URLDecoder: Illegal hex characters in escape (%) pattern - For input string:

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来自：http://stackoverflow.com/questions/11257509/urldecoder-illegal-hex-characters-in-escape-pattern-for-input-string-p

Whoever created the URL should have percent encoded the % by writing %25.

Example invalid URL

http://example.com/test?q=%.P

Example valid URL

http://example.com/test?q=%25.P

The answer provided by Mark Byers will work just fine if there're only % chars that need to be escaped but will fail if url contains percent-encoded chars. To avoid this there's a little bit more work needed.

In percent-encoding (url-encoding) only reserved and unreserved chars won't be percent-encoded.

Reserved chars:
╔═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╗
║ ! ║ # ║ $ ║ & ║ ' ║ ( ║ ) ║ * ║ + ║ , ║ / ║ : ║ ; ║ = ║ ? ║ @ ║ [ ║ ] ║
╚═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╝Unreserved chars:
╔═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╗
║ A ║ B ║ C ║ D ║ E ║ F ║ G ║ H ║ I ║ J ║ K ║ L ║ M ║ N ║ O ║ P ║ Q ║ R ║
╚═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╝
╔═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╗
║ S ║ T ║ U ║ V ║ W ║ X ║ Y ║ Z ║ a ║ b ║ c ║ d ║ e ║ f ║ g ║ h ║ i ║ j ║
╚═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╝
╔═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╗
║ k ║ l ║ m ║ n ║ o ║ p ║ q ║ r ║ s ║ t ║ u ║ v ║ w ║ x ║ y ║ z ║
╚═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╝
╔═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╦═══╗
║ 0 ║ 1 ║ 2 ║ 3 ║ 4 ║ 5 ║ 6 ║ 7 ║ 8 ║ 9 ║ - ║ _ ║ . ║ ~ ║
╚═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╩═══╝

According to RFC 3986 percent-encoded character has following format: % + hex. So if you want to properly escape url that has unescaped % chars without breaking the whole url before actually decoding it, you need to replace only those % signs that are not followed by hex.

Finding substring that violates some pattern is pretty easy task with regex. In this case pattern will look like this:

%(?![0-9a-fA-F]{2})

Sample:

class Main
{public static void main (String[] args) throws java.lang.Exception{String url = "http://example.com/test?q=%.P%20some%20other%20Text";url = url.replaceAll("%(?![0-9a-fA-F]{2})", "%25");System.out.println(url);}
}

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