Given a shell variable whose value is a semantic version, how can I create another shell variable whose value is (tuple 1 × 1000000) + (tuple 2 × 1000) + (tuple 3) ?
E.g.
$ FOO=1.2.3
$ BAR=#shell magic that, given ${FOO} returns `1002003`
# Shell-native string-manipulation? sed? ...?
I'm unclear about how POSIX-compliance vs. shell-specific syntax comes into play here, but I think a solution not bash-specific is preferred.
Update: To clarify: this isn't as straightforward as replacing "." with zero(es), which was my initial thought.
E.g. The desired output for 1.12.30 is 1012030, not 100120030, which is what a .-replacement approach might provide.
Bonus if the answer can be a one-liner variable-assignment.
A perl one-liner:
echo $FOO | perl -pne 's/\.(\d+)/sprintf "%03d", $1/eg'
How it works:
perl -pne does a REPL with the supplied program
The program contains a replacement function s///
The search string is the regex \.(\d+) which matches a string beginning with dot and ends with digits and capture those digits
The e modifier of the s/// function evaluates the right-hand side of the s/// replacement as an expression. Since we captured the digits, they'll be converted into int and formatted into leading zeros with sprintf
The g modifier replaces all instances of the regex in the input string
Demo
Split on dots, then loop and multiply/add:
version="1.12.30"
# Split on dots instead of spaces from now on
IFS="."
# Loop over each number and accumulate
int=0
for n in $version
do
int=$((int*1000 + n))
done
echo "$version is $int"
Be aware that this treats 1.2 and 0.1.2 the same. If you want to always treat the first number as major/million, consider padding/truncating beforehand.
This should do it
echo $foo | sed 's/\./00/g'
How about this?
$ ver=1.12.30
$ foo=$(bar=($(echo $ver|sed 's/\./ /g')); expr ${bar[0]} \* 1000000 + ${bar[1]} \* 1000 + ${bar[2]})
$ echo $foo
1012030
Related
I need help with this its been busting my mind.
I have a read with a variable with integers 10 20 -30.
All separated by white space. I try to change the minus to plus and save it onto another variable but it's not saving. If I can't change to plus I would like to remove it so then I can do:
var=$((${num// /+/}))
So it can add all integers.
This is what I have:
read num
echo $num
sum=$num | sed -e 's/-/+/g'
echo $sum
Using standard POSIX variable expansion and arithmetic:
#!/usr/bin/env sh
# Computes the sum of all arguments
sum () {
# Save the IFS value
_OIFS=$IFS
# Set the IFS to + sign
IFS=+
# Expand the arguments with the IFS + sign
# inside an arithmetic expression to get
# the sum of all arguments.
echo "$(($*))"
# Restore the original IFS
IFS=$_OIFS
}
num='10 20 -30'
# shellcheck disable=SC2086 # Intended word splitting of string into arguments
sum $num
More featured version with a join function:
#!/usr/bin/env sh
# Join arguments with the provided delimiter character into a string
# $1: The delimiter character
# $#: The arguments to join
join () {
# Save the IFS value
_OIFS=$IFS
# Set the IFS to the delimiter
IFS=$1
# Shift out the delimiter from the arguments
shift
# Output the joined string
echo "$*"
# Restore the original IFS
IFS=$_OIFS
}
# Computes the sum of all arguments
sum () {
# Join the arguments with a + sign to form a sum expression
sum_expr=$(join + "$#")
# Submit the sum expression to a shell's arithmetic expression
# shellcheck disable=SC2004 # $sum_expr needs its $ to correctly split terms
echo "$(($sum_expr))"
}
num='10 20 -30'
# shellcheck disable=SC2086 # Intended word splitting of string into arguments
sum $num
Simply: whipe last slash:
num="10 20 -30"
echo $((${num// /+}))
0
Some details
*Bash battern substitution has nothing common with so called regular expression. Correct syntax is:
${parameter/pattern/string}
... If pattern begins with /, all matches of pattern are replaced with string.
Normally only the first match is replaced. ...
See: man -Pless\ +/parameter.pattern.string bash
If you try your syntax:
echo ${num// /+/}
10+/20+/-30
Then
echo ${num// /+}
10+20+-30
Or even, to make this pretty:
echo ${num// / + }
10 + 20 + -30
But result will stay same:
echo $(( ${num// / + } ))
0
sum=$num | sed -e 's/-/+/g'
With respect to what's present above, sum=$num and sed become two different commands. It's not grouped together as you wanted, which makes the sed ineffective.
Also, you'd need to echo $num
Solution is to group them together, like:
sum=`echo $num | sed -e 's/-/+/g`
OR
sum=$(echo $num | sed -e 's/-/+/g')
OR Rather, an alternate approach
sum=${num//-/+}
I have this variable:
A="Some variable has value abc.123"
I need to extract this value i.e abc.123. Is this possible in bash?
Simplest is
echo "$A" | awk '{print $NF}'
Edit: explanation of how this works...
awk breaks the input into different fields, using whitespace as the separator by default. Hardcoding 5 in place of NF prints out the 5th field in the input:
echo "$A" | awk '{print $5}'
NF is a built-in awk variable that gives the total number of fields in the current record. The following returns the number 5 because there are 5 fields in the string "Some variable has value abc.123":
echo "$A" | awk '{print NF}'
Combining $ with NF outputs the last field in the string, no matter how many fields your string contains.
Yes; this:
A="Some variable has value abc.123"
echo "${A##* }"
will print this:
abc.123
(The ${parameter##word} notation is explained in §3.5.3 "Shell Parameter Expansion" of the Bash Reference Manual.)
Some examples using parameter expansion
A="Some variable has value abc.123"
echo "${A##* }"
abc.123
Longest match on " " space
echo "${A% *}"
Some variable has value
Longest match on . dot
echo "${A%.*}"
Some variable has value abc
Shortest match on " " space
echo "${A%% *}"
some
Read more Shell-Parameter-Expansion
The documentation is a bit painful to read, so I've summarised it in a simpler way.
Note that the '*' needs to swap places with the ' ' depending on whether you use # or %. (The * is just a wildcard, so you may need to take off your "regex hat" while reading.)
${A% *} - remove shortest trailing * (strip the last word)
${A%% *} - remove longest trailing * (strip the last words)
${A#* } - remove shortest leading * (strip the first word)
${A##* } - remove longest leading * (strip the first words)
Of course a "word" here may contain any character that isn't a literal space.
You might commonly use this syntax to trim filenames:
${A##*/} removes all containing folders, if any, from the start of the path, e.g.
/usr/bin/git -> git
/usr/bin/ -> (empty string)
${A%/*} removes the last file/folder/trailing slash, if any, from the end:
/usr/bin/git -> /usr/bin
/usr/bin/ -> /usr/bin
${A%.*} removes the last extension, if any (just be wary of things like my.path/noext):
archive.tar.gz -> archive.tar
How do you know where the value begins? If it's always the 5th and 6th words, you could use e.g.:
B=$(echo "$A" | cut -d ' ' -f 5-)
This uses the cut command to slice out part of the line, using a simple space as the word delimiter.
As pointed out by Zedfoxus here. A very clean method that works on all Unix-based systems. Besides, you don't need to know the exact position of the substring.
A="Some variable has value abc.123"
echo "$A" | rev | cut -d ' ' -f 1 | rev
# abc.123
More ways to do this:
(Run each of these commands in your terminal to test this live.)
For all answers below, start by typing this in your terminal:
A="Some variable has value abc.123"
The array example (#3 below) is a really useful pattern, and depending on what you are trying to do, sometimes the best.
1. with awk, as the main answer shows
echo "$A" | awk '{print $NF}'
2. with grep:
echo "$A" | grep -o '[^ ]*$'
the -o says to only retain the matching portion of the string
the [^ ] part says "don't match spaces"; ie: "not the space char"
the * means: "match 0 or more instances of the preceding match pattern (which is [^ ]), and the $ means "match the end of the line." So, this matches the last word after the last space through to the end of the line; ie: abc.123 in this case.
3. via regular bash "indexed" arrays and array indexing
Convert A to an array, with elements being separated by the default IFS (Internal Field Separator) char, which is space:
Option 1 (will "break in mysterious ways", as #tripleee put it in a comment here, if the string stored in the A variable contains certain special shell characters, so Option 2 below is recommended instead!):
# Capture space-separated words as separate elements in array A_array
A_array=($A)
Option 2 [RECOMMENDED!]. Use the read command, as I explain in my answer here, and as is recommended by the bash shellcheck static code analyzer tool for shell scripts, in ShellCheck rule SC2206, here.
# Capture space-separated words as separate elements in array A_array, using
# a "herestring".
# See my answer here: https://stackoverflow.com/a/71575442/4561887
IFS=" " read -r -d '' -a A_array <<< "$A"
Then, print only the last elment in the array:
# Print only the last element via bash array right-hand-side indexing syntax
echo "${A_array[-1]}" # last element only
Output:
abc.123
Going further:
What makes this pattern so useful too is that it allows you to easily do the opposite too!: obtain all words except the last one, like this:
array_len="${#A_array[#]}"
array_len_minus_one=$((array_len - 1))
echo "${A_array[#]:0:$array_len_minus_one}"
Output:
Some variable has value
For more on the ${array[#]:start:length} array slicing syntax above, see my answer here: Unix & Linux: Bash: slice of positional parameters, and for more info. on the bash "Arithmetic Expansion" syntax, see here:
https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Arithmetic-Expansion
https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Shell-Arithmetic
You can use a Bash regex:
A="Some variable has value abc.123"
[[ $A =~ [[:blank:]]([^[:blank:]]+)$ ]] && echo "${BASH_REMATCH[1]}" || echo "no match"
Prints:
abc.123
That works with any [:blank:] delimiter in the current local (Usually [ \t]). If you want to be more specific:
A="Some variable has value abc.123"
pat='[ ]([^ ]+)$'
[[ $A =~ $pat ]] && echo "${BASH_REMATCH[1]}" || echo "no match"
echo "Some variable has value abc.123"| perl -nE'say $1 if /(\S+)$/'
I have a string that contains
fastcgi_cache_path /var/run/nginx-cache15 levels=1:2 keys_zone=MYSITEP:100m inactive=60m;
One of the goals of this script is to increment nginx-cache two digits based on the value find on previous file. For doing that I used this code:
# Replace cache_path
PREV=$(ls -t /etc/nginx/sites-available | head -n1) #find the previous cache_path number
CACHE=$(grep fastcgi_cache_path $PREV | awk '{print $2}' |cut -d/ -f4) #take the string to change
SUB=$(echo $CACHE |sed "s/nginx-cache[0-9]*[0-9]/&#/g;:a {s/0#/1/g;s/1#/2/g;s/2#/3/g;s/3#/4/g;s/4#/5/g;s/5#/6/g;s/6#/7/g;s/7#/8/g;s/8#/9/g;s/9#/#0/g;t a};s/#/1/g") #increment number
sed -i "s/nginx-cache[0-9]*/$SUB/g" $SITENAME #replace number
Maybe not so elegant, but it works.
The other goal is to increment last letter of all occurrences of MYSITEx (MYSITEP, in that case, should become MYSITEQ, after MYSITEP, etc. etc and once MYSITEZ will be reached add another letter, like MYSITEAA, MYSITEAB, etc. etc.
I thought something like:
sed -i "s/MYSITEP[A-Z]*/MYSITEGG/g" $SITENAME
but it can't works cause MYSITEGG is a static value and can't be used.
How can I calculate the last letter, increment it to the next one and once the last Z letter will be reached, add another letter?
Thank you!
Perl's autoincrement will work on letters as well as digits, in exactly the manner you describe
We may as well tidy your nginx-cache increment as well while we're at it
I assume SITENAME holds the name of the file to be modified?
It would look like this. I have to assign the capture $1 to an ordinary variable $n to increment it, as $1 is read-only
perl -i -pe 's/nginx-cache\K(\d+)/ ++($n = $1) /e; s/MYSITE\K(\w+)/ ++($n = $1) /e;' $SITENAME
If you wish, this can be done in a single substitution, like this
perl -i -pe 's/(?:nginx-cache|MYSITE)\K(\w+)/ ++($n = $1) /ge' $SITENAME
Note: The solution below is needlessly complicated, because as Borodin's helpful answer demonstrates (and #stevesliva's comment on the question hinted at), Perl directly supports incrementing letters alphabetically in the manner described in the question, by applying the ++ operator to a variable containing a letter (sequence); e.g.:
$ perl -E '$letters = "ZZ"; say ++$letters'
AAA
The solution below may still be of interest as an annotated showcase of how Perl's power can be harnessed from the shell, showing techniques such as:
use of s///e to determine the replacement string with an expression.
splitting a string into a character array (split //, "....")
use of the ord and chr functions to get the codepoint of a char., and convert a(n incremented) codepoint back to a char.
string replication (x operator)
array indexing and slices:
getting an array's last element ($chars[-1])
getting all but the last element of an array (#chars[0..$#chars-1])
A perl solution (in effect a re-implementation of what ++ can do directly):
perl -pe 's/\bMYSITE\K([A-Z]+)/
#chars = split qr(), $1; $chars[-1] eq "Z" ?
"A" x (1 + scalar #chars)
:
join "", #chars[0..$#chars-1], chr (1 + ord $chars[-1])
/e' <<'EOF'
...=MYSITEP:...
...=MYSITEZP:...
...=MYSITEZZ:...
EOF
yields:
...=MYSITEQ:... # P -> Q
...=MYSITEZQ:... # ZP -> ZQ
...=MYSITEAAA:... # ZZ -> AAA
You can use perl's -i option to replace the input file with the result
(perl -i -pe '...' "$SITENAME").
As Borodin's answer demonstrates, it's not hard to solve all tasks in the question using perl alone.
The s function's /e option allows use of a Perl expression for determining the replacement string, which enables sophisticated replacements:
$1 references the current MYSITE suffix in the expression.
#chars = split qr(), $1 splits the suffix into a character array.
$chars[-1] eq "Z" tests if the last suffix char. is Z
If so: The suffix is replaced with all As, with an additional A appended
("A" x (1 + scalar #chars)).
Otherwise: The last suffix char. is replaced with the following letter in the alphabet
(join "", #chars[0..$#chars-1], chr (1 + ord $chars[-1]))
To get what is between "aa=" and either % or empty
string = "aa=value%bb"
string2 = "bb=%aa=value"
The rule must work on both strings to get the value of "aa="
I would like a BASH LANGUAGE solution if possible.
Use this:
result=$(echo "$string" | grep -o 'aa=[^%]*')
result=${result:3} # remove aa=
[^%]* matches any sequence of characters that doesn't contain %, so it will stop when it gets to % or the end of the string. $(result:3} expands to the substring starting from character 3, which removes aa= from the beginning.
for example qa_sharutils-2009-04-22-15-20-39, want chop last 20 bytes, and get 'qa_sharutils'.
I know how to do it in sed, but why $A=${A/.\{20\}$/} does not work?
Thanks!
If your string is stored in a variable called $str, then this will get you give you the substring without the last 20 digits in bash
${str:0:${#str} - 20}
basically, string slicing can be done using
${[variableName]:[startIndex]:[length]}
and the length of a string is
${#[variableName]}
EDIT:
solution using sed that works on files:
sed 's/.\{20\}$//' < inputFile
similar to substr('abcdefg', 2-1, 3) in php:
echo 'abcdefg'|tail -c +2|head -c 3
using awk:
echo $str | awk '{print substr($0,1,length($0)-20)}'
or using strings manipulation - echo ${string:position:length}:
echo ${str:0:$((${#str}-20))}
In the ${parameter/pattern/string} syntax in bash, pattern is a path wildcard-style pattern, not a regular expression. In wildcard syntax a dot . is just a literal dot and curly braces are used to match a choice of options (like the pipe | in regular expressions), so that line will simply erase the literal string ".20".
There are several ways to accomplish the basic task.
$ str="qa_sharutils-2009-04-22-15-20-39"
If you want to strip the last 20 characters. This substring selection is zero based:
$ echo ${str::${#str}-20}
qa_sharutils
The "%" and "%%" to strip from the right hand side of the string. For instance, if you want the basename, minus anything that follows the first "-":
$ echo ${str%%-*}
qa_sharutils
only if your last 20 bytes is always date.
$ str="qa_sharutils-2009-04-22-15-20-39"
$ IFS="-"
$ set -- $str
$ echo $1
qa_sharutils
$ unset IFS
or when first dash and beyond are not needed.
$ echo ${str%%-*}
qa_sharutils