I want to get two latest words from a string variable.
Total number of words in the string variable is not constant.
This is what I try:
LIST=`some command`
LATEST1=""
LATEST2=""
for ITEM in $LIST
do
LATEST2="$LATEST1"
LATEST1="$ITEM"
done
echo "Latest: $LATEST1"
echo "2nd latest: $LATEST2"
But it is slow. Is there any better way to do it?
sh shell of busybox is used. Other scripting languages are not available.
You can just use parameter substitution:
words="one two three four"
last=${words##* }
echo $last # => four
tmp=${words% *}
last2=${tmp##* }
echo $last2 # => three
Regex which will match the last 2 words (including whitespaces in this case)
(\s\w+){2}$
$ echo -e 'hello world\nhow are you' |
tr '\n' ' ' |
awk '
END{if(NF>1)printf("Latest:\t\t%s\n2nd latest:\t%s\n", $NF, $(NF-1)); else print "ERROR"}'
Latest: you
2nd latest: are
You can do this in a purely shell way using something like this:
words="one two three four"
words=($words)
echo ${words[${#words}]} # prints 'four'
echo ${words[((${#words} - 1))]} # prints 'three'
This works by assigning the variable containing the strings into an array and then accessing it by key.
Related
I am trying to split the string by double back slashes in bash and somehow it's not working.
My string is as follow:
abc\\xyz
My Code:
my_str='abc\\xyz'
IFS='\\' read -r -a arr <<< "$my_str"
echo "${#arr[#]}"
I am expecting that the output would be '2' as the length of the array would be 2.
However, I get 3 as a result and when I try to print the array values, I only get 'abc', and the second index remains empty. It seems like something is wrong with my code but I am unable to identify it.
Can anyone please help to resolve the issue?
If there are no spaces in your string you could use bash pattern substitutions to replace pairs of backslashes by a space and assign the result to an indexed array:
$ my_str='abc\\xyz\uvw\\rst\\\012\\\\345'
$ declare -a arr=( ${my_str//\\\\/ } )
$ echo "${#arr[#]}"
5
$ printf '%s\n' "${arr[#]}"
abc
xyz\uvw
rst
\012
345
Perhaps you could try to replace the backslashes on the string fist as showcased in this previous question. However, this would be inefficient for very large strings.
A slightly different take -
$: my_str='abc\\123\\ghi\\\012\\jkl\\foo bar\\\\xyz'
$: IFS='|' read -r -a arr <<< "${my_str//\\\\/\|}"
$: printf "[%s]\n" "${arr[#]}"
[abc]
[123]
[ghi]
[\012]
[jkl]
[foo bar]
[]
[xyz]
I have a variable MyVar with values stored in it. For example:
MyVar="123, 234, 345, 456"
Each entry in the variable is separated by a coma as in the example above.
I want to be able to pick the first and last entry from this variable, i.e 123 and 456 respectively.
Any idea how I can achieve this from the command prompt terminal ?
Thanks!
Using bash substring removal:
$ echo ${MyVar##*,}
456
$ echo ${MyVar%%,*}
123
Also:
$ echo ${MyVar/,*,/,}
123, 456
More for example here:
https://tldp.org/LDP/abs/html/parameter-substitution.html
Edit: Above kind of expects the substrings to be separated by commas only. See comments where #costaparas gloriously demonstrates a case with , .
Try using sed:
MyVar="123, 234, 345, 456"
first=$(echo "$MyVar" | sed 's/,.*//')
last=$(echo "$MyVar" | sed 's/.*, //')
echo $first $last
Explanation:
To obtain the first string, we replace everything after & including
the first comma with nothing (empty string).
To obtain the last string, we replace everything before & including the last comma with nothing (empty string).
Using bash array:
IFS=', ' arr=($MyVar)
echo ${arr[0]} ${arr[-1]}
Where ${arr[0]} and ${arr[-1]} are your first and last respective values. Negative index requires bash 4.2 or later.
You could try following also with latest BASH version, by sending variable values into an array and then retrieve first and last element, keeping all either values in it saved in case you need them later in program etc.
IFS=', ' read -r -a array <<< "$MyVar"
echo "${array[0]}"
123
echo "${array[-1]}"
456
Awk alternative:
awk -F "(, )" '{ print $1" - "$NF }' <<< $MyVar
Set the field separator to command and a space. Print the first field and the last field (NF) with " - " in between.
I have this variable:
A="Some variable has value abc.123"
I need to extract this value i.e abc.123. Is this possible in bash?
Simplest is
echo "$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 $1 if /(\S+)$/'
I run a script with the param -A AA/BB . To get an array with AA and BB, i can do this.
INPUT_PARAM=(${AIRLINE_OPTION//-A / }) #get rid of the '-A ' in the begining
LIST=(${AIRLINES_PARAM//\// }) # split by '/'
Can we achieve this in a single line?
Thanks in advance.
One way
IFS=/ read -r -a LIST <<< "${AIRLINE_OPTION//-A /}"
This places the output from the parameter substitution ${AIRLINE_OPTION//-A /} into a "here-string" and uses the bash read built-in to parse this into an array. Splitting by / is achieved by setting the value of IFS to / for the read command.
LIST=( $(IFS=/; for x in ${AIRLINE_OPTION#-A }; do printf "$x "; done) )
This is a portable solution, but if your read supports -a and you don't mind portability then you should go for #1_CR's solution.
With awk, for example, you can create an array and store it in LIST variable:
$ LIST=($(awk -F"[\/ ]" '{print $2,$3}' <<< "-A AA/BB"))
Result:
$ echo ${LIST[0]}
AA
$ echo ${LIST[1]}
BB
Explanation
-F"[\/ ]" defines two possible field separators: a space or a slash /.
'{print $2$3}' prints the 2nd and 3rd fields based on those separators.
Greetings!
I have a text file with parameter set as follows:
NameOfParameter Value1 Value2 Value3 ...
...
I want to find needed parameter by its NameOfParameter using regexp pattern and return a selected Value to my Bash script.
I tried to do this with grep, but it returns a whole line instead of Value.
Could you help me to find as approach please?
It was not clear if you want all the values together or only one specific one. In either case, use the power of cut command to cut the columns you want from a file (-f 2- will cut columns 2 and on (so everything except parameter name; -d " " will ensure that the columns are considered to be space-separated as opposed to default tab-separated)
egrep '^NameOfParameter ' your_file | cut -f 2- -d " "
Bash:
values=($(grep '^NameofParameter '))
echo ${values[0]} # NameofParameter
echo ${values[1]} # Value1
echo ${values[2]} # Value2
# etc.
for value in ${values[#:1]} # iterate over values, skipping NameofParameter
do
echo "$value"
done