't. I dont' understand is it true that .strip function already remove all leading and trailing white space?
also when i tried to print out key.length, it gave me these weird number:
020098-10 length 18
020098-10 length 17
which should be 9....
If you will inspect the strings you receive, you will probably get something like:
" \x000\x002\x000\x000\x009\x008\x00-\x001\x000\x00"
This happens because of the strings encoding. Excel works with unicode while ruby uses ISO-8859-1 by default. The encodings will differ on various platforms.
You need to convert the data you receive from excel to a printable encoding.
However when you should not encode strings created in ruby as you will end with garbage.
Consider this code:
#enc = Encoding::Converter.new("UTF-16LE", "UTF-8")
def convert(cell)
if cell.numeric
cell.value
else
#enc.convert(cell.value).strip
end
end
parts = {}
worksheet.each do |row|
continue unless row
key = convert row.at(1)
value = convert row.at(0)
parts[key] = value unless parts.has_key?(key) or key.empty?
end
You may want change the encodings to a different ones.
The newer Spreadsheet-gem handles charset conversion automatically for you, to UTF-8 I think as standard but you can change it, so I'd recommend using it instead.