I have a file containing on each line a string of the form
string1.string2:\string3{string4}{number}
and what I want to extract is the number. I've searched and tried for a while to get this done using sed or bash, but failed. Any help would be much appreciated.
Edit 1: The strings may contains numbers.
$ echo 'string1.string2:\string3{string4}{number}' |\
cut -d'{' -f3 | cut -d'}' -f 1
number
Using sed:
sed 's/[^}]*}{\([0-9]*\)}/\1/' input_file
Description:
[^}]*} : match anything that is not } and the following }
{\([0-9]*\)}: capture the following digits within {...}
/\1/ : substitute all with the captured number
Use grep:
grep -o '\{[0-9]\+\}' | tr -d '[{}]'
In bash:
sRE='[[:alnum:]]+'
nRE='[[:digit:]]+'
[[ $str =~ $sRE\.$sRE:\\$sRE\{$sRE\}\{($nRE)\} ]] && number=${BASH_REMATCH[1]}
You can drop the first part of the regular expression, if your text file is sufficiently uniform:
[[ $str =~ \\$sRE{$sRE}{($nRE)} ]] && number=${BASH_REMATCH[1]}
or even
[[ $str =~ {$sRE}{($nRE)} ]] && number=${BASH_REMATCH[1]}
Related
I have a string such as plantford1775.274.284b63.11.
I have been using identity=$( echo "$identity" | cut -d'.' -f3) to cut at each dot, and then choose the third section. I am left with 284b63.
The format of this part is always a letter, sandwiched by varying amounts of numbers. I would like to take the first few numbers before the letter. An example code line would be this:
identity=$( echo "$identity" | cut -d'anyletter' -f1)
What do I replace anyletter with to cut at whatever letter is listed there, so that I end with a string of 284?
This could be done in single awk, please try following written and tested with your shown samples.
echo "$identity" | awk -F'.' '{sub(/[^0-9].*/,"",$3);print $3}'
Explanation: simple explanation would be, passing echo command's output as a standard input to awk code. In awk program, setting field separator as . for values. Then in 3rd field substituting(using sub function of awk) everything apart from digits with NULL in 3rd field, then printing it.
Try:
echo plantford1775.274.284b63.11 | cut -d. -f3 | sed 's/[a-z].*//'
Or a slight variation on the REGEX, with [[...]] in bash:
v="plantford1775.274.284b63.11"
[[ $v =~ ^[^.]+.[^.]+.([^.]+).*$ ]] && echo ${BASH_REMATCH[1]}
Output
284b63
Or if you are only interested in the digits before the letter:
[[ $v =~ ^[^.]+.[^.]+.([[:digit:]]+)[^.]+.*$ ]] && echo ${BASH_REMATCH[1]}
Output
284
With bash, using the =~ operator :
[[ $identity =~ [^.]*.[^.]*.([0-9]+) ]] && identity=${BASH_REMATCH[1]}
or, in POSIX shell:
identity=${identity#*.*.}
identity=${identity%%[^0-9]*}
or, using sed:
identity=$(sed 's/[^.]*.[^.]*.\([0-9]*\).*/\1/' <<< "$identity")
Maybe you can use a bash regex and get the result from $BASH_REMATCH.
[[ "$identity" =~ ([0-9]+)[a-z][0-9]+ ]] && identity="${BASH_REMATCH[1]}"
Say we have
identity=284b63
then you can do a
lead=${identity%[a-z]*}
to set lead to 284. Feel free to adapt the pattern to upper case letters and/or other separators.
If the format of this part is always a letter, sandwiched by varying amounts of numbers, and you want to match this format, you might also use gnu awk, setting the field separator to . and use a pattern with a capture group for the 3rd field.
The pattern captures 1 or more digits from the start of the string, and match one of more chars [a-z] after it followed by a digit.
echo "$identity" | awk -F'.' 'match($3, /^([0-9]+)[a-z]+[0-9]/, ary) {print ary[1]}'
Output
284
Or using sed with a pattern matching the first 2 dots and the capture group after the 2nd dot:
identity=$(sed 's/^[^.]\+\.[^\.]\+\.\([0-9]\+\)[a-z]\+[0-9].*/\1/' <<< "$identity")
I am trying to collect the lines from a file which doesn't start with a # as its first caracter.
I have this code I am able to get them:
while IFS= read -r line
do
[[ -z "$line" ]] && continue
[[ "$line" =~ ^# ]] && continue
#echo "LINEREADED: $line"
done < $file
So the output I have is something like this:
modules/core_as/xxxx/xxxxxxxxxxxxxxxxxxxxxxxxxxx [100]
My question is how can I get only the string without the [100]?
I know there is some commands like sed or trim but the problem is that the string is not always that length, sometimes is different like:
cross_modules/core_as/xxxx/xxxxxxxxx [100-103]
or
cross_modules/core_as/xxxxxxxxxxxx/xxxxxxxxx [100-103]
or anything like that...
And in all this cases I only need the string without the [....] and without the last blank space at the end of last x, whichever the length of the string is, like cross_modules/core_as/xxxxxxxxxxxx/xxxxxxxxx
echo ${caseReaded:1:${#caseReaded}-7}
This also do the job but is not generic for any length.
Does anyone knows how I can get this?
You can strip a certain part of a string in bash
echo "${line% [*}"
cross_modules/core_as/xxxx/xxxxxxxxx
modules/core_as/xxxx/xxxxxxxxxxxxxxxxxxxxxxxxxxx
cross_modules/core_as/xxxxxxxxxxxx/xxxxxxxxx
If the spaces are only before [:
while IFS= read -r line _
do
[[ -z $line ]] && continue
[[ $line =~ ^# ]] && continue
done < "$file"
grep to match all lines not starting with # and then display the first field using cut, which works if the first field doesn't contain spaces:
grep -v ^# "$file" | cut -f1 -d' '
If the thing before [100] contains spaces, this may be the way to go:
grep -v ^# "$file" | sed -E 's/^(.*) .*$/\1/'
The last one works because the .* match in sed is greedy so only the last space will be left to match the outer condition .*$.
I have a string like that
1-a-bc-dxyz
I'd want to get 1-a-bc-d ( first 5 characters, only number and alphabet)
Thanks
With gawk:
awk '{ for ( i=1;i<=length($0);i++) { if ( match(substr($0,i,1),/[[:alnum:]]/)) { cnt++;if ( cnt==5) { print substr($0,1,i) } } } }' <<< "1-a-bc-dxyz"
Read each character one by one and then if there is a pattern match for an alpha-numeric character (using the match function), increment a variable cnt. When cnt gets to 5, print the string we have seen so far (using the substr function)
Output:
1-a-bc-d
a='1-a-bc-dxyz'
count=0
for ((i=0;i<${#a};i++)); do
if [[ "${a:$i:1}" =~ [0-9]|[a-Z] ]] && [[ $((++count)) -eq 5 ]]; then
echo "${a:0:$((i+1))}"
exit
fi
done
You can further shrink this as;
a='1-a-bc-dxyz'
count=0
for ((i=0;i<${#a};i++)); do [[ "${a:$i:1}" =~ [0-9]|[a-Z] ]] && [[ $((++count)) -eq 5 ]] && echo "${a:0:$((i+1))}"; done
Using GNU awk:
$ echo 1-a-bc-dxyz | \
awk -F '' '{b=i="";while(gsub(/[0-9a-z]/,"&",b)<5)b=b $(++i);print b}'
1-a-bc-d
Explained:
awk -F '' '{ # separate each char to its own field
b=i="" # if you have more than one record to process
while(gsub(/[0-9a-z]/,"&",b)<5) # using gsub for counting (adjust regex if needed)
b=b $(++i) # gather buffer
print b # print buffer
}'
GNU sed supports an option to replace the k-th occurrence and all after that.
echo "1-a-bc-dxyz" | sed 's/[^a-zA-Z0-9]*[a-zA-Z0-9]//g6'
Using Combination of sed & AWK
echo 1-a-bc-dxyz | sed 's/[-*%$##]//g' | awk -F '' {'print $1$2$3$4$5'}
You can use for loop for printing character as well.
echo '1-a-bc-dxyz' | grep -Eo '^[[:print:]](-*[[:print:]]){4}'
That is pretty simple.
Neither sed nor awk.
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
I cannot get this to work. I only want to get the string between 2 others in bash. Like this:
FOUND=$(echo "If show <start>THIS WORK<end> then it work" | **the magic**)
echo $FOUND
It seems so simple...
sed -n 's/.*<start>\(.*\)<end>.*/\1/p'
This can be done in bash without any external commands such as awk and sed. When doing a regex match in bash, the results of the match are put into a special array called BASH_REMATCH. The second element of this array contains the match from the first capture group.
data="If show <start>THIS WORK<end> then it work"
regex="<start>(.*)<end>"
[[ $data =~ $regex ]] && found="${BASH_REMATCH[1]}"
echo $found
This can also be done using perl regex in grep (GNU specific):
found=$(grep -Po '(?<=<start>).*(?=<end>)' <<< "If show <start>THIS WORK<end> then it work")
echo "$found"
If you have < start > and < end > in your string then this will work. Set the FS to < and >.
[jaypal:~/Temp] FOUND=$(echo "If show <start>THIS WORK<end> then it work" |
awk -v FS="[<>]" '{print $3}')
[jaypal:~/Temp] echo $FOUND
THIS WORK