i have some questions about pattern matching operators.
what's difference between these examples
$ VAR=/usr/bin/iecset
$ echo ${VAR#*/}
usr/bin/iecset
and
$ VAR=/usr/bin/iecset
$ echo ${VAR##*/}
iecset
and
$ VAR=/usr/bin/iecset
$ echo ${VAR%*/}
/bin/iecset
and
$ VAR=/usr/bin/iecset
$ echo ${VAR%%*/}
The "pattern" here is a glob or extended glob pattern - most people call them wildcards. The characters have a different meaning to those used in Regular Expressions. So * means "zero or more of any character".
${var#pattern} # delete shortest match of pattern from left
${var##pattern} # delete longest match of pattern from left
${var%pattern} # delete shortest match of pattern from right
${var%%pattern} # delete longest match of pattern from right
Your examples:
(Bad idea to use uppercase variable names)
Hints:
to delete from the left, the * goes on the left.
# delete from left, as # is on the left in "We are #1"
% delete from the right, as the % is on the right in 50%
var=/usr/bin/iecset
Remove the shortest string on the left ending in /
echo ${var#*/}
usr/bin/iecset
There are no characters to the left of the first /, remember that * means zero or more. So the leftmost / is removed.
Remove the longest string on the left ending in /
echo ${var##*/}
iecset
The next two are wrong in your post! To delete from the right, the * should be on the right of the /.
$ echo ${VAR%*/} # WRONG
$ echo ${VAR%%*/} # WRONG
I think you mean:
Delete the shortest string on the right starting with /
var=/usr/bin/iecset
echo ${var%/*}
/usr/bin
Delete the longest string on the right starting with /
echo ${var%%/*}
(blank line)
There are many other meta-characters other than * you can use.
Related
I got string as followed:
text=\"abcdef\"gfijk\"lmno\"
How can I extract the text between last two " - so I will get only lmno ?
I tried to use & but without success
sub_text=${text&\"*}
echo ${sub_text&\"*}
This is easily done with parameter expansion.
First, delete that final quote with ${var%pattern}, which removes the shortest match for pattern from the end of $var:
result=${text%'"'} # result=\"abcdef\"gfijk\"lmno
Then, delete everything from the beginning up to the last remaining quote with ${var##pattern}, which removes the longest match for pattern from the beginning of $var:
result=${result##*'"'} # result=lmno
...and then you're there:
echo "$result"
I have this variable:
A="Some variable has value abc.123"
I need to extract this value i.e abc.123. Is this possible in bash?
Simplest is
echo "$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 $1 if /(\S+)$/'
Here is a script I use:
for dir in $(find . -type d -name "single_copy_busco_sequences"); do
sppname=$(dirname $(dirname $(dirname $dir))| sed 's#./##g');
for file in ${dir}/*.faa; do name=$(basename $file); cp $file /Users/admin/Documents/busco_aa/${sppname}_${name}; sed -i '' 's#>#>'${sppname}'|#g' /Users/admin/Documents/busco_aa/${sppname}_${name}; cut -f 1 -d ":" /Users/admin/Documents/busco_aa/${sppname}_${name} > /Users/admin/Documents/busco_aa/${sppname}_${name}.1;
done;
done
The sppname variable is something like Gender_species
do you know how could I add a line in my script to creat a new variable called abbrev which transformes Gender_species into Genspe, the 3 first letters cat with the 3 first letters after _
exemples:
Homo_sapiens gives Homsap
Canis_lupus gives Canlup
etc
Thank for your help :)
You can achieve this using a regular expression with sed:
echo "Homo_sapiens" | sed -e s'/^\(...\).*_\(...\).*/\1\2/'
Homsap
start, get 3 chars (to keep in \1), anything, _, anything, get 3 chars (to keep in \2), anything
Replace echo "Homo_sapiens" by your $dir thing
PS: will fail if you have less than 3 chars in one word
You can do it all with bash built-in parameter expansions. Specifically, string indexes and substring removal.
$ a=Homo_sapiens; prefix=${a:0:3}; a=${a#*_}; postfix=${a:0:3}; echo $prefix$postfix
Homsap
$ a=Canis_lupus; prefix=${a:0:3}; a=${a#*_}; postfix=${a:0:3}; echo $prefix$postfix
Canlup
Using bash built-ins is always more efficient than spawning separate subshell(s) to invoke utilities to accomplish the same thing.
Explanation
Your string index form (bash only) allows you to index characters from within a string, e.g.
* ${parameter:offset:length} ## indexes are zero based, ${a:0:2} is 1st 2 chars
Where parameter is simply the variable name holding the string.
(you can index from the end of a string by using a negative offset preceded by a space or enclosed in parenthesis, e.g. a=12345; echo ${a: -3:2} outputs "34")
prefix=${a:0:3} ## save the first 3 characters in prefix
a=${a#*_} ## remove the front of the string through '_' (see below)
postfix=${a:0:3} ## save the first 3 characters after '_'
Your substring removal forms (POSIX) are:
${parameter#word} trim to 1st occurrence of word from parameter from left
${parameter##word} trim to last occurrence of word from parameter from left
and
${parameter%word} trim to 1st occurrence of word from parameter from right
${parameter%%word} trim to last occurrence of word from parameter from right
(word can contain globbing to expand to a pattern as well)
a=${a#*_} ## trim from left up to (and including) the first '_'
See bash(1) - Linux manual page for full details.
I have files with names like "words_transfer1_morewords.txt". I would like to ensure that the number after "transfer" is five digits, as in "words_transfer00001_morewords.txt". How would I do this with a ksh script? Thanks.
This will work in any Bourne-type/POSIX shell as long as your words and morewords don't contain numbers:
file=words_transfer1_morewords.txt
prefix=${file%%[0-9]*} # words_transfer
suffix=${file##*[0-9]} # _morewords.txt
num=${file#$prefix} # 1_morewords.txt
num=${num%$suffix} # 1
file=$(printf "%s%05d%s" "$prefix" "$num" "$suffix")
echo "$file"
Use ksh's regular expression matching operation to break the filename down into separate parts, them put them back together again after formatting the number.
pre="[^[:digit:]]+" # What to match before the number
num="[[:digit:]]+" # The number to match
post=".*" # What to match after the number
[[ $file =~ ($pre)($num)($post) ]]
new_file=$(printf "%s%05d%s\n" "${.sh.match[#]:1:3}")
Following a successful match with =~, the special array parameter .sh.match contains the full match in element 0, and each capture group in order starting with element 1.
I'm reading in values from an .ini file, and sometimes may get trailing or leading whitespace.
How do I amend this first line to prevent that?
db=$(sed -n 's/.*DB_USERNAME *= *\([^ ]*.*\)/\1/p' < config.ini);
echo -"$db"-
Result;
-myinivar -
I need;
-myinivar-
Use parameter expansion.
echo "=${db% }="
You don't need the .* inside the capturing group (or the semicolon at the end of line):
db="$(sed -n 's/.*DB_USERNAME *= *\([^ ]*\).*/\1/p' < config.ini)"
To elaborate:
.* matches anything at all
DB_USERNAME matches that literal string
* (a single space followed by an asterisk) matches any number of spaces
= matches that literal string
* (a single space followed by an asterisk) matches any number of spaces
\( starts the capturing group that is used for \1 later
[^ ] matches anything which is not a space character
* repeats that zero or more times
\) ends the capturing group
.* matches anything at all
Therefore, the result will be all the characters after DB_USERNAME = and any number of spaces, up to the next space or end of line, whichever comes first.
You can use echo to trim whitespace:
db='myinivar '
echo -"$(echo $db)"-
-myinivar-
Use crudini which handles these ini file edge cases transparently
db=$(crudini --get config.ini '' DB_USERNAME)
To get rid of more than one trailing space, use %% which removes the longest matching pattern from the end of the string
echo "=${db%% *}="