I want to extract the strings from file name - ksh

one_two_three_four_five.rtf
I need five in A variable
I need four in B variable
And remaining in C variable
Should read from the last character
Note after 2 underscore from the last. There could be many underscores but should take has C variable.
Is it possible?

For example using parameter expansion
#!/bin/ksh
string="one_two_three_four_five.rtf"
base=${string%.rtf}
a=${base##*_}; base=${base%_$a}
b=${base##*_}; base=${base%_$b}
c=$base
echo "$a - $b - $c"

s="one_two_three_four_five.rtf"
source <(sed -r 's/(.*)_([^_]*)_([^_]*)[.].*/C="\1"; B="\2";A="\3"/' <<< "${s}")
# Result:
echo "A=$A, B=$B, C=$C"
A=five, B=four, C=one_two_three
Explanation:
sed -r No need for escaping backslashes
(.*)_ Matches largest string until underscore with the condition that there are underscores left for matching the remaining string
([^_]*) String without underscore
[.] A dot without special meaning
"\1" First remembered string
<<< "${s}" Input for sed is like echo "${s}" | sed ...
<(..) Simulates a file, so sourcing these will execute the commands.

Related

In bash how can I get the last part of a string after the last hyphen [duplicate]

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 ${parameter##word} notation is explained in ยง3.5.3 "Shell Parameter Expansion" of the Bash Reference Manual.)
Some examples using parameter expansion
A="Some variable has value abc.123"
echo "${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 $1 if /(\S+)$/'

sed Capital_Case not working

I'm trying to convert a string that has either - (hyphen) or _ (underscore) to Capital_Case string.
#!/usr/bin/env sh
function cap_case() {
[ $# -eq 1 ] || return 1;
_str=$1;
_capitalize=${_str//[-_]/_} | sed -E 's/(^|_)([a-zA-Z])/\u\2/g'
echo "Capitalize:"
echo $_capitalize
return 0
}
read string
echo $(cap_case $string)
But I don't get anything out.
First I am replacing any occurrence of - and _ with _ ${_str//[-_]/_}, and then I pipe that string to sed which finds the first letter, or _ as the first group, and then the letter after the first group in the second group, and I want to uppercase the found letter with \u\2. I tried with \U\2 but that didn't work as well.
I want the string some_string to become
Some_String
And string some-string to become
Some_String
I'm on a mac, using zsh if that is helpful.
EDIT: More generic solution here to make each field's first letter Capital.
echo "some_string_other" | awk -F"_" '{for(i=1;i<=NF;i++){$i=toupper(substr($i,1,1)) substr($i,2)}} 1' OFS="_"
Following awk may help you.
echo "some_string" | awk -F"_" '{$1=toupper(substr($1,1,1)) substr($1,2);$2=toupper(substr($2,1,1)) substr($2,2)} 1' OFS="_"
Output will be as follows.
echo "some_string" | awk -F"_" '{$1=toupper(substr($1,1,1)) substr($1,2);$2=toupper(substr($2,1,1)) substr($2,2)} 1' OFS="_"
Some_String
This being zsh, you don't need sed (or even a function, really):
$ s=some-string-bar
$ print ${(C)s:gs/-/_}
Some_String_Bar
The (C) flag capitalizes words (where "words" are defined as sequences of alphanumeric characters separated by other characters); :gs/-/_ replaces hyphens with underscores.
If you really want a function, it's cap_case () { print ${(C)1:gs/-/_} }.
pure bash:
#!/bin/bash
camel_case(){
local d display string
declare -a strings # = scope local
[ "$2" ] && d="$2" || d=" " # optional output delimiter
ifs_ini="$IFS"
IFS+='_-' # we keep initial IFS
strings=( "$1" ) # array
for string in ${strings[#]} ; do
display+="${string^}$d"
done
echo "${display%$d}"
IFS="$ifs_ini"
}
camel_case "some-string_here" "_"
camel_case "some-string_here some strings here" "+"
camel_case "some-string_here some strings here"
echo "$BASH_VERSION"
exit
output:
Some_String_Here
Some+String+Here+Some+Strings+Here
Some String Here Some Strings Here
4.4.18(1) release
You can try this gnu sed
echo 'some_other-string' | sed -E 's/(^.)/\u&/;s/[_-](.)/_\u\1/g'
Explains :
s/(^.)/\u&/
(^.) match the first char and \u& put the match in capital letter.
s/[_-](.)/_\u\1/g
[_-](.) capture a char preceded by _ or - and replace it by _ and the matched char in capital letter.
The g at the end tell sed to make the replacement for each char which meet the criteria
You didn't assign to _capitalize - you set a _capitalize environment variable for the empty command that you piped into sed.
You probably meant
_capitalize=$(<<<"${_str//[-_]/_}" sed -E 's/(^|_)([a-zA-Z])/\1\u\2/g')
Note also that ${//} isn't standard shell, so you really ought to specify an interpreter other than sh.
A simpler approach would be simply:
#!/bin/sh
cap_case() {
printf "Capitalize: "
echo "$*" | sed -e 'y/-/_/' -e 's/\(^\|_\)[[:alpha:]]/\U&/g'
}
echo $(cap_case "snake_case")
Note that the \u / \U replacement is a GNU extension to sed - if you're using a non-GNU implementation, check whether it supports this feature.

Adding a comma after $variable

I'm writing a for loop in bash to run a command and I need to add a comma after one of my variables. I can't seem to do this without an extra space added. When I move "," right next to $bams then it outputs *.sorted,
#!/bin/bash
bams=*.sorted
for i in $bams
do echo $bams ","
done;
Output should be this:
'file1.sorted','file2.sorted','file3.sorted'
The eventual end goal is to be able to insert a list of files into a --flag in the format above. Not sure how to do that either.
First, a literal answer (if your goal were to generate a string of the form 'foo','bar','baz', rather than to run a program with a command line equivalent to somecommand --flag='foo','bar','baz', which is quite different):
shopt -s nullglob # generate a null result if no matches exist
printf -v var "'%s'," *.sorted # put list of files, each w/ a comma, in var
echo "${var%,}" # echo contents of var, with last comma removed
Or, if you don't need the literal single quotes (and if you're passing your result to another program on its command line with the single quotes being syntactic rather than literal, you absolutely don't want them):
files=( *.sorted ) # put *.sorted in an array
IFS=, # set the comma character as the field separator
somecommand --flag "${files[*]}" # run your program with the comma-separated list
try this -
lst=$( echo *.sorted | sed 's/ /,/g' ) # stack filenames with commas
echo $lst
if you really need the single-ticks around each filename, then
lst="'$( echo *.sorted | sed "s/ /','/g" )'" # commas AND quotes
#!/bin/bash
bams=*.sorted
for i in $bams
do flag+="${flag:+,}'$i'"
done
echo $flag

shell split a string using variable delimiters

My problem is quite simple but I do not manage to solve it:
I have a string that looks like this:
-3445.51692 -7177.16664 -9945.11057
the tricky part is that there could be zero or more withe space between each number and the latter can be either negative or positive, meaning that the string could also be like:
-3445.51692-7177.16664 -9945.11057
or
-3445.51692 7177.16664-9945.11057
(in case of a positive value there is at least one white space that precedes)
and I would like to split this string into three variables that contains each number, e.g.:
a=-3445.51692
b=-7177.16664
c=-9945.11057
Thus, I wanted to use something like
IFS=' -' read -a array <<< "$string"
but I don't know how to specify "zero or more white space". And using "-" as a delimiter removes it from the final result, while I want to keep the sign.
Any ideas ?
Canonicalize the input before you do the IFS splitting, i.e. any minus gets a space prepended:
canonicalized_string=$(echo "$string" | sed 's/-/ -/g')
set -- $canonicalized_string # No need to mess with IFS.
a=$1
b=$2
c=$3
This assumes exactly 3 numbers. In super-compact form:
set -- $(echo "$string" | sed 's/-/ -/g')
a=$1 b=$2 c=$3
Simply use sed to add a space infront of every -, Something like:
echo $string | sed 's/-/ -/g'
You can use read -a by injecting a space first:
s='-3445.51692-7177.16664 -9945.11057'
IFS=' ' read -ra arr <<< "${s//-/ -}"
printf "[%s]\n" "${arr[#]}"
[-3445.51692]
[-7177.16664]
[-9945.11057]

how to chop last n bytes of a string in bash string choping?

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

Resources