Question

下面的格罗维派指挥官说明了我的问题。

首先,这项工作(,如预期(0061/>t61/>)。



>>> print "a".matches(/u0061/)

true


Now let s say that we want to match 
, using the Unicode escape u000A. The following, using "pattern" as a string, behaves as expected:

>>> print "
".matches("u000A");

Interpreter exception: com.google.lotrepls.shared.InterpreterException:
org.codehaus.groovy.control.MultipleCompilationErrorsException: startup failed,
Script1.groovy: 1: expecting anything but   
  ; got it anyway
@ line 1, column 21. 1 error

这是因为,至少在 Java,统法协会的越狱程序早处理(JLS3.3),因此:

print "
".matches("u000A")


实际情况与:

print "
".matches("
")


固定装置是为了逃脱统法协会的密码,让列车的发动机进行如下操作:

>>> print "
".matches("\u000A")

true

现在这里的问题是部分:我们如何能够与格罗夫y <代码>/pattern/<合作。 syntax, 而不是使用脚石?

一些失败的尝试:

>>> print "
".matches(/u000A/)

Interpreter exception: com.google.lotrepls.shared.InterpreterException:
org.codehaus.groovy.control.MultipleCompilationErrorsException: startup failed,
Script1.groovy: 1: expecting EOF, found  (  @ line 1, column 19.
1 error

>>> print "
".matches(/\u000A/)

false

>>> print "\u000A".matches(/\u000A/);

true

Answer 1

~"[u0000-u0008u000Bu000Cu000E-u001Fu007F-u009F]

期望按需要开展工作。据我所看到的,需要双向反弹,这样我就不知道编辑们为什么不满意。

Answer 2

Firstly, it seems Groovy changed in this regard in the meantime, at least on https://groovyconsole.appspot.com/ and a local Groovy shell, " ".matches(/u000A/) works perfectly fine, evaluating to true.

In case you have a similar situation again, just encode the backslash with a unicode escape like in " ".matches(/u005Cu000A/) as then the unicode escape to character conversion makes it a backslash again and then the sequence for the regex parser is kept.

Another option would be to separate the backslash from the u for example by using " ".matches(/${ \ }u000A/) or " ".matches( \ + /u000A/)

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