CODEWITHSUNDEEP
javaxmlregex
cyroxx
cyroxx

Reputation: 3877

Strip escape sequences for invalid XML characters

According to the XML spec, only the following charcters are legal:

Char ::= #x9 | #xA | #xD | [#x20-#xD7FF] | [#xE000-#xFFFD] | [#x10000-#x10FFFF] /* any Unicode character, excluding the surrogate blocks, FFFE, and FFFF. */

I have a string named foo containing the JSON representation of an object. Some strings of the JSON object contain escape sequences for characters that are illegal in XML, e.g. \u0002 and \u000b.

I want to strip those escape sequences from foo before throwing it to a JSON to XML converter, because the converter is a black box that provides no ability to handle those invalid characters.

Example for what I would like to do:

MAGIC_REGEX = "<here's what needs to be found>"  # TODO

String foo = "\\u0002bar b\\u000baz qu\\u000fx"
String clean_foo = foo.replace(MAGIC_REGEX, "�")  # � Unicode replacement character

System.out.println(clean_foo)  # Output is "bar baz qux"

How can I achieve that? Bonus points for solutions that use a regex instead of parsing the string and comparing Unicode codepoints.

I am aware of this SO question. However, my problem here are the escape sequences of the illegal characters, not the real characters themselves.

Upvotes: 0

Views: 2176

Answers (1)

cyroxx
cyroxx

Reputation: 3877

I finally came up with this regex, which matches almost all illegal characters according to the XML spec, except the ones above #x10000 (#x11000 and onwards):

# case-sensitive version
\\\\u(00(0[^9ADad]|1[0-9A-Fa-f])|D[8-9A-Fa-f][0-9A-Fa-f]{2}|[Ff]{3}[EFef])

# case-insensitive version
\\\\u(00(0[^9ad]|1[0-9a-f])|D[8-9a-f][0-9a-f]{2}|fff[ef])

Upvotes: 1

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