Alias for perl script
Clash Royale CLAN TAG#URR8PPP
So far while surfing web I've discovered nice addition to my gpg password generator and I've decided to adjust my alias with these commands.
Here is the code
gpg --gen-random 1 20 | perl -ne'print "Your password: ";s/[x00-x20]/chr(ord($^N)+50)/ge;s/([x7E-xDB])/chr(ord($^N)-93)/ge;s/([xDC-xFF])/chr(ord($^N)-129)/ge;print $_, "n"'
Here is what I've tried so far:
alias genpass()
gpg --gen-random 1 $1
but error occurs when spawning new instance of bash:
bash: /home/user/.bashrc: line 18: syntax error near unexpected token `('
bash: /home/user/.bashrc: line 18: `alias genpass()'
I cannot figure out how this token should be properly escaped.
perl password alias bashrc gpg
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
add a comment |
So far while surfing web I've discovered nice addition to my gpg password generator and I've decided to adjust my alias with these commands.
Here is the code
gpg --gen-random 1 20 | perl -ne'print "Your password: ";s/[x00-x20]/chr(ord($^N)+50)/ge;s/([x7E-xDB])/chr(ord($^N)-93)/ge;s/([xDC-xFF])/chr(ord($^N)-129)/ge;print $_, "n"'
Here is what I've tried so far:
alias genpass()
gpg --gen-random 1 $1
but error occurs when spawning new instance of bash:
bash: /home/user/.bashrc: line 18: syntax error near unexpected token `('
bash: /home/user/.bashrc: line 18: `alias genpass()'
I cannot figure out how this token should be properly escaped.
perl password alias bashrc gpg
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
You should double-quote$1.
– 200_success
Dec 21 '14 at 0:54
add a comment |
So far while surfing web I've discovered nice addition to my gpg password generator and I've decided to adjust my alias with these commands.
Here is the code
gpg --gen-random 1 20 | perl -ne'print "Your password: ";s/[x00-x20]/chr(ord($^N)+50)/ge;s/([x7E-xDB])/chr(ord($^N)-93)/ge;s/([xDC-xFF])/chr(ord($^N)-129)/ge;print $_, "n"'
Here is what I've tried so far:
alias genpass()
gpg --gen-random 1 $1
but error occurs when spawning new instance of bash:
bash: /home/user/.bashrc: line 18: syntax error near unexpected token `('
bash: /home/user/.bashrc: line 18: `alias genpass()'
I cannot figure out how this token should be properly escaped.
perl password alias bashrc gpg
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
So far while surfing web I've discovered nice addition to my gpg password generator and I've decided to adjust my alias with these commands.
Here is the code
gpg --gen-random 1 20 | perl -ne'print "Your password: ";s/[x00-x20]/chr(ord($^N)+50)/ge;s/([x7E-xDB])/chr(ord($^N)-93)/ge;s/([xDC-xFF])/chr(ord($^N)-129)/ge;print $_, "n"'
Here is what I've tried so far:
alias genpass()
gpg --gen-random 1 $1
but error occurs when spawning new instance of bash:
bash: /home/user/.bashrc: line 18: syntax error near unexpected token `('
bash: /home/user/.bashrc: line 18: `alias genpass()'
I cannot figure out how this token should be properly escaped.
perl password alias bashrc gpg
perl password alias bashrc gpg
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
edited Jan 6 at 21:43
Rui F Ribeiro
39.6k1479132
39.6k1479132
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
asked Dec 20 '14 at 16:22
im_infamousim_infamous
1711211
1711211
You should double-quote$1.
– 200_success
Dec 21 '14 at 0:54
add a comment |
You should double-quote$1.
– 200_success
Dec 21 '14 at 0:54
You should double-quote
$1.– 200_success
Dec 21 '14 at 0:54
You should double-quote
$1.– 200_success
Dec 21 '14 at 0:54
add a comment |
1 Answer
1
active
oldest
votes
You have the alias statement when you don't need it, what you're actually creating is a function, replace the word alias with function and it will work as expected.
You also don't actually need function either, you could just have the following and it will work as expected;
genpass()
gpg --gen-random 1 $1
edited Dec 20 '14 at 16:57
jherran
2,25931227
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have the alias statement when you don't need it, what you're actually creating is a function, replace the word alias with function and it will work as expected.
You also don't actually need function either, you could just have the following and it will work as expected;
genpass()
gpg --gen-random 1 $1
edited Dec 20 '14 at 16:57
jherran
2,25931227
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
add a comment |
You have the alias statement when you don't need it, what you're actually creating is a function, replace the word alias with function and it will work as expected.
You also don't actually need function either, you could just have the following and it will work as expected;
genpass()
gpg --gen-random 1 $1
edited Dec 20 '14 at 16:57
jherran
2,25931227
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
add a comment |
You have the alias statement when you don't need it, what you're actually creating is a function, replace the word alias with function and it will work as expected.
You also don't actually need function either, you could just have the following and it will work as expected;
genpass()
gpg --gen-random 1 $1
edited Dec 20 '14 at 16:57
jherran
2,25931227
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
You have the alias statement when you don't need it, what you're actually creating is a function, replace the word alias with function and it will work as expected.
You also don't actually need function either, you could just have the following and it will work as expected;
genpass()
gpg --gen-random 1 $1
edited Dec 20 '14 at 16:57
jherran
2,25931227
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
edited Dec 20 '14 at 16:57
jherran
2,25931227
edited Dec 20 '14 at 16:57
jherran
2,25931227
edited Dec 20 '14 at 16:57
jherran
2,25931227
2,25931227
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
answered Dec 20 '14 at 16:36
Chris DavidsonChris Davidson
1,157610
1,157610
add a comment |
add a comment |
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You should double-quote
$1.– 200_success
Dec 21 '14 at 0:54