[ACCEPTED]-Strange Base64 encode/decode problem-base64

Accepted answer

Score: 13

Whatever populates params expects the request 6 to be a URL-encoded form (specifically, application/x-www-form-urlencoded, where 5 "+" means space), but you didn't URL-encode 4 it. I don't know what functions your language 3 provides, but in pseudo code, queryString should be 2 constructed from

concat(uri_escape("data"), "=", uri_escape(base64_encode(rawBytes)))

which simplifies to

concat("data=", uri_escape(base64_encode(rawBytes)))

The "+" characters 1 will be replaced with "%2B".

Score: 7

You have to use a special base64encode which 5 is also url-safe. The problem is that standard 4 base64encode includes +, / and = characters 3 which are replaced by the percent-encoded 2 version.

http://en.wikipedia.org/wiki/Base64#URL_applications

I'm using the following code in 1 php:

    /**
     * Custom base64 encoding. Replace unsafe url chars
     *
     * @param string $val
     * @return string
     */
    static function base64_url_encode($val) {

        return strtr(base64_encode($val), '+/=', '-_,');

    }

    /**
     * Custom base64 decode. Replace custom url safe values with normal
     * base64 characters before decoding.
     *
     * @param string $val
     * @return string
     */
    static function base64_url_decode($val) {

        return base64_decode(strtr($val, '-_,', '+/='));

    }

Score: 3

Because it is a parameter to a POST you 1 must URL encode the data.

See http://en.wikipedia.org/wiki/Percent-encoding

Score: 1

paraquote from the wikipedia link

The encoding 7 used by default is based on a very early 6 version of the general URI percent-encoding 5 rules, with a number of modifications 4 such as newline normalization and replacing spaces 3 with "+" instead of "%20"

another hidden 2 pitfall everyday web developers like myself 1 know little about

Source: stackoverflow.com

[ACCEPTED]-Strange Base64 encode/decode problem-base64

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