val data = "val re = \"\"\"^[^/]*://[^/]*/[^/]*$\"\"\".r" val source = """here LATEX_THING{abc} there""" val re = "LATEX_THING\\{abc\\}".r println(re.replaceAllIn(source,data))
java.lang.IllegalArgumentException: Illegal group reference
如果我将数据从简单的内容更改为简单的内容:@H_502_4@
val data = "This will work"
一切都很好.@H_502_4@
看起来像replaceAllIn以某种方式查看第二个字符串并将其用作另一个RE来引用从第一个RE记住的内容……但是文档对此没有任何说明.@H_502_4@
我错过了什么?@H_502_4@
编辑:好的,所以在查看java.util.regex.Matcher类之后,似乎预期的修复是:@H_502_4@
re.replaceAllIn(source,java.util.regex.Matcher.quoteReplacement(data))
val data = "val re = \"\"\"^[^/]*://[^/]*/[^/]*\\$\"\"\".r"
否则,它被解释为组引用的开头(仅当$后跟一个或多个数字时才有效).有关更多详细信息,请参阅the documentation for java.util.regex.Matcher:@H_502_4@
The replacement string may contain references to subsequences captured
during the prevIoUs match: Each occurrence of$g
will be replaced by
the result of evaluatinggroup(g)
… A dollar sign ($
) may be
included as a literal in the replacement string by preceding it with a
backslash (\$
).@H_502_4@
更新以解决您的评论并在上面编辑:是的,如果您不使用字符串文字(或者如果您是,我猜,但是在这种情况下逃避$似乎更容易),您可以使用Matcher.quoteReplacement,并且至少a chance将来,indexReplacement将作为scala.util.matching.Regex上的方法提供.@H_502_4@