how to split scala string with regular expression - scala

Question

I come up a pattern like
val pattern = "(\\w+)\\|(.*)\\|\\[(.*)\\]\\|\"(.*)\"\\|\"(.*)\"\\|\\[(.*)\\]\\|\\[(.*)\\]\\|(.*)\\|\\[(.*)\\]\\|\\[(.*)\\]".r
and I have a original string
var str = """AuthLogout|vmlxapp21a|[13/Jan/2016:16:33:15 +0100]|"66.77.444.44 uid=XXXXX,ou=People,o=Bank,o=External,dc=xxxx,dc=com"|"abcd_123_portalweb_w "|[]|[41]||[]|[]"""
then apply pattern to the string, but it is always empty.
val items = pattern.findAllIn(str).toList

Answer 1

If I understand what you're trying to do, perhaps using a giant regex isn't the easiest way: You can split by | and get rid of the unwanted separators ([, ], ") using replaceAll:
val str = """AuthLogout|vmlxapp21a|[13/Jan/2016:16:33:15 +0100]|"66.77.444.44 uid=XXXXX,ou=People,o=Bank,o=External,dc=xxxx,dc=com"|"abcd_123_portalweb_w "|[]|[41]||[]|[]"""
val withoutBoundaries = str.replaceAll("[\"\\]\\[]","")
val result = withoutBoundaries.split("\\|")
result.foreach(println)
Which prints:
AuthLogout
vmlxapp21a
13/Jan/2016:16:33:15 +0100
66.77.444.44 uid=XXXXX,ou=People,o=Bank,o=External,dc=xxxx,dc=com
abcd_123_portalweb_w
41
If you do want to use a regex here, I'd create sub-regex vars representing the different text parts that you're after, to make this somewhat manageable:
val plain = "(.*)" // no boundary characters
val boxed = s"\\[$plain\\]" // same, encapsulated by square brackets
val quoted = '"' + plain + '"' // same, encapsulated by double quotes
// the whole thing, separated by pipes:
val r = s"$plain\\|$plain\\|$boxed\\|$quoted\\|$quoted\\|$boxed\\|$boxed\\|$plain\\|$boxed\\|$boxed".r
val result = r.findAllIn(str).toList // this list has one item, as expected.
Now, if you want to see how this regex looks like, here it is - but I don't recommend having this in your code...:
val r = """(.*)\|(.*)\|\[(.*)\]\|"(.*)"\|"(.*)"\|\[(.*)\]\|\[(.*)\]\|(.*)\|\[(.*)\]\|\[(.*)\]""".r