Re: interpoliation within regexp
Mumia W. wrote:
On 09/29/2006 01:44 PM, Rob Dixon wrote:
Derek B. Smith wrote:
--- Mumia W. [EMAIL PROTECTED]
wrote:
What is the purpose of this program?
To generate a random 6 character string.
If the first character starts with a # then I just
ignore the new string and tell it to goto LABLE, b/c
for 0-32 on the ASCII table cannot be used as a 1st
character in a user password.
## generate random 8 char password.
PASSWD: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z');
my $password = join '', map { $a[int rand @a] } 0 ..
5;
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack (A1 x length($password),
$password);
if ($chars[0] =~ /^\D/) {
print Your new password is:\t,@chars,\n;
}
else {
#print string starts with number:\t,@chars,
\trestarting\n;
#substr($chars[0],0,1) =~ s/$chars[0]/chr ($1)/e);
## execute with regexp substitute ?/? goto PASSWD;
}
This will do what you want. It shuffles all of the possible characters
and joins
them into a string, and then finds the first substring of six
characters that
starts with a non-numeric character. The only proviso is that a
password can
never have the same character twice, which isn't true of the general
solution.
use List::Util qw/shuffle/;
my $chars = join '', shuffle (0..9, 'a'..'z', 'A'..'Z');
my ($password) = $a =~ /(\D.)/;
HTH,
Rob
Aren't you concerned that this is random in a completely wrong way? Only
one of each character can appear in the password, and that's kind of weak.
No. I mentioned it in my post, but since it's meant for password generation and
not for creating data that's 'random' in the mathematical sense I think it's
plenty good enough. Even if you take truly random samples of six characters out
of a 62-character set, about three-quarters of them will be six distinct
characters.
It's also a /very/ minor issue that you assign to $chars but then use $a
after; the main problem is that this password generating algorithm is
that it is, to steal a phrase from Lo Wang, weaker than a baby's
[passing gas].
I fixed that in a later post. I wrote a working version and then posted an
earlier incarnation. Call me an old [passing gas] if you will. :(
Rob
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Re: interpoliation within regexp
Derek B. Smith wrote: --- Rob Dixon [EMAIL PROTECTED] wrote: Rob Dixon wrote: This will do what you want. It shuffles all of the possible characters and joins them into a string, and then finds the first substring of six characters that starts with a non-numeric character. The only proviso is that a password can never have the same character twice, which isn't true of the general solution. [snip faulty code] use List::Util qw/shuffle/; my $chars = join '', shuffle (0..9, 'a'..'z', 'A'..'Z'); my ($password) = $chars =~ /(\D.)/; so this solution DOES NOT allow two of the same characters twice? cool thanks! derek Yes. And it doesn't allow even one of the same character twice either! :) Rob -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
Rob Dixon schreef: [6 random characters from 0-9A-Za-z] it's meant for password generation and not for creating data that's 'random' in the mathematical sense I think it's plenty good enough. Even if you take truly random samples of six characters out of a 62-character set, about three-quarters of them will be six distinct characters. Yes: (X * 61 * 60 * 59 * 58 * 57) / (X * 62 ** 5) = 78% -- Affijn, Ruud Gewoon is een tijger. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
I reread the docs and I am still unclear with the code: $a[ 10 + rand( @a - 10 ) I do understand everything but $a[ 10 + rand( @a - 10 ) b/c you say subtract 10 from each element occurrance = @a - 10 then add 10 to the result of rand (@a - 10) To me this is offsets itself which is why I am confused. Will you explain again? I think you missed an explanation step between a and b? Reagardless it work so thank you. --- D. Bolliger [EMAIL PROTECTED] wrote: Derek B. Smith am Donnerstag, 28. September 2006 22:28: Why not just specify a non-digit for the first character: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z'); my $password = join '', $a[ 10 + rand( @a - 10 ) ], map $a[ rand @a ], 1 .. 5; John Ok great, but I do not fully understand this. Will you explain in English? Join with '': a) a randomly selected entry from @a excluding the digits at positions 0..9 [the part between the 1st and 2nd comma] b) five randomly selected entries from @a [the map part] Note: @a is used in scalar context both times, meaning the number of entries in @a. perldoc -f map perldoc -f join perldoc -f rand My tip for cases where you get a working solution you don't understand fully: a) Try to find out what belongs together (imagine '()'s) b) Try to break the solution apart according to the findings in a) c) examine the parts: print them out, dump them with Data::Dumper, modify them, read the man pages d) put them together again, eventually one by one part, using examination techniques as in c) Hope this helps! Dani __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
On 09/29/2006 12:15 PM, Derek B. Smith wrote: --- D. Bolliger [EMAIL PROTECTED] wrote: Derek B. Smith am Donnerstag, 28. September 2006 22:28: Why not just specify a non-digit for the first character: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z'); my $password = join '', $a[ 10 + rand( @a - 10 ) ], map $a[ rand @a ], 1 .. 5; John Ok great, but I do not fully understand this. Will you explain in English? Join with '': a) a randomly selected entry from @a excluding the digits at positions 0..9 [the part between the 1st and 2nd comma] b) five randomly selected entries from @a [the map part] Note: @a is used in scalar context both times, meaning the number of entries in @a. perldoc -f map perldoc -f join perldoc -f rand My tip for cases where you get a working solution you don't understand fully: a) Try to find out what belongs together (imagine '()'s) b) Try to break the solution apart according to the findings in a) c) examine the parts: print them out, dump them with Data::Dumper, modify them, read the man pages d) put them together again, eventually one by one part, using examination techniques as in c) Hope this helps! Dani I reread the docs and I am still unclear with the code: $a[ 10 + rand( @a - 10 ) I do understand everything but $a[ 10 + rand( @a - 10 ) b/c you say subtract 10 from each element occurrance = @a - 10 then add 10 to the result of rand (@a - 10) To me this is offsets itself which is why I am confused. Will you explain again? I think you missed an explanation step between a and b? Reagardless it work so thank you. Does this slice help demonstrate it? my @a = (0..9,'a'..'z','A'..'Z'); my @b = @[EMAIL PROTECTED]; print @a, \n; print @b, \n; # OUTPUT: #0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ #abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQ # BTW, placing print statements in the right places is a great way to learn how someone's program works. Anyway, I hope this is easier to digest: my @a = (0..9,'a'..'z','A'..'Z'); my @b = @[EMAIL PROTECTED]; my $password = $b[int rand (@b)]; $password .= join '', map $a[int rand (@a)], (1..5); print $password, \n; __HTH__ -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
Derek B. Smith wrote:
--- Mumia W. [EMAIL PROTECTED]
wrote:
What is the purpose of this program?
To generate a random 6 character string.
If the first character starts with a # then I just
ignore the new string and tell it to goto LABLE, b/c
for 0-32 on the ASCII table cannot be used as a 1st
character in a user password.
## generate random 8 char password.
PASSWD: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z');
my $password = join '', map { $a[int rand @a] } 0 ..
5;
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack (A1 x length($password),
$password);
if ($chars[0] =~ /^\D/) {
print Your new password is:\t,@chars,\n;
}
else {
#print string starts with number:\t,@chars,
\trestarting\n;
#substr($chars[0],0,1) =~ s/$chars[0]/chr ($1)/e);
## execute with regexp substitute ?/?
goto PASSWD;
}
This will do what you want. It shuffles all of the possible characters and joins
them into a string, and then finds the first substring of six characters that
starts with a non-numeric character. The only proviso is that a password can
never have the same character twice, which isn't true of the general solution.
use List::Util qw/shuffle/;
my $chars = join '', shuffle (0..9, 'a'..'z', 'A'..'Z');
my ($password) = $a =~ /(\D.)/;
HTH,
Rob
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Re: interpoliation within regexp
Rob Dixon wrote: This will do what you want. It shuffles all of the possible characters and joins them into a string, and then finds the first substring of six characters that starts with a non-numeric character. The only proviso is that a password can never have the same character twice, which isn't true of the general solution. use List::Util qw/shuffle/; my $chars = join '', shuffle (0..9, 'a'..'z', 'A'..'Z'); my ($password) = $a =~ /(\D.)/; My apologies: I posted an early version of that. Here is the working one: use List::Util qw/shuffle/; my $chars = join '', shuffle (0..9, 'a'..'z', 'A'..'Z'); my ($password) = $chars =~ /(\D.)/; Rob -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
-- Mumia W. [EMAIL PROTECTED] wrote: On 09/29/2006 12:15 PM, Derek B. Smith wrote: --- D. Bolliger [EMAIL PROTECTED] wrote: Derek B. Smith am Donnerstag, 28. September 2006 22:28: Why not just specify a non-digit for the first character: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z'); my $password = join '', $a[ 10 + rand( @a - 10 ) ], map $a[ rand @a ], 1 .. 5; John Ok great, but I do not fully understand this. Will you explain in English? Join with '': a) a randomly selected entry from @a excluding the digits at positions 0..9 [the part between the 1st and 2nd comma] b) five randomly selected entries from @a [the map part] Note: @a is used in scalar context both times, meaning the number of entries in @a. perldoc -f map perldoc -f join perldoc -f rand My tip for cases where you get a working solution you don't understand fully: a) Try to find out what belongs together (imagine '()'s) b) Try to break the solution apart according to the findings in a) c) examine the parts: print them out, dump them with Data::Dumper, modify them, read the man pages d) put them together again, eventually one by one part, using examination techniques as in c) Hope this helps! Dani I reread the docs and I am still unclear with the code: $a[ 10 + rand( @a - 10 ) I do understand everything but $a[ 10 + rand( @a - 10 ) b/c you say subtract 10 from each element occurrance = @a - 10 then add 10 to the result of rand (@a - 10) To me this is offsets itself which is why I am confused. Will you explain again? I think you missed an explanation step between a and b? Reagardless it work so thank you. Does this slice help demonstrate it? my @a = (0..9,'a'..'z','A'..'Z'); my @b = @[EMAIL PROTECTED]; print @a, \n; print @b, \n; # OUTPUT: #0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ #abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQ # BTW, placing print statements in the right places is a great way to learn how someone's program works. Anyway, I hope this is easier to digest: my @a = (0..9,'a'..'z','A'..'Z'); my @b = @[EMAIL PROTECTED]; my $password = $b[int rand (@b)]; $password .= join '', map $a[int rand (@a)], (1..5); print $password, \n; __HTH__ yes this explains it... thank you. __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
On 09/29/2006 01:44 PM, Rob Dixon wrote:
Derek B. Smith wrote:
--- Mumia W. [EMAIL PROTECTED]
wrote:
What is the purpose of this program?
To generate a random 6 character string.
If the first character starts with a # then I just
ignore the new string and tell it to goto LABLE, b/c
for 0-32 on the ASCII table cannot be used as a 1st
character in a user password.
## generate random 8 char password.
PASSWD: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z');
my $password = join '', map { $a[int rand @a] } 0 ..
5;
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack (A1 x length($password),
$password);
if ($chars[0] =~ /^\D/) {
print Your new password is:\t,@chars,\n;
}
else {
#print string starts with number:\t,@chars,
\trestarting\n;
#substr($chars[0],0,1) =~ s/$chars[0]/chr ($1)/e);
## execute with regexp substitute ?/? goto PASSWD;
}
This will do what you want. It shuffles all of the possible characters
and joins
them into a string, and then finds the first substring of six characters
that
starts with a non-numeric character. The only proviso is that a password
can
never have the same character twice, which isn't true of the general
solution.
use List::Util qw/shuffle/;
my $chars = join '', shuffle (0..9, 'a'..'z', 'A'..'Z');
my ($password) = $a =~ /(\D.)/;
HTH,
Rob
Aren't you concerned that this is random in a completely wrong way? Only
one of each character can appear in the password, and that's kind of weak.
It's also a /very/ minor issue that you assign to $chars but then use $a
after; the main problem is that this password generating algorithm is
that it is, to steal a phrase from Lo Wang, weaker than a baby's
[passing gas].
:-)
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Re: interpoliation within regexp
--- Derek B. Smith [EMAIL PROTECTED]
wrote:
I need to substitute a conversion using chr, but
have
failed on multiple attempts. Basically if the first
element contains a # then convert it. Will anyone
advise?
thank you
derek
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack
(A1 x length($password),$password);
if ($chars[0] =~ /^\D/) {
print Your new string is:\t,@chars,\n;
}
else {
print string starts with number, now
converting\n,
@chars, \n;
substr($chars[0],0,1) =~
s/\Q{$chars[0]}/{chr($chars[0]}\E/;
print @chars;
}
I will try to use the /e modifier as such:
s/(\$\w+)/$1/e;
An explanation is below:
/e
Righthand side of a s/// is code to eval
/ee
Righthand side of a s/// is a string to eval, then
run as code, and its return value eval'led again.
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Re: interpoliation within regexp
On 09/28/2006 08:16 AM, Derek B. Smith wrote:
--- Derek B. Smith [EMAIL PROTECTED]
wrote:
I need to substitute a conversion using chr, but
have
failed on multiple attempts. Basically if the first
element contains a # then convert it. Will anyone
advise?
thank you
derek
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack
(A1 x length($password),$password);
if ($chars[0] =~ /^\D/) {
print Your new string is:\t,@chars,\n;
}
else {
print string starts with number, now
converting\n,
@chars, \n;
substr($chars[0],0,1) =~
s/\Q{$chars[0]}/{chr($chars[0]}\E/;
print @chars;
}
I will try to use the /e modifier as such:
s/(\$\w+)/$1/e;
An explanation is below:
/e
Righthand side of a s/// is code to eval
/ee
Righthand side of a s/// is a string to eval, then
run as code, and its return value eval'led again.
What does your input data look like, and what do you want the output to
look like?
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Re: interpoliation within regexp
-- Derek B. Smith [EMAIL PROTECTED]
wrote:
--- Derek B. Smith [EMAIL PROTECTED]
wrote:
I need to substitute a conversion using chr, but
have
failed on multiple attempts. Basically if the
first
element contains a # then convert it. Will anyone
advise?
thank you
derek
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack
(A1 x length($password),$password);
if ($chars[0] =~ /^\D/) {
print Your new string is:\t,@chars,\n;
}
else {
print string starts with number, now
converting\n,
@chars, \n;
substr($chars[0],0,1) =~
s/\Q{$chars[0]}/{chr($chars[0]}\E/;
print @chars;
}
I will try to use the /e modifier as such:
s/(\$\w+)/$1/e;
An explanation is below:
/e
Righthand side of a s/// is code to eval
/ee
Righthand side of a s/// is a string to eval, then
run as code, and its return value eval'led again.
*
The results I am getting when using data::dumper is:
string starts with number, now converting
6FhJ9Z
DUMP: $VAR1 = 1;
DUMP2: $VAR1 = ' ';
$VAR2 = 'F';
$VAR3 = 'h';
$VAR4 = 'J';
$VAR5 = '9';
$VAR6 = 'Z';
$VAR7 = '
';
from code:
## generate random 8 char string.
my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z');
my $password = join '', map { $a[int rand @a] } 0 ..
5;
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack (A1 x length($password),
$password);
if ($chars[0] =~ /^\D/) {
print Your new string is:\t,@chars,\n;
}
else {
print string starts with number, now converting\n,
@chars, \n;
print DUMP:\t, Dumper(substr($chars[0],0,1) =~
s/$chars[0]/chr ($1)/e);
print DUMP2:\t, Dumper @chars,\n;
}
Any help please on as to why chr is not working?
thx
derek
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Re: interpoliation within regexp
--- Mumia W. [EMAIL PROTECTED]
wrote:
On 09/28/2006 08:16 AM, Derek B. Smith wrote:
--- Derek B. Smith [EMAIL PROTECTED]
wrote:
I need to substitute a conversion using chr, but
have
failed on multiple attempts. Basically if the
first
element contains a # then convert it. Will anyone
advise?
thank you
derek
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack
(A1 x length($password),$password);
if ($chars[0] =~ /^\D/) {
print Your new string is:\t,@chars,\n;
}
else {
print string starts with number, now
converting\n,
@chars, \n;
substr($chars[0],0,1) =~
s/\Q{$chars[0]}/{chr($chars[0]}\E/;
print @chars;
}
I will try to use the /e modifier as such:
s/(\$\w+)/$1/e;
An explanation is below:
/e
Righthand side of a s/// is code to eval
/ee
Righthand side of a s/// is a string to eval,
then
run as code, and its return value eval'led again.
What does your input data look like, and what do you
want the output to
look like?
**
The input data is a 6 character randomized string that
could start with a # such as 6FhJ9Z. If it does start
with a number then I need to convert this character
into its cooresponding alpha char, [a-z,A-Z].
The output data from the string 6FhJ9Z should be
[a-z,A-Z]FhJ9Z I guess what I am asking for is not
plausible after looking at the ASCII table b/c there
is no cooresponding letter for number 6. Silly me, ok
then maybe grab any [a-z,A-Z] character.
my $foo = pack(C*, 6); print $foo;
or
chr ('6');
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Re: interpoliation within regexp
On 09/28/2006 12:04 PM, Derek B. Smith wrote:
**
The input data is a 6 character randomized string that
could start with a # such as 6FhJ9Z. If it does start
with a number then I need to convert this character
into its cooresponding alpha char, [a-z,A-Z].
The output data from the string 6FhJ9Z should be
[a-z,A-Z]FhJ9Z I guess what I am asking for is not
plausible after looking at the ASCII table b/c there
is no cooresponding letter for number 6. Silly me, ok
then maybe grab any [a-z,A-Z] character.
my $foo = pack(C*, 6); print $foo;
or
chr ('6');
What is the purpose of this program?
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Re: interpoliation within regexp
--- Mumia W. [EMAIL PROTECTED]
wrote:
On 09/28/2006 12:04 PM, Derek B. Smith wrote:
**
The input data is a 6 character randomized string
that
could start with a # such as 6FhJ9Z. If it does
start
with a number then I need to convert this
character
into its cooresponding alpha char, [a-z,A-Z].
The output data from the string 6FhJ9Z should be
[a-z,A-Z]FhJ9Z I guess what I am asking for is not
plausible after looking at the ASCII table b/c
there
is no cooresponding letter for number 6. Silly me,
ok
then maybe grab any [a-z,A-Z] character.
my $foo = pack(C*, 6); print $foo;
or
chr ('6');
What is the purpose of this program?
To generate a random 6 character string.
If the first character starts with a # then I just
ignore the new string and tell it to goto LABLE, b/c
for 0-32 on the ASCII table cannot be used as a 1st
character in a user password.
## generate random 8 char password.
PASSWD: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z');
my $password = join '', map { $a[int rand @a] } 0 ..
5;
#if first char is a-z then print it else warn
#chop string into individual characters
my @chars = unpack (A1 x length($password),
$password);
if ($chars[0] =~ /^\D/) {
print Your new password is:\t,@chars,\n;
}
else {
#print string starts with number:\t,@chars,
\trestarting\n;
#substr($chars[0],0,1) =~ s/$chars[0]/chr ($1)/e);
## execute with regexp substitute ?/?
goto PASSWD;
}
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Re: interpoliation within regexp
Derek B. Smith wrote:
--- Mumia W. [EMAIL PROTECTED]
wrote:
What is the purpose of this program?
To generate a random 6 character string.
If the first character starts with a # then I just
ignore the new string and tell it to goto LABLE, b/c
for 0-32 on the ASCII table cannot be used as a 1st
character in a user password.
## generate random 8 char password.
PASSWD: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z');
my $password = join '', map { $a[int rand @a] } 0 ..
5;
Why not just specify a non-digit for the first character:
my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z');
my $password = join '', $a[ 10 + rand( @a - 10 ) ], map $a[ rand @a ], 1 .. 5;
John
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Re: interpoliation within regexp
Why not just specify a non-digit for the first character: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z'); my $password = join '', $a[ 10 + rand( @a - 10 ) ], map $a[ rand @a ], 1 .. 5; John Ok great, but I do not fully understand this. Will you explain in English? thx __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
Derek B. Smith am Donnerstag, 28. September 2006 22:28: Why not just specify a non-digit for the first character: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z'); my $password = join '', $a[ 10 + rand( @a - 10 ) ], map $a[ rand @a ], 1 .. 5; John Ok great, but I do not fully understand this. Will you explain in English? Join with '': a) a randomly selected entry from @a excluding the digits at positions 0..9 [the part between the 1st and 2nd comma] b) five randomly selected entries from @a [the map part] Note: @a is used in scalar context both times, meaning the number of entries in @a. perldoc -f map perldoc -f join perldoc -f rand My tip for cases where you get a working solution you don't understand fully: a) Try to find out what belongs together (imagine '()'s) b) Try to break the solution apart according to the findings in a) c) examine the parts: print them out, dump them with Data::Dumper, modify them, read the man pages d) put them together again, eventually one by one part, using examination techniques as in c) Hope this helps! Dani -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: interpoliation within regexp
Derek B. Smith wrote: Why not just specify a non-digit for the first character: my @a = ( 0 .. 9, 'a' .. 'z', 'A' .. 'Z'); my $password = join '', $a[ 10 + rand( @a - 10 ) ], map $a[ rand @a ], 1 .. 5; Ok great, but I do not fully understand this. Will you explain in English? The first ten elements of @a are the digits 0-9 so the first list item picks a random element that ignores the first ten elements of @a and the next five list items pick a random element from the entire array. John -- Perl isn't a toolbox, but a small machine shop where you can special-order certain sorts of tools at low cost and in short order. -- Larry Wall -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
