Get lines of file matching a regex
Clash Royale CLAN TAG#URR8PPP
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Is there a more compact way to get lines of a file that match a regex than the following?
open my $fh, '<', $file or die "Cannot open $file: $!";
while ( my $line = <$fh> )
next unless $line =~ m/^foo/([^/]+/?)*==/;
chomp($line);
$line =~ s/==$//g;
$search_result_set$line = 1;
files regular-expression perl
asked Nov 30 at 8:55
Jim
390213
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show 13 more comments
up vote
0
down vote
favorite
Is there a more compact way to get lines of a file that match a regex than the following?
open my $fh, '<', $file or die "Cannot open $file: $!";
while ( my $line = <$fh> )
next unless $line =~ m/^foo/([^/]+/?)*==/;
chomp($line);
$line =~ s/==$//g;
$search_result_set$line = 1;
files regular-expression perl
asked Nov 30 at 8:55
Jim
390213
Can you please explain what you want to accomplish? Do not expect we will put effort to understand your script.
– Romeo Ninov
Nov 30 at 8:59
@RomeoNinov: How is the regex relevant? I was under the impression that it can be more compact in the read line by line part after opening the file
– Jim
Nov 30 at 9:08
@Inian: Same comment as above
– Jim
Nov 30 at 9:08
@RomeoNinov: E.g. I think there is something likeslurpbut that is considered insecure or something?
– Jim
Nov 30 at 9:09
1
s^(foo/[^/]+/?)==s*$$1 && $sr$_++ while <$fh>; in any case, do not write/../../../in perl, ever; writem.../...orm#../..#instead.
– mosvy
Nov 30 at 9:31
|
show 13 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is there a more compact way to get lines of a file that match a regex than the following?
open my $fh, '<', $file or die "Cannot open $file: $!";
while ( my $line = <$fh> )
next unless $line =~ m/^foo/([^/]+/?)*==/;
chomp($line);
$line =~ s/==$//g;
$search_result_set$line = 1;
files regular-expression perl
asked Nov 30 at 8:55
Jim
390213
Is there a more compact way to get lines of a file that match a regex than the following?
open my $fh, '<', $file or die "Cannot open $file: $!";
while ( my $line = <$fh> )
next unless $line =~ m/^foo/([^/]+/?)*==/;
chomp($line);
$line =~ s/==$//g;
$search_result_set$line = 1;
files regular-expression perl
files regular-expression perl
asked Nov 30 at 8:55
Jim
390213
asked Nov 30 at 8:55
Jim
390213
edited Nov 30 at 9:21
asked Nov 30 at 8:55
Jim
390213
asked Nov 30 at 8:55
Jim
390213
asked Nov 30 at 8:55
Jim
390213
390213
Can you please explain what you want to accomplish? Do not expect we will put effort to understand your script.
– Romeo Ninov
Nov 30 at 8:59
@RomeoNinov: How is the regex relevant? I was under the impression that it can be more compact in the read line by line part after opening the file
– Jim
Nov 30 at 9:08
@Inian: Same comment as above
– Jim
Nov 30 at 9:08
@RomeoNinov: E.g. I think there is something likeslurpbut that is considered insecure or something?
– Jim
Nov 30 at 9:09
1
s^(foo/[^/]+/?)==s*$$1 && $sr$_++ while <$fh>; in any case, do not write/../../../in perl, ever; writem.../...orm#../..#instead.
– mosvy
Nov 30 at 9:31
|
show 13 more comments
Can you please explain what you want to accomplish? Do not expect we will put effort to understand your script.
– Romeo Ninov
Nov 30 at 8:59
@RomeoNinov: How is the regex relevant? I was under the impression that it can be more compact in the read line by line part after opening the file
– Jim
Nov 30 at 9:08
@Inian: Same comment as above
– Jim
Nov 30 at 9:08
@RomeoNinov: E.g. I think there is something likeslurpbut that is considered insecure or something?
– Jim
Nov 30 at 9:09
1
s^(foo/[^/]+/?)==s*$$1 && $sr$_++ while <$fh>; in any case, do not write/../../../in perl, ever; writem.../...orm#../..#instead.
– mosvy
Nov 30 at 9:31
Can you please explain what you want to accomplish? Do not expect we will put effort to understand your script.
– Romeo Ninov
Nov 30 at 8:59
Can you please explain what you want to accomplish? Do not expect we will put effort to understand your script.
– Romeo Ninov
Nov 30 at 8:59
@RomeoNinov: How is the regex relevant? I was under the impression that it can be more compact in the read line by line part after opening the file
– Jim
Nov 30 at 9:08
@RomeoNinov: How is the regex relevant? I was under the impression that it can be more compact in the read line by line part after opening the file
– Jim
Nov 30 at 9:08
@Inian: Same comment as above
– Jim
Nov 30 at 9:08
@Inian: Same comment as above
– Jim
Nov 30 at 9:08
@RomeoNinov: E.g. I think there is something like
slurp but that is considered insecure or something?– Jim
Nov 30 at 9:09
@RomeoNinov: E.g. I think there is something like
slurp but that is considered insecure or something?– Jim
Nov 30 at 9:09
1
1
s^(foo/[^/]+/?)==s*$$1 && $sr$_++ while <$fh>; in any case, do not write /../../../ in perl, ever; write m.../... or m#../..# instead.– mosvy
Nov 30 at 9:31
s^(foo/[^/]+/?)==s*$$1 && $sr$_++ while <$fh>; in any case, do not write /../../../ in perl, ever; write m.../... or m#../..# instead.– mosvy
Nov 30 at 9:31
|
show 13 more comments
1 Answer
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1
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@lines = grep s^foo/([^/]+/?)*K==n?z, <$fh>;
Would store in the @lines array the lines that match the pattern with a trailing ==n removed. $ matches both at the end of the subject and before a trailing n in the subject, here we're using n?z (an optional n followed by the end of the subject (z)) for that newline to be removed if present, effectively doing chomp's job. K marks the start of the portion to be substituted.
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
So what about the removal part of the second regex? Should I go over the@linesa second time?
– Jim
Nov 30 at 9:19
@Jim, I missed that part of the requirement. Then, it's even simpler. See edit.
– Stéphane Chazelas
Nov 30 at 10:14
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
@lines = grep s^foo/([^/]+/?)*K==n?z, <$fh>;
Would store in the @lines array the lines that match the pattern with a trailing ==n removed. $ matches both at the end of the subject and before a trailing n in the subject, here we're using n?z (an optional n followed by the end of the subject (z)) for that newline to be removed if present, effectively doing chomp's job. K marks the start of the portion to be substituted.
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
So what about the removal part of the second regex? Should I go over the@linesa second time?
– Jim
Nov 30 at 9:19
@Jim, I missed that part of the requirement. Then, it's even simpler. See edit.
– Stéphane Chazelas
Nov 30 at 10:14
add a comment |
up vote
1
down vote
@lines = grep s^foo/([^/]+/?)*K==n?z, <$fh>;
Would store in the @lines array the lines that match the pattern with a trailing ==n removed. $ matches both at the end of the subject and before a trailing n in the subject, here we're using n?z (an optional n followed by the end of the subject (z)) for that newline to be removed if present, effectively doing chomp's job. K marks the start of the portion to be substituted.
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
So what about the removal part of the second regex? Should I go over the@linesa second time?
– Jim
Nov 30 at 9:19
@Jim, I missed that part of the requirement. Then, it's even simpler. See edit.
– Stéphane Chazelas
Nov 30 at 10:14
add a comment |
up vote
1
down vote
up vote
1
down vote
@lines = grep s^foo/([^/]+/?)*K==n?z, <$fh>;
Would store in the @lines array the lines that match the pattern with a trailing ==n removed. $ matches both at the end of the subject and before a trailing n in the subject, here we're using n?z (an optional n followed by the end of the subject (z)) for that newline to be removed if present, effectively doing chomp's job. K marks the start of the portion to be substituted.
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
@lines = grep s^foo/([^/]+/?)*K==n?z, <$fh>;
Would store in the @lines array the lines that match the pattern with a trailing ==n removed. $ matches both at the end of the subject and before a trailing n in the subject, here we're using n?z (an optional n followed by the end of the subject (z)) for that newline to be removed if present, effectively doing chomp's job. K marks the start of the portion to be substituted.
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
edited Nov 30 at 10:19
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
answered Nov 30 at 9:02
Stéphane Chazelas
296k54559904
296k54559904
So what about the removal part of the second regex? Should I go over the@linesa second time?
– Jim
Nov 30 at 9:19
@Jim, I missed that part of the requirement. Then, it's even simpler. See edit.
– Stéphane Chazelas
Nov 30 at 10:14
add a comment |
So what about the removal part of the second regex? Should I go over the@linesa second time?
– Jim
Nov 30 at 9:19
@Jim, I missed that part of the requirement. Then, it's even simpler. See edit.
– Stéphane Chazelas
Nov 30 at 10:14
So what about the removal part of the second regex? Should I go over the
@lines a second time?– Jim
Nov 30 at 9:19
So what about the removal part of the second regex? Should I go over the
@lines a second time?– Jim
Nov 30 at 9:19
@Jim, I missed that part of the requirement. Then, it's even simpler. See edit.
– Stéphane Chazelas
Nov 30 at 10:14
@Jim, I missed that part of the requirement. Then, it's even simpler. See edit.
– Stéphane Chazelas
Nov 30 at 10:14
add a comment |
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Can you please explain what you want to accomplish? Do not expect we will put effort to understand your script.
– Romeo Ninov
Nov 30 at 8:59
@RomeoNinov: How is the regex relevant? I was under the impression that it can be more compact in the read line by line part after opening the file
– Jim
Nov 30 at 9:08
@Inian: Same comment as above
– Jim
Nov 30 at 9:08
@RomeoNinov: E.g. I think there is something like
slurpbut that is considered insecure or something?– Jim
Nov 30 at 9:09
1
s^(foo/[^/]+/?)==s*$$1 && $sr$_++ while <$fh>; in any case, do not write/../../../in perl, ever; writem.../...orm#../..#instead.– mosvy
Nov 30 at 9:31