scala, string variable processed as string not regexp

Question

Is it seamlessly possible to do?

 scala> val p = "$"
 scala> "hello, I have 65 dollars".replaceFirst("dollars", p)

Current result is

 java.lang.StringIndexOutOfBoundsException: String index out of range: 1
 ....

The expected result in scala 2.10:

 hello, I have 65 $

Problem is with variable p which stores symbol $, I need to process it as a string not regexp.

Note: I can't modify (e.g. replace all non-letter symbols) the p variable (only standard functions, e.g. .toString)

Note2: The given example is rather toy-example. I'd appreciate a more general solution. I.e. variable p can contain any type of content (symbols, numbers, text,...), therefore replacing "$" for "\\$" doesn.t make much sense

(this is improved version of similar problem: scala string, raw string )

@om-nom-nom yes, I read it already, but there is replacement... Doesn't exist some method which would do the task without guessing what variable p can or can not contain in order not to get error? — xhudik
– xhudik, Commented Apr 9, 2013 at 10:17

som-snytt · Accepted Answer · 2014-04-25 12:11:46Z

2

Use Regex.quote for patterns and quoteReplacement for replacement strings. (These just call into Pattern.)

scala> import util.matching._
import util.matching._

scala> "hello, I have 65 dollars".replaceFirst("dollars", Regex quoteReplacement p)
res7: String = hello, I have 65 $

scala> "dollars".r replaceFirstIn ("hello, I have 65 dollars", Regex quoteReplacement p)
res8: String = hello, I have 65 $

scala> "hello, I have 65 dollars".replaceAllLiterally("dollars", p) // quotes both
res9: String = hello, I have 65 $

answered Apr 25, 2014 at 12:11

som-snytt

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Comments

Hanno · Accepted Answer · 2013-04-09 10:27:30Z

2

You need to escape the dollar sign literal, because Java uses it in its implementation of regular expressions as a group reference.

You noted that you can't modify the string literal in the p variable, so you need to resort to replacing the dollar sign and other special characters like this:

Pattern.quote(p);

edited Apr 9, 2013 at 10:27

answered Apr 9, 2013 at 10:15

Hanno

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5 Comments

xhudik Over a year ago

isn't there some other method how to do it? If not, what symbols do I have to replace? I'm aware of $ and \ -anything else? Variable p can contain everything(text, symbols, numbers,..) therefore I need to get rid of all "special characters"

Hanno Over a year ago

Oh, I understood your question the wrong way. I thought you were just concerned with the dollar symbol. Updated my answer.

xhudik Over a year ago

I wanted to put simple example - probably I'll edit it. Thanks for answer anyway

xhudik Over a year ago

quote is interesting point, but I'll wait if something else will appear

xhudik Over a year ago

I've tried your proposal, but it doesn't work: "i have 65 dollars".replaceFirst("dollars", Pattern.quote("$")) . gives Illegal group reference error

Keith Pinson · Accepted Answer · 2013-04-11 10:22:06Z

1

The problem is replaceFirst() uses regular expressions so:

"65 dollars".replaceFirst("dollars","$0") // compiles
"65 dollars".replaceFirst("dollars","$")  // throws "StringIndexOutOfBoundsException"

If,

val dollars = "$"

You could escape the $ symbol,

"65 dollars".replaceFirst( "dollars", if(dollars == "$") "\\$" else dollars )

or use string interpolation,

s"65 $dollars"

or go old school string manipulation,

val t = "65 dollars".split("dollars"); if(t.size>1) t.mkString(dollars) else t(0) + dollars

or with a map,

val ff = "dollars"
val r1 = "$"
"65 dollars, 3 dollars, 50 dollars".split(ff).zipWithIndex.map{case (t,0) => t+r1; case (t,_) => t+ff}.mkString

edited Apr 11, 2013 at 10:22

answered Apr 9, 2013 at 12:59

Keith Pinson

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6 Comments

xhudik Over a year ago

pls see my note2, i'm looking for some general solution, not just for $. That means I'm not willing to replace all possible escape and special characters ($,\,&,@, and who knows what else) but to treat the variable p like string instead of regexp

Keith Pinson Over a year ago

That's why I provided multiple solutions. You might try string interpolation. I upvoted your question because the $ symbol can be a "gotcha" unexpected error in regex.

xhudik Over a year ago

thanks Keith, but still it is not the expected answer. e.g. if your variable dollars contains "\ " it will throw an error. Yes, you solved $ issue, but there can be similar problems with other symbols e.g. "\". As you stated replaceFirst takes its arugument as regexp - is it possible to change this "feature" into accepting string instead regexp. Or, use other "replaceFirst" function which would do the job?

Keith Pinson Over a year ago

I am not aware of a function that will parse a string to sanitize it for regex string replacements. If you don't use regex you still need do something about the lone backslash "\", as you discovered. Incidentally, replace() doesn't use regular expressions even though replaceFirst() does.

xhudik Over a year ago

yes, replace and replaceAllLitarally works as needed. The only problem is that they replace all occurrences instead of just first one which I need. Thanks Keith!

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