If the length of a string is 5 then how can I print upto 4th character of the string using shell scripting.I have stored the string in a variable and length in other variable.but how can i print upto length -1.
If you are using BASH then it is fairly straight forward to remove last character:
s="string1,string2,"
echo "${s%?}"
? matches any single character and %? removes any character from right hand side.
That will output:
string1,string2
Otherwise you can use this sed to remove last character:
echo "$s" | sed 's/.$//'
string1,string2
You can do it with bash "parameter substitution":
string=12345
new=${string:0:$((${#string}-1))}
echo $new
1234
where I am saying:
new=${string:a:b}
where:
a=0 (meaning starting from the first character)
and:
b=${#string} i.e. the length of the string minus 1, performed in an arithmetic context, i.e. inside `$((...))`
str="something"
echo $str | cut -c1-$((${#str}-1))
will give result as
somethin
If you have two different variables, then you can try this also.
str="something"
strlen=9
echo $str | cut -c1-$((strlen-1))
cut -c1-8 will print from first character to eighth.
Just for fun:
When you have the string and length in vars already,
s="example"
slen=${#s}
you can use
printf "%.$((slen-1))s\n" "$s"
As #anubhava showed, you can also have a clean solution.
So do not try
rev <<< "${s}" | cut -c2- | rev
Related
I have the following file:
APP_VERSION.ts
export const APP_VERSION = 1;
This is the only content of that file, and the APP_VERSION variable will be incremented as needed.
So, the APP_VERSION could be a single digit number or multiple digit number, like 15 or 999, etc.
I need to use that value in one of my bash scripts.
use-app-version.sh
APP_VERSION=`cat src/constants/APP_VERSION.ts`
echo $APP_VERSION
I know I can read it with cat. But how can I parse that string so I can get exactly the APP_VERSION value, whether it's 1 or 999, for example.
sed -En 's/(^.*APP_VERSION.*)([[:digit:]]+.*)(\;.*$)/\2/p' src/constants/APP_VERSION
Using sed, split the line into three sections defined by opening and closing brackets. Substitute the line for second section on ( the version value) and print.
You may use this awk:
app_ver=$(awk -F '[[:blank:];=]+' '$(NF-2) == "APP_VERSION" {print $(NF-1)}' src/constants/APP_VERSION.ts)
echo "$app_ver"
1
You can concat some commands to remove everything else:
APP_VERSION=`cat src/constants/APP_VERSION.ts | awk -F '=' '{print $2}' | tr -d ' ' | tr -d ';'`
1 - Cat get all file content
2 - AWK gets all content after '='
3 - Remove space
4 - Remove ;
A simple
APP_VERSION=$(grep --text -Eo '[0-9]+' src/constants/APP_VERSION.ts)
should be enough
With bash only:
APP_VERSION=$(cat src/constants/APP_VERSION.ts)
APP_VERSION=${APP_VERSION%;}
APP_VERSION=${APP_VERSION/*= }
Line 2 removes the trailing ';', line 3 removes everything before "= ".
Alternatively, you could set APP_VERSION as an array, take 5th element, and remove trailing ';'.
Or, another solution, using IFS:
IFS='=;' read a APP_VERSION < src/constants/APP_VERSION.ts
In this version, the space will remain before version number.
Assuming that the task can be rephrased to "extract the digits from a file", there are a few options:
Delete all characters that aren't digits with tr:
version=$(tr -cd '[:digit:]' < infile)
Use grep to match all digits and retain nothing but the match:
version=$(grep -Eo '[[:digit:]]+' infile)
Read file into string and delete all non-digits with just Bash:
contents=$(< infile)
version=${contents//[![:digit:]]}
I have this variable:
A="Some variable has value abc.123"
I need to extract this value i.e abc.123. Is this possible in bash?
Simplest is
echo "$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 have to format the width of a substring within a string using a bash script, but without using tokens or loops. A single character between two colons should be prepended by a 0 in order to match the standard width of 2 for each field.
For e.g
from:
6:0:36:35:30:30:72:6c:73:0:c:52:4c:30:31:30:31:30:30:30:31:36:39:0:1:3
to
06:00:36:35:30:30:72:6c:73:00:0c:52:4c:30:31:30:31:30:30:30:31:36:39:00:01:03
How can I do this?
sed -r 's/\<([0-9a-f])\>/0\1/g'
Search and replace with a regex. Use \< and \> to match word boundaries so [0-9a-f] only matches single digits.
$ sed -r 's/\<([0-9a-f])\>/0\1/g' <<< "6:0:36:35:30:30:72:6c:73:0:c:52:4c:30:31:30:31:30:30:30:31:36:39:0:1:3"
06:00:36:35:30:30:72:6c:73:00:0c:52:4c:30:31:30:31:30:30:30:31:36:39:00:01:03
awk -F: -v OFS=: '{for(i=1;i<=NF;i++) if(length($i)==1)gsub($i,"0&",$i)}1' file
Output:
06:00:36:35:30:30:72:6c:73:00:0c:52:4c:30:31:30:31:30:30:30:31:36:39:00:01:03
This will divide the whole line into fields separated by : , if the length of any of the field is == 1. then it will replace that field with 0field.
Bash solution:
IFS=:; for i in $string; do echo -n 0$i: | tail -c 3; done
With
str="06:00:36:35:30:30:72:6c:73:00:0c:52:4c:30:31:30:31:30:30:30:31:36:39:00:01:03"
you can add a '0' to all tokens and remove those that are unwanted:
sed -r 's/0([0-9a-f]{2})/\1/g' <<< "0${str//:/:0}"
That doesn't feel right, making errors and repairing them.
A better alternative is
echo $(IFS=:; printf "%2s:" ${str} | tr " " "0")
How can I reverse a four length of letters with sed?
For example:
the year was 1815.
Reverse to:
the raey was 5181.
This is my attempt:
cat filename | sed's/\([a-z]*\) *\([a-z]*\)/\2, \1/'
But it does not work as I intended.
not sure it is possible to do it with GNU sed for all cases. If _ doesn't occur immediately before/after four letter words, you can use
sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
\b is word boundary, word definition being any alphabet or digit or underscore character. So \b will ensure to match only whole words not part of words
$ echo 'the year was 1815.' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
the raey was 5181.
$ echo 'two time five three six good' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
two emit evif three six doog
$ # but won't work if there are underscores around the words
$ echo '_good food' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
_good doof
tool with lookaround support would work for all cases
$ echo '_good food' | perl -pe 's/(?<![a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])(?!=[a-z0-9])/$4$3$2$1/gi'
_doog doof
(?<![a-z0-9]) and (?!=[a-z0-9]) are negative lookbehind and negative lookahead respectively
Can be shortened to
perl -pe 's/(?<![a-z0-9])[a-z0-9]{4}(?!=[a-z0-9])/reverse $&/gie'
which uses the e modifier to place Perl code in substitution section. This form is suitable to easily change length of words to be reversed
Possible shortest sed solution even if a four length of letters contains _s.
sed -r 's/\<(.)(.)(.)(.)\>/\4\3\2\1/g'
Following awk may help you in same. Tested this in GNU awk and only with provided sample Input_file
echo "the year was 1815." |
awk '
function reverse(val){
num=split(val, array,"");
i=array[num]=="."?num-1:num;
for(;i>q;i--){
var=var?var array[i]:array[i]
};
printf (array[num]=="."?var".":var);
var=""
}
{
for(j=1;j<=NF;j++){
printf("%s%s",j==NF||j==2?reverse($j):$j,j==NF?RS:FS)
}}'
This might work for you (GNU sed):
sed -r '/\<\w{4}\>/!b;s//\n&\n/g;s/^[^\n]/\n&/;:a;/\n\n/!s/(.*\n)([^\n])(.*\n)/\2\1\3/;ta;s/^([^\n]*)(.*)\n\n/\2\1/;ta;s/\n//' file
If there are no strings of the length required to reverse, bail out.
Prepend and append newlines to all required strings.
Insert a newline at the start of the pattern space (PS). The PS is divided into two parts, the first line will contain the current word being reversed. The remainder will contain the original line.
Each character of the word to be reversed is inserted at the front of the first line and removed from the original line. When all the characters in the word have been processed, the original word will have gone and only the bordering newlines will exist. These double newlines are then replaced by the word in the first line and the process is repeated until all words have been processed. Finally the newline introduced to separate the working line and the original is removed and the PS is printed.
N.B. This method may be used to reverse strings of varying string length i.e. by changing the first regexp strings of any number can be reversed. Also strings between two lengths may also be reversed e.g. /\<w{2,4}\>/ will change all words between 2 and 4 character length.
It's a recurrent problem so somebody created a bash command called "rev".
echo "$(echo the | rev) $(echo year | rev) $(echo was | rev) $(echo 1815 | rev)".
OR
echo "the year was 1815." | rev | tr ' ' '\n' | tac | tr '\n' ' '
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