我有一些长队,我想用我想要的颜色突出显示匹配.一个函数使用替换,另一个函数使用递归.有些是作品,有些则没有;我正在寻找一种插入到正则表达式的一致方法.非常感谢你 ！！！
sub colorMatch ($aStr,$aRegex,$aColor) {
my $colorOff = '\e[0m';
my $a =$aStr.subst(/(<{$aRegex}>)/,$aColor ~ "# $/ #" ~ $colorOff,:g);
# not working,$/ not interpolating corresponding matches,all Nil
say $a;
}
colorMatch("a-12-b-3-c-4-def-567",'\d+','\e[1;31m');
colorMatch("a-12-b-3-c-4-def-567",'<alpha>+','\e[1;31m');
say "\e[1;31m" ~ " color1 " ~ "\e[0m" ~
"\e[1;36m" ~ " color2 " ~ "\e[0m";
sub colorMatch2 ($aStr,$colorNumber) {
my $colorOff = "\e[0m";
if $aStr.chars > 0 {
my $x1 = ($aStr ~~ m/<{$aRegex}>/);
my $x2 = $x1.prematch;
my $x3 = $x1.Str;
my $x4 = $x1.postmatch;
return ($x2 ~ "\e[1;{$colorNumber}m" ~ $x3 ~ $colorOff)
~ colorMatch2($x4,$colorNumber);
}
}
say colorMatch2("a-12-b-3-c-4-def-567",31); # works,red color
say colorMatch2("a-12-b-3-c-4-def-567",'567',36); # works,green color
say colorMatch2("a-12-b-3-c-4-def-567",'\w+','[a..z]+',31); # fails with [] and ..
say colorMatch2("a-12-b-3-c-4-def-567","<alpha>+",31); # fails with <>
Use of Nil in string context
in sub colorMatch at colorMatch.pl line 4
a-\e[1;31m# #\e[0m-b-\e[1;31m# #\e[0m-c-\e[1;31m# #\e[0m-def-\e[1;31m# #\e[0m
# seems to do substitution,but $/ is empty and color not shown;
Use of Nil in string context
in sub colorMatch at colorMatch.pl line 4
\e[1;31m# #\e[0m-12-\e[1;31m# #\e[0m-3-\e[1;31m# #\e[0m-4-\e[1;31m# #\e[0m-567
# seems to do substitution,but $/ is empty and color not shown;
color1 color2 # both colors work and shown as expected,# color1 red and color 2 green; why inconsistent with above???
a-12-b-3-c-4-def-567 # works,red color
a-12-b-3-c-4-def-567 # works,green color
a-12-b-3-c-4-def-567 # works,red color
No such method 'prematch' for invocant of type 'Bool'
in sub colorMatch2 at colorMatch.pl line 17
in block <unit> at colorMatch.pl line 28
解决方法
这不是插值问题.字符类和范围的语法略有不同.你需要：
'text' ~~ / <[a..z]>+ /
至于你在bool上调用prematch()的其他错误,结果是False,因为在最终的递归级别,它不再匹配.在假设匹配之前,您需要检查结果.
最后,对于“在字符串上下文中使用Nil”,这是因为您使用的是Str.subst,并且与大多数语言中的大多数函数调用一样,在函数开始之前会对参数进行求值.你在参数中使用$/,但它还没有设置,因为函数的主体甚至没有开始执行. s / match / replacement /运算符不会遇到这种困难,所以我建议你将代码更改为：
$aStr ~~ s:g/(<$aRegex>)/$aColor# $/ #$colorOff/;
(或者为了更好的可读性:)
$aStr ~~ s:g {(<$aRegex>)} = "$aColor# $/ #$colorOff";
这假设您已将$aRegex变为正则表达式而不是字符串.此外,由于新代码修改了$aStr,您需要将$aStr更改为$aStr是函数签名中的副本.