I 'd like to find a bash only (no sed, awk, perl, ...) for finding out if a word is in alphabetical order, in other words every letter is.
example:
bdjkz is true,
ahjmno is true,
sdgla is false.
I'm already struggling just comparing ascii values for characters, so if anyone could point me in the right direction for that it would help a lot!
Thanks
Pure bash solution (no external tool used), using Parameter Expansion to address characters inside strings:
function compare () {
word=$1
for (( pos=0; pos<${#word}-1; pos++ )) ; do
[[ ${word:pos:1} < ${word:pos+1:1} ]] || return 1
done
return 0
}
Tested with
for word in bdjkz ahjmno sdgla ; do
if compare $word ; then
echo $word ordered
else
echo $word not ordered
fi
done
If you can utilize other command line tools (but not awk, sed, perl), you can try:
[[ "YOURSTRING" = "$(echo "YOURSTRING" | grep -o '.' | sort -n |tr -d '\n')" ]] && \
echo "Alphabetic order"
[[ ... ]] is testing the expresion
"YOURSTRING" = string comparison
"$( ... )" capture the inner workings output in a string
echo "YOURSTRING" | grep -o '.' print every character on a line from "YOURSTRING" (-o '.': print only the matches for any single character - NOTE: you might need a new version of grep for this option)
... sort -n | sort the output from 4.
... tr -d '\n' rejoin the characters from 5. (by deleting the trailing new line characters)
You can use:
p='bdjkz'
q=$(fold -w1 <<< "$p"|sort|tr -d "\n")
[[ "$p" == "$q" ]] && echo "in alphabetical order" || echo "not in alphabetical order"
s=($(echo "existingString" | grep -o .)) # put each character of input string in an array.
k=($(printf '%s\n' "${s[#]}" | sort)) # sorts the input string
if [[ "${s[*]}" == "${k[*]}" ]]; then # comparing the input string array with sorted array
echo "alphabetical"
else
echo "not alphabetical"
fi
Related
I've been trying to write bash script, that validates user input with given rules: length > 8, at least one digit, and at least one of these: [#, #, $]
So regex for that is this:
((?=.*\d)(?=.*[##$%&*+-=]).{8,})
I've tried this, but with no result:
result=$(echo $1 | egrep "((?=.*\d)(?=.*[##$%&*+-=]).{8,})")
echo $result
with $1 being input parameter. Also, I'd like to wrap it in IF clause, but echo never outputs anything. What am i doing wrong?
This might help:
[[ ${#1} -ge 8 && $1 =~ [0-9] && $1 =~ [##$] ]] && result="$1"
or with three grep:
result=$(grep -E '.{8}' <<< "$1" | grep '[0-9]' | grep '[##$]')
I'm not use to the syntax of bash script. I'm trying to read a file. For each line I want to keep only the part of the string before the delimiter '/' and put it back into a new file if the word respect a perticular length. I've download a dictionary, but the format does not meet my expectation. Since there is 84000 words, I don't really want to manualy remove what after the '/' for each word. I though it would be an easy thing and I follow couple of idea in other similar question on this site, but it seem that I'm missing something somewhere because it still doesn't work. I can't get the length right. The file Test_Input contains one word per line. Here's the code:
#!/usr/bin/bash
filename="Test_Input.txt"
while read -r line
do
sub= echo $line | cut -d '/' -f1
length= echo ${#sub}
if $length >= 4 && $length <= 10;
then echo $sub >> Test_Output.txt
fi
done < "$filename"
Several items:
I assume that you have been using single back-quotes in the assignments, and not literally sub= echo $line | cut -d '/' -f1, as this would have certainly failed. Alternatively, you can also use sub=$(), as in $(echo $line | cut -d '/' -f1)
The conditions in an if clause need to be encompassed by single or double [], like this: if [[ $length -ge 4 ]] && [[ $length -le 10 ]];
Which brings me to the next point: <= doesn't reliably work in bash. Just use -ge for "greater or equal" and -le for "less or equal".
If your line does not contain any / characters, in your version sub will contain the whole line. This might not be what you want, so I'd advise to also add the -s flag to cut.
You don't need somevar=$(echo $someothervar). Just use somevar=$someothervar
Here's a version that works:
#!/usr/bin/env bash
filename="Test_Input.txt"
while read -r line
do
sub=$(echo $line | cut -s -d '/' -f 1)
length=${#sub}
if [[ $length -ge 4 ]] && [[ $length -le 10 ]];
then echo $sub >> Test_Output.txt
fi
done < "$filename"
Of course, you could also just use sed:
sed -n -r '/^[^/]{4,10}\// s;/.*$;;p' Test_Input.txt > Test_Output.txt
Explanation:
-n Don't print anything unless explicitly marked for printing.
-r Use the extended regex
/<searchterm>/ <operation> Search for lines that match a certain criteria, and perform this operation:
Searchterm is: ^[^/]{4,10}\/ From the beginning of the line, there should be between 4 and 10 non-slash characters, followed by the slash
Operation is: s;/.*$;;p replace everything between the first slash and the end of the line with nothing, then print.
awk is the best tool for this
awk -F/ 'length($1) >= 4 && length($1) <= 10 {print $1} > newfile
I have the next two regex in Bash:
1.^[-a-zA-Z0-9\,\.\;\:]*$
2.^[]a-zA-Z0-9\,\.\;\:]*$
The first matches when the string contains a "-" and the other values.
The second when contains a "]".
I put this values at the beginning of my regex because I can't scape them.
How I can get match the two values at the same time?
You can also place the - at the end of the bracket expression, since a range must be closed on both ends.
^[]a-zA-Z0-9,.;:-]*$
You don't have to escape any of the other characters, either. Colons, semicolons, and commas have no special meaning in any part of a regular expression, and while a period loses its special meaning inside a bracket expression.
Basically you can use this:
grep -E '^.*\-.*\[|\[.*\-.*$'
It matches either a - followed by zero or more arbitrary chars and a [ or a [ followed by zero or more chars and a -
However since you don't accept arbitrary chars, you need to change it to:
grep -E '^[a-zA-Z0-9,.;:]*\-[a-zA-Z0-9,.;:]*\[|\[[a-zA-Z0-9,.;:]*\-[a-zA-Z0-9,.;:]*$'
Maybe, this can help you
#!/bin/bash
while read p; do
echo $p | grep -E '\-.*\]|\].*\-' | grep "^[]a-zA-Z0-9,.;:-]*$"
done <$1
user-host:/tmp$ cat test
-i]string
]adfadfa-
string-
]string
str]ing
]123string
123string-
?????
++++++
user-host:/tmp$ ./test.sh test
-i]string
]adfadfa-
There are two questions in your post.
One is in the description:
How I can get match the two values at the same time?
That is an OR match, which could be done with a range that mix your two ranges:
pattern='^[]a-zA-Z0-9,.;:-]*$'
That will match a line that either contains one (or several) -…OR…]…OR any of the included characters. That would be all the lines (except ?????, ++++++ and as df gh) in the test script below.
Two is in the title:
… a string contains “-” and “]” at the same time
That is an AND match. The simplest (and slowest) way to do it is:
echo "$line" | grep '-' | grep ']' | grep '^[-a-zA-Z0-9,.;:]*$'
The first two calls to grep select only the lines that:
contain both (one or several) - and (one or several) ]
Test script:
#!/bin/bash
printlines(){
cat <<-\_test_lines_
asdfgh
asdfgh-
asdfgh]
as]df
as,df
as.df
as;df
as:df
as-df
as]]]df
as---df
asAS]]]DFdf
as123--456DF
as,.;:-df
as-dfg]h
as]dfg-h
a]s]d]f]g]h
a]s]d]f]g]h-
s-t-r-i-n-g]
as]df-gh
123]asdefgh
123asd-fgh-
?????
++++++
as df gh
_test_lines_
}
pattern='^[]a-zA-Z0-9,.;:-]*$'
printf '%s\n' "Testing the simple pattern of $pattern"
while read line; do
resultgrep="$( echo "$line" | grep "$pattern" )"
printf '%13s %-13s\n' "$line" "$resultgrep"
done < <(printlines)
echo "#############################################################"
echo
p1='-'; p2=']'; p3='^[]a-zA-Z0-9,.;:-]*$'
printf '%s\n' "Testing a 'grep AND' of '$p1', '$p2' and '$p3'."
while read line; do
resultgrep="$( echo "$line" | grep "$p1" | grep "$p2" | grep "$p3" )"
[[ $resultgrep ]] && printf '%13s %-13s\n' "$line" "$resultgrep"
done < <(printlines)
echo "#############################################################"
echo
printf '%s\n' "Testing an 'AWK AND' of '$p1', '$p2' and '$p3'."
while read line; do
resultawk="$( echo "$line" |
awk -v p1="$p1" -v p2="$p2" -v p3="$p3" '$0~p1 && $0~p2 && $0~p3' )"
[[ $resultawk ]] && printf '%13s %-13s\n' "$line" "$resultawk"
done < <(printlines)
echo "#############################################################"
echo
printf '%s\n' "Testing a 'bash AND' of '$p1', '$p2' and '$p3'."
while read line; do
rgrep="$( echo "$line" | grep "$p1" | grep "$p2" | grep "$p3" )"
[[ ( $line =~ $p1 ) && ( $line =~ $p2 ) && ( $line =~ $p3 ) ]]
rbash=${BASH_REMATCH[0]}
[[ $rbash ]] && printf '%13s %-13s %-13s\n' "$line" "$rgrep" "$rbash"
done < <(printlines)
echo "#############################################################"
echo
I have this
if [[ ! $newstring == *['!'##\$%^\&*()_+]* ]]
then
echo Error - Does not contain One Special Character - $newstring
i=$((i+1))
fi
Which checks if the string only has one single character from the bank, i want to check if it has more than one?
What would be the best way?
Either add a second class
if [[ "$newstring" != *['!'##\$%^\&*\(\)_+]*['!'##\$%^\&*\(\)_+]* ]]
or strip anything else out and check length
t="${newstring//[^!##\$%^\&*()_+]}"
if [ ${#t} -lt 2 ]
We can use tr to solve it.
$ string='Hello-World_12#$##*&%)(!####'
$ number=$(( $(tr -d '[[:alnum:]]' <<< "$string"|wc -m) - 1 ))
$ echo "We have $number of special characters"
$ 16
This should be short and faster.
#!/bin/bash
a='!*#%6789';
if [[ `echo $a | sed "s/\(.\)/\1\n/g"|grep -c "[[:punct:]]"` -gt 1 ]]; then echo shenzi; else echo koba; fi
grep can be useful to provide the match
grep -oP "^[^'\!'##\$%^\&*()_+]*['\!'##\$%^\&*()_+][^'\!'##\$%^\&*()_+]+$"
test
$ echo "#asdfasdf234" | grep -oP "^[^'\!'##\$%^\&*()_+]*['\!'##\$%^\&*()_+][^'\!'##\$%^\&*()_+]+$"
will match the string as
#asdfasdf234
$ echo "#asdf#asdf234" | grep -oP "^[^'\!'##\$%^\&*()_+]*['\!'##\$%^\&*()_+][^'\!'##\$%^\&*()_+]+$"
will not match the string
The if construct can be
echo $newstring| grep -oP "^[^'\!'##\$%^\&*()_+]*['\!'##\$%^\&*()_+][^'\!'##\$%^\&*()_+]+$"
if [[ $? -eq 0 ]] > /dev/null
then
echo Error - Does not contain One Special Character - $newstring
i=$((i+1))
fi
Here the regex
^[^'\!'##\$%^\&*()_+]*['\!'##\$%^\&*()_+][^'\!'##\$%^\&*()_+]+$
matches all strings with exact one occurence of the special character
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.