String must contain at least 3 uppercase, 3 lowercase, 3 digits and at least 3 special characters.
I searched a lot, but not able to get the required solution.
head /dev/urandom | tr -dc 'A-Za-z0-9-_' | head -c15
This is what I found, but I am able to get the random combination of all.
What i needed is it must output at least 3 characters from each set.
Try this script:
#!/bin/sh
export LC_ALL=C
upp=$(tr -dc 'A-Z' </dev/urandom | head -c3)
low=$(tr -dc 'a-z' </dev/urandom | head -c3)
dig=$(tr -dc '0-9' </dev/urandom | head -c3)
spe=$(tr -dc '!-/' </dev/urandom | head -c3)
res=$(tr -dc '!-}' </dev/urandom | head -c3)
echo "$upp$low$dig$spe$res"
First, the statement export LC_ALL=C makes sure that we are using just plain ASCII. This eliminates potential issues associated with unicode characters.
Next, the variable upp is assigned to 3 upper-case characters. Similarly, low gets three lower-case, $dig gets three digits, spe gets three special characters, and res gets 3 random characters. The echo statement combines all four variables ands prints them.
The above prints the upper case characters first, lower case second, etc. If you want this order mixed up, replace the last line above with:
echo "$upp$low$dig$spe$res" | sed 's/./&\n/g' | shuf | tr -d '\n'
This script uses openssl-rand to get strings of 15 letters/numbers/special characters until the loop breaks when all conditions are met and the password is printed.
#!/bin/bash
until [ -z "$pw+x" ]
do
pw=$(openssl rand -base64 32 | cut -c1-15) &&\
[[ $(sed "s/[^[:upper:]]//g" <<< $pw | wc -c) -gt 3 ]] &&\
[[ $(sed "s/[^[:lower:]]//g" <<< $pw | wc -c) -gt 3 ]] &&\
[[ $(sed "s/[^0-9]//g" <<< $pw | wc -c) -gt 3 ]] &&\
[[ $(sed "s/[[:alnum:]]//g" <<< $pw | wc -c) -gt 3 ]] && break
done
echo "$pw"
Related
I'm trying to find all words in a file that don't exist in the dictionary. If I look for a single word the following works
b=ther; look $b | grep -i "^$b$" | ifne -n echo $b => ther
b=there; look $b | grep -i "^$b$" | ifne -n echo $b => [no output]
However if I try to run a "while read" loop
while read a; do look $a | grep -i "^$a$" | ifne -n echo "$a"; done < <(tr -s '[[:punct:][:space:]]' '\n' <lotr.txt |tr '[:upper:]' '[:lower:]')
The output seems to contain all (?) words in the file. Why doesn't this loop only output non-dictionary words?
Regarding ifne
If stdin is non-empty, ifne -n reprints stdin to stdout. From the manpage:
-n Reverse operation. Run the command if the standard input is empty
Note that if the standard input is not empty, it is passed through
ifne in this case.
strace on ifne confirms this behavior.
Alternative
Perhaps, as an alternative:
1 #!/bin/bash -e
2
3 export PATH=/bin:/sbin:/usr/bin:/usr/sbin
4
5 while read a; do
6 look "$a" | grep -qi "^$a$" || echo "$a"
7 done < <(
8 tr -s '[[:punct:][:space:]]' '\n' < lotr.txt \
9 | tr '[A-Z]' '[a-z]' \
10 | sort -u \
11 | grep .
12 )
wc -l file.txt
outputs number of lines and file name.
I need just the number itself (not the file name).
I can do this
wc -l file.txt | awk '{print $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$FOO" | wc -c # " 3" (x)
echo -n "$FOO" | wc -c | bc # "3" (√)
echo -n "$FOO" | wc -c | tr -d ' ' # "3" (√)
echo -n "$FOO" | wc -c | awk '{print $1}' # "3" (√)
echo -n "$FOO" | wc -c | cut -d ' ' -f1 # "" for -f < 8 (x)
echo -n "$FOO" | wc -c | cut -d ' ' -f8 # "3" (√)
echo -n "$FOO" | wc -c | perl -pe 's/^\s+//' # "3" (√)
echo -n "$FOO" | wc -c | grep -ch '^' # "1" (x)
echo $( printf '%s' "$FOO" | wc -c ) # "3" (√)
I wouldn't rely on the cut -f* method in general since it requires that you know the exact number of leading spaces that any given output may have. And the grep one works for counting lines, but not characters.
bc is the most concise, and awk and perl seem a bit overkill, but they should all be relatively fast and portable enough.
Also note that some of these can be adapted to trim surrounding whitespace from general strings, as well (along with echo `echo $FOO`, another neat trick).
How about
wc -l file.txt | cut -d' ' -f1
i.e. pipe the output of wc into cut (where delimiters are spaces and pick just the first field)
How about
grep -ch "^" file.txt
Obviously, there are a lot of solutions to this.
Here is another one though:
wc -l somefile | tr -d "[:alpha:][:blank:][:punct:]"
This only outputs the number of lines, but the trailing newline character (\n) is present, if you don't want that either, replace [:blank:] with [:space:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
Best way would be first of all find all files in directory then use AWK NR (Number of Records Variable)
below is the command :
find <directory path> -type f | awk 'END{print NR}'
example : - find /tmp/ -type f | awk 'END{print NR}'
This works for me using the normal wc -l and sed to strip any char what is not a number.
wc -l big_file.log | sed -E "s/([a-z\-\_\.]|[[:space:]]*)//g"
# 9249133
I have the following code:
</dev/urandom tr -dc 'A-Za-z0-9##$%&_+=' | head -c 16
which is randomly generating passwords perfectly.
I want two changes:
It should only contain one special character listed above
It should choose a random length
I tried with length = $(($RANDOM%8+9))
then putting length as
</dev/urandom tr -dc 'A-Za-z0-9##$%&_+=' | head -c$length
but got no positive result.
#! /bin/bash
chars='##$%&_+='
{ </dev/urandom LC_ALL=C grep -ao '[A-Za-z0-9]' \
| head -n$((RANDOM % 8 + 9))
echo ${chars:$((RANDOM % ${#chars})):1} # Random special char.
} \
| shuf \
| tr -d '\n'
LC_ALL=C prevents characters like ř from appearing.
grep -o outputs just the matching substring, i.e. a single character.
shuf shuffles the lines. I originally used sort -R, but it kept the same characters together (ff1#22MvbcAA).
## Objective: Generating Random Password:
function random_password () {
[[ ${#1} -gt 0 ]] && { local length=${1}; } || { local length=16; }
export DEFAULT_PASSWORDLENGTH=${length};
export LC_CTYPE=C;
local random="$(
tr -cd "[:graph:]" < /dev/urandom \
| head -c ${length} \
| sed -e 's|\`|~|g' \
-e 's|\$(|\\$(|g';
)";
echo -e "${random}";
return 0;
}; alias random-password='random_password';
$ random-password 32 ;
)W#j*deZ2#eMuhU4TODO&eu&r)&.#~3F
# Warning: Do not consider these other options
# date +%s | sha256sum | base64 | head -c 32 | xargs -0;
# Output: MGFjNDlhMTE2ZWJjOTI4OGI4ZTFiZmEz
# dd if=/dev/urandom count=200 bs=1 2>/dev/null \
# | tr -cd "[:graph:]" \
# | cut -c-${length} \
# | xargs -0;
# Output: AuS*D=!wkHR.4DZ_la
how easily determine how much is in a variable number of such characters
VAR="ddfasf♣dsdad ♣ as dsdsd ♣ sadsd ♣ df"
echo $VAR # result 4
VAR="♣♣ as dsd♣sd ♣ s♣adsd ♣ df"
echo $VAR # result 5
I tried this but it does not work :-(
echo "$(echo $VAR | tr -dc ♣ | wc -c)"
thank you very much
Try:
chars=${VAR//[!♣]}
echo "${#chars}"
The option to delete all but the specified chars in tr is -C
$(echo $VAR | tr -dC ♣ | wc -c)
Supposed to be a simple bash script, but turned into a monster. This is the 5th try. You don't even want to see the 30 line monstrosity that was attempt #4.. :)
Here's what I want to do: Script generates a random password, with $1=password length, and $2=amount of special characters present in the output.
Or at least, verify before sending to standard out, that at least 1 special character exists. I would prefer the former, but settle for the latter.
Here's my very simple 5th version of this script. It has no verification, or $2:
#!/bin/bash
cat /dev/urandom | tr -dc [=!=][=#=][=#=][=$=][=%=][=^=][:alnum:] | head -c $1
This works just fine, and it's a sufficiently secure password with Usage:
$ passgen 12
2ZuQacN9M#6!
But it, of course, doesn't always print special characters, and it's become an obsession for me now to be able to allow selection of how many special characters are present in the output. It's not as easy as I thought.
Make sense?
By the way, I don't mind a complete rework of the code, I'd be very interested to see some creative solutions!
(By the way: I've tried to pipe it into egrep/grep in various ways, to no avail, but I have a feeling that is a possible solution...)
Thanks
Kevin
How about this:
HASRANDOM=0
while [ $HASRANDOM -eq 0 ]; do
PASS=`cat /dev/urandom | tr -dc [=!=][=#=][=#=][=$=][=%=][=^=][:alnum:] | head -c $1`
if [[ "$PASS" =~ "[~\!#\#\$%^&\*\(\)\-\+\{\}\\\/=]{$2,}" ]]; then
HASRANDOM=1
fi
done
echo $PASS
Supports specifying characters in the output. You could add characters in the regex though I couldn't seem to get square brackets to work even when escaping them.
You probably would want to add some kind of check to make sure it doesn't loop infinitely (though it never went that far for me but I didn't ask for too many special characters either)
Checking for special characters is easy:
echo "$pass" | grep -q '[^a-zA-Z0-9]'
Like this:
while [ 1 ]; do
pass=`cat /dev/urandom | tr -dc [=!=][=#=][=#=][=$=][=%=][=^=][:alnum:] | head -c $1`
if echo "$pass" | grep -q '[^a-zA-Z0-9]'; then
break;
fi
done
And finally:
normal=$(($1 - $2))
(
for ((i=1; i <= $normal; i++)); do
cat /dev/urandom | tr -dc [:alnum:] | head -c 1
echo
done
for ((i=1; i <= $2; i++)); do
cat /dev/urandom | tr -dc [=!=][=#=][=#=][=$=][=%=][=^=] | head -c 1
echo
done
) | shuf | sed -e :a -e '$!N;s/\n//;ta'
Keep it simple... Solution in awk that return the number of "special characters" in input
BEGIN {
FS=""
split("!##$%^",special,"")
}
{
split($0,array,"")
}
END {
for (i in array) {
for (s in special) {
if (special[s] == array[i])
tot=tot+1
}
}
print tot
}
Example output for a2ZuQacN9M#6! is
2
Similar approach in bash:
#!/bin/bash
MyString=a2ZuQacN9M#6!
special=!##$%^
i=0
while (( i++ < ${#MyString} ))
do
char=$(expr substr "$MyString" $i 1)
n=0
while (( n++ < ${#special} ))
do
s=$(expr substr "$special" $n 1)
if [[ $s == $char ]]
then
echo $s
fi
done
done
You may also use a character class in parameter expansion to delete all special chars in a string and then apply some simple Bash string length math to check if there was a minimum (or exact) number of special chars in the password.
# example: delete all punctuation characters in string
str='a!#%3"'
echo "${str//[[:punct:]]/}"
# ... taking Cfreak's approach we could write ...
(
set -- 12 3
strlen1=$1
strlen2=0
nchars=$2
special_chars='[=!=][=#=][=#=][=$=][=%=][=^=]'
HASRANDOM=0
while [ $HASRANDOM -eq 0 ]; do
PASS=`cat /dev/urandom | LC_ALL=C tr -dc "${special_chars}[:alnum:]" | head -c $1`
PASS2="${PASS//[${special_chars}]/}"
strlen2=${#PASS2}
#if [[ $((strlen1 - strlen2)) -eq $nchars ]]; then # set exact number of special chars
if [[ $((strlen1 - strlen2)) -ge $nchars ]]; then # set minimum number of special chars
echo "$PASS"
HASRANDOM=1
fi
done
)
You can count the number of special chars using something like:
number of characters - number of non special characters
Try this:
$ # define a string
$ string='abc!d$'
$ # extract non special chars to letters
$ letters=$(echo $string | tr -dc [:alnum:] )
$ # substract the number on non special chars from total
$ echo $(( ${#string} - ${#letters} ))
2
The last part $(( ... )) evaluate a mathematical expression.