Perl6 getting a regex from a string
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I want to put a regex in a YAML config file (say: config/filename.yaml), eg.
-
section: Begining
regex: '/ ^ <[ a..b, A..B ]> /'
-
section: Next
regex: '/ ^ <[ c..z, C..Z ]> /'
When I read this into a hash (eg. using YAMLish), eg.,
use YAMLish;
my @h = load-yaml('config/filename.yaml'.IO.slurp);
naturally I have a string in @h[0]<regex>
So how do I recover a regex from a string for use in a match?
I want something like the following, but the following doesn't work:
say 'Its a beginning' if 'A beginning' ~~ @h[0]<regex>
It doesn't work as desired because @h[0]<regex> is a Str, so the smart match is testing a Str in @h[0]<regex> against a Str literal. So how to get the regex out of the Str?
regex perl6
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
add a comment |
I want to put a regex in a YAML config file (say: config/filename.yaml), eg.
-
section: Begining
regex: '/ ^ <[ a..b, A..B ]> /'
-
section: Next
regex: '/ ^ <[ c..z, C..Z ]> /'
When I read this into a hash (eg. using YAMLish), eg.,
use YAMLish;
my @h = load-yaml('config/filename.yaml'.IO.slurp);
naturally I have a string in @h[0]<regex>
So how do I recover a regex from a string for use in a match?
I want something like the following, but the following doesn't work:
say 'Its a beginning' if 'A beginning' ~~ @h[0]<regex>
It doesn't work as desired because @h[0]<regex> is a Str, so the smart match is testing a Str in @h[0]<regex> against a Str literal. So how to get the regex out of the Str?
regex perl6
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
add a comment |
I want to put a regex in a YAML config file (say: config/filename.yaml), eg.
-
section: Begining
regex: '/ ^ <[ a..b, A..B ]> /'
-
section: Next
regex: '/ ^ <[ c..z, C..Z ]> /'
When I read this into a hash (eg. using YAMLish), eg.,
use YAMLish;
my @h = load-yaml('config/filename.yaml'.IO.slurp);
naturally I have a string in @h[0]<regex>
So how do I recover a regex from a string for use in a match?
I want something like the following, but the following doesn't work:
say 'Its a beginning' if 'A beginning' ~~ @h[0]<regex>
It doesn't work as desired because @h[0]<regex> is a Str, so the smart match is testing a Str in @h[0]<regex> against a Str literal. So how to get the regex out of the Str?
regex perl6
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
I want to put a regex in a YAML config file (say: config/filename.yaml), eg.
-
section: Begining
regex: '/ ^ <[ a..b, A..B ]> /'
-
section: Next
regex: '/ ^ <[ c..z, C..Z ]> /'
When I read this into a hash (eg. using YAMLish), eg.,
use YAMLish;
my @h = load-yaml('config/filename.yaml'.IO.slurp);
naturally I have a string in @h[0]<regex>
So how do I recover a regex from a string for use in a match?
I want something like the following, but the following doesn't work:
say 'Its a beginning' if 'A beginning' ~~ @h[0]<regex>
It doesn't work as desired because @h[0]<regex> is a Str, so the smart match is testing a Str in @h[0]<regex> against a Str literal. So how to get the regex out of the Str?
regex perl6
regex perl6
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
asked Nov 24 '18 at 14:44
Richard HainsworthRichard Hainsworth
22216
22216
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You interpolate a string held in a variable $var into a regex via / <$var> /, or, in case of more complex expressions like yours, / <{ @h[0]<regex> }> /.
Note, however, that you first have to remove the enclosing slashes from your string. For trusted input, there's of course always EVAL...
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
add a comment |
I think it's a good idea to compile external regexes, so you know right away if they fail (and they will be faster afterwards):
Perl6 REPL:
> my $str = 'd+'; my $rx = rx/ <$str> /; say "a" ~~ $rx; say "10" ~~ $rx
Nil
｢10｣
Of course it's easier providing a "pure" regex, not one that also has syntax (/ /).
Perl documentation
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You interpolate a string held in a variable $var into a regex via / <$var> /, or, in case of more complex expressions like yours, / <{ @h[0]<regex> }> /.
Note, however, that you first have to remove the enclosing slashes from your string. For trusted input, there's of course always EVAL...
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
add a comment |
You interpolate a string held in a variable $var into a regex via / <$var> /, or, in case of more complex expressions like yours, / <{ @h[0]<regex> }> /.
Note, however, that you first have to remove the enclosing slashes from your string. For trusted input, there's of course always EVAL...
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
add a comment |
You interpolate a string held in a variable $var into a regex via / <$var> /, or, in case of more complex expressions like yours, / <{ @h[0]<regex> }> /.
Note, however, that you first have to remove the enclosing slashes from your string. For trusted input, there's of course always EVAL...
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
You interpolate a string held in a variable $var into a regex via / <$var> /, or, in case of more complex expressions like yours, / <{ @h[0]<regex> }> /.
Note, however, that you first have to remove the enclosing slashes from your string. For trusted input, there's of course always EVAL...
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
answered Nov 24 '18 at 14:52
ChristophChristoph
131k32161219
131k32161219
add a comment |
add a comment |
I think it's a good idea to compile external regexes, so you know right away if they fail (and they will be faster afterwards):
Perl6 REPL:
> my $str = 'd+'; my $rx = rx/ <$str> /; say "a" ~~ $rx; say "10" ~~ $rx
Nil
｢10｣
Of course it's easier providing a "pure" regex, not one that also has syntax (/ /).
Perl documentation
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
add a comment |
I think it's a good idea to compile external regexes, so you know right away if they fail (and they will be faster afterwards):
Perl6 REPL:
> my $str = 'd+'; my $rx = rx/ <$str> /; say "a" ~~ $rx; say "10" ~~ $rx
Nil
｢10｣
Of course it's easier providing a "pure" regex, not one that also has syntax (/ /).
Perl documentation
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
add a comment |
I think it's a good idea to compile external regexes, so you know right away if they fail (and they will be faster afterwards):
Perl6 REPL:
> my $str = 'd+'; my $rx = rx/ <$str> /; say "a" ~~ $rx; say "10" ~~ $rx
Nil
｢10｣
Of course it's easier providing a "pure" regex, not one that also has syntax (/ /).
Perl documentation
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
I think it's a good idea to compile external regexes, so you know right away if they fail (and they will be faster afterwards):
Perl6 REPL:
> my $str = 'd+'; my $rx = rx/ <$str> /; say "a" ~~ $rx; say "10" ~~ $rx
Nil
｢10｣
Of course it's easier providing a "pure" regex, not one that also has syntax (/ /).
Perl documentation
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
edited Dec 9 '18 at 11:46
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
answered Dec 9 '18 at 11:38
nxadmnxadm
1,9711016
1,9711016
add a comment |
add a comment |
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