CODEWITHSUNDEEP
xmlscala
user972946
user972946

Reputation:

How to correctly use xml.Utility.unescape?

After I read xml.Utility.unescape's SDK document, I would think that it is the reverse action of xml.Utility.escape, however it does not seem to do anything at all:

scala> xml.Utility.escape("& <")
res0: String = &amp; &lt;

scala> val sb = new StringBuilder
sb: StringBuilder = 

scala> xml.Utility.unescape("&amp; &lt;", sb)
res1: StringBuilder = null

scala> sb.toString
res2: String = ""

How to correctly use xml.Utility.unescape?

Upvotes: 4

Views: 2080

Answers (1)

Dylan
Dylan

Reputation: 13922

I checked out the unescape documentation and I'm pretty disappointed...

Appends unescaped string to s, amp becomes &, lt becomes < etc..

returns null if ref was not a predefined entity.

So it seems like unescape is meant to append a single character to the StringBuilder. It thought that "&amp; &lt;" was "not a predefined entity" so it returned null for your res1.

Some testing in the REPL:

scala> unescape("amp", sb)
res9: StringBuilder = &

scala> unescape("lt", sb)
res10: StringBuilder = &<

scala> unescape(" ", sb)
res11: StringBuilder = null

Upvotes: 5

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