I have a string in a Bash shell script that I want to split into an array of characters, not based on a delimiter but just one character per array index. How can I do this? Ideally it would not use any external programs. Let me rephrase that. My goal is portability, so things like sed that are likely to be on any POSIX compatible system are fine.
Try
echo "abcdefg" | fold -w1
Edit: Added a more elegant solution suggested in comments.
echo "abcdefg" | grep -o .
You can access each letter individually already without an array conversion:
$ foo="bar"
$ echo ${foo:0:1}
b
$ echo ${foo:1:1}
a
$ echo ${foo:2:1}
r
If that's not enough, you could use something like this:
$ bar=($(echo $foo|sed 's/\(.\)/\1 /g'))
$ echo ${bar[1]}
a
If you can't even use sed or something like that, you can use the first technique above combined with a while loop using the original string's length (${#foo}) to build the array.
Warning: the code below does not work if the string contains whitespace. I think Vaughn Cato's answer has a better chance at surviving with special chars.
thing=($(i=0; while [ $i -lt ${#foo} ] ; do echo ${foo:$i:1} ; i=$((i+1)) ; done))
As an alternative to iterating over 0 .. ${#string}-1 with a for/while loop, there are two other ways I can think of to do this with only bash: using =~ and using printf. (There's a third possibility using eval and a {..} sequence expression, but this lacks clarity.)
With the correct environment and NLS enabled in bash these will work with non-ASCII as hoped, removing potential sources of failure with older system tools such as sed, if that's a concern. These will work from bash-3.0 (released 2005).
Using =~ and regular expressions, converting a string to an array in a single expression:
string="wonkabars"
[[ "$string" =~ ${string//?/(.)} ]] # splits into array
printf "%s\n" "${BASH_REMATCH[#]:1}" # loop free: reuse fmtstr
declare -a arr=( "${BASH_REMATCH[#]:1}" ) # copy array for later
The way this works is to perform an expansion of string which substitutes each single character for (.), then match this generated regular expression with grouping to capture each individual character into BASH_REMATCH[]. Index 0 is set to the entire string, since that special array is read-only you cannot remove it, note the :1 when the array is expanded to skip over index 0, if needed.
Some quick testing for non-trivial strings (>64 chars) shows this method is substantially faster than one using bash string and array operations.
The above will work with strings containing newlines, =~ supports POSIX ERE where . matches anything except NUL by default, i.e. the regex is compiled without REG_NEWLINE. (The behaviour of POSIX text processing utilities is allowed to be different by default in this respect, and usually is.)
Second option, using printf:
string="wonkabars"
ii=0
while printf "%s%n" "${string:ii++:1}" xx; do
((xx)) && printf "\n" || break
done
This loop increments index ii to print one character at a time, and breaks out when there are no characters left. This would be even simpler if the bash printf returned the number of character printed (as in C) rather than an error status, instead the number of characters printed is captured in xx using %n. (This works at least back as far as bash-2.05b.)
With bash-3.1 and printf -v var you have slightly more flexibility, and can avoid falling off the end of the string should you be doing something other than printing the characters, e.g. to create an array:
declare -a arr
ii=0
while printf -v cc "%s%n" "${string:(ii++):1}" xx; do
((xx)) && arr+=("$cc") || break
done
If your string is stored in variable x, this produces an array y with the individual characters:
i=0
while [ $i -lt ${#x} ]; do y[$i]=${x:$i:1}; i=$((i+1));done
The most simple, complete and elegant solution:
$ read -a ARRAY <<< $(echo "abcdefg" | sed 's/./& /g')
and test
$ echo ${ARRAY[0]}
a
$ echo ${ARRAY[1]}
b
Explanation: read -a reads the stdin as an array and assigns it to the variable ARRAY treating spaces as delimiter for each array item.
The evaluation of echoing the string to sed just add needed spaces between each character.
We are using Here String (<<<) to feed the stdin of the read command.
I have found that the following works the best:
array=( `echo string | grep -o . ` )
(note the backticks)
then if you do: echo ${array[#]} ,
you get: s t r i n g
or: echo ${array[2]} ,
you get: r
Pure Bash solution with no loop:
#!/usr/bin/env bash
str='The quick brown fox jumps over a lazy dog.'
# Need extglob for the replacement pattern
shopt -s extglob
# Split string characters into array (skip first record)
# Character 037 is the octal representation of ASCII Record Separator
# so it can capture all other characters in the string, including spaces.
IFS= mapfile -s1 -t -d $'\37' array <<<"${str//?()/$'\37'}"
# Strip out captured trailing newline of here-string in last record
array[-1]="${array[-1]%?}"
# Debug print array
declare -p array
string=hello123
for i in $(seq 0 ${#string})
do array[$i]=${string:$i:1}
done
echo "zero element of array is [${array[0]}]"
echo "entire array is [${array[#]}]"
The zero element of array is [h]. The entire array is [h e l l o 1 2 3 ].
Yet another on :), the stated question simply says 'Split string into character array' and don't say much about the state of the receiving array, and don't say much about special chars like and control chars.
My assumption is that if I want to split a string into an array of chars I want the receiving array containing just that string and no left over from previous runs, yet preserve any special chars.
For instance the proposed solution family like
for (( i=0 ; i < ${#x} ; i++ )); do y[i]=${x:i:1}; done
Have left overs in the target array.
$ y=(1 2 3 4 5 6 7 8)
$ x=abc
$ for (( i=0 ; i < ${#x} ; i++ )); do y[i]=${x:i:1}; done
$ printf '%s ' "${y[#]}"
a b c 4 5 6 7 8
Beside writing the long line each time we want to split a problem, so why not hide all this into a function we can keep is a package source file, with a API like
s2a "Long string" ArrayName
I got this one that seems to do the job.
$ s2a()
> { [ "$2" ] && typeset -n __=$2 && unset $2;
> [ "$1" ] && __+=("${1:0:1}") && s2a "${1:1}"
> }
$ a=(1 2 3 4 5 6 7 8 9 0) ; printf '%s ' "${a[#]}"
1 2 3 4 5 6 7 8 9 0
$ s2a "Split It" a ; printf '%s ' "${a[#]}"
S p l i t I t
If the text can contain spaces:
eval a=( $(echo "this is a test" | sed "s/\(.\)/'\1' /g") )
$ echo hello | awk NF=NF FS=
h e l l o
Or
$ echo hello | awk '$0=RT' RS=[[:alnum:]]
h
e
l
l
o
I know this is a "bash" question, but please let me show you the perfect solution in zsh, a shell very popular these days:
string='this is a string'
string_array=(${(s::)string}) #Parameter expansion. And that's it!
print ${(t)string_array} -> type array
print $#string_array -> 16 items
This is an old post/thread but with a new feature of bash v5.2+ using the shell option patsub_replacement and the =~ operator for regex. More or less same with #mr.spuratic post/answer.
str='There can be only one, the Highlander.'
regexp="${str//?/(&)}"
[[ "$str" =~ $regexp ]] &&
printf '%s\n' "${BASH_REMATCH[#]:1}"
Or by just: (which includes the whole string at index 0)
declare -p BASH_REMATCH
If that is not desired, one can remove the value of the first index (index 0), with
unset -v 'BASH_REMATCH[0]'
instead of using printf or echo to print the value of the array BASH_REMATCH
One can check/see the value of the variable "$regexp" with either
declare -p regexp
Output
declare -- regexp="(T)(h)(e)(r)(e)( )(c)(a)(n)( )(b)(e)( )(o)(n)(l)(y)( )(o)(n)(e)(,)( )(t)(h)(e)( )(H)(i)(g)(h)(l)(a)(n)(d)(e)(r)(.)"
or
echo "$regexp"
Using it in a script, one might want to test if the shopt is enabled or not, although the manual says it is on/enabled by default.
Something like.
if ! shopt -q patsub_replacement; then
shopt -s patsub_replacement
fi
But yeah, check the bash version too! If you're not sure which version of bash is in use.
if ! ((BASH_VERSINFO[0] >= 5 && BASH_VERSINFO[1] >= 2)); then
printf 'No dice! bash version 5.2+ is required!\n' >&2
exit 1
fi
Space can be excluded from regexp variable, change it from
regexp="${str//?/(&)}"
To
regexp="${str//[! ]/(&)}"
and the output is:
declare -- regexp="(T)(h)(e)(r)(e) (c)(a)(n) (b)(e) (o)(n)(l)(y) (o)(n)(e) (t)(h)(e) (H)(i)(g)(h)(l)(a)(n)(d)(e)(r)(.)"
Maybe not as efficient as the other post/answer but it is still a solution/option.
If you want to store this in an array, you can do this:
string=foo
unset chars
declare -a chars
while read -N 1
do
chars[${#chars[#]}]="$REPLY"
done <<<"$string"x
unset chars[$((${#chars[#]} - 1))]
unset chars[$((${#chars[#]} - 1))]
echo "Array: ${chars[#]}"
Array: f o o
echo "Array length: ${#chars[#]}"
Array length: 3
The final x is necessary to handle the fact that a newline is appended after $string if it doesn't contain one.
If you want to use NUL-separated characters, you can try this:
echo -n "$string" | while read -N 1
do
printf %s "$REPLY"
printf '\0'
done
AWK is quite convenient:
a='123'; echo $a | awk 'BEGIN{FS="";OFS=" "} {print $1,$2,$3}'
where FS and OFS is delimiter for read-in and print-out
For those who landed here searching how to do this in fish:
We can use the builtin string command (since v2.3.0) for string manipulation.
↪ string split '' abc
a
b
c
The output is a list, so array operations will work.
↪ for c in (string split '' abc)
echo char is $c
end
char is a
char is b
char is c
Here's a more complex example iterating over the string with an index.
↪ set --local chars (string split '' abc)
for i in (seq (count $chars))
echo $i: $chars[$i]
end
1: a
2: b
3: c
zsh solution: To put the scalar string variable into arr, which will be an array:
arr=(${(ps::)string})
If you also need support for strings with newlines, you can do:
str2arr(){ local string="$1"; mapfile -d $'\0' Chars < <(for i in $(seq 0 $((${#string}-1))); do printf '%s\u0000' "${string:$i:1}"; done); printf '%s' "(${Chars[*]#Q})" ;}
string=$(printf '%b' "apa\nbepa")
declare -a MyString=$(str2arr "$string")
declare -p MyString
# prints declare -a MyString=([0]="a" [1]="p" [2]="a" [3]=$'\n' [4]="b" [5]="e" [6]="p" [7]="a")
As a response to Alexandro de Oliveira, I think the following is more elegant or at least more intuitive:
while read -r -n1 c ; do arr+=("$c") ; done <<<"hejsan"
declare -r some_string='abcdefghijklmnopqrstuvwxyz'
declare -a some_array
declare -i idx
for ((idx = 0; idx < ${#some_string}; ++idx)); do
some_array+=("${some_string:idx:1}")
done
for idx in "${!some_array[#]}"; do
echo "$((idx)): ${some_array[idx]}"
done
Pure bash, no loop.
Another solution, similar to/adapted from Léa Gris' solution, but using read -a instead of readarray/mapfile :
#!/usr/bin/env bash
str='azerty'
# Need extglob for the replacement pattern
shopt -s extglob
# Split string characters into array
# ${str//?()/$'\x1F'} replace each character "c" with "^_c".
# ^_ (Control-_, 0x1f) is Unit Separator (US), you can choose another
# character.
IFS=$'\x1F' read -ra array <<< "${str//?()/$'\x1F'}"
# now, array[0] contains an empty string and the rest of array (starting
# from index 1) contains the original string characters :
declare -p array
# Or, if you prefer to keep the array "clean", you can delete
# the first element and pack the array :
unset array[0]
array=("${array[#]}")
declare -p array
However, I prefer the shorter (and easier to understand for me), where we remove the initial 0x1f before assigning the array :
#!/usr/bin/env bash
str='azerty'
shopt -s extglob
tmp="${str//?()/$'\x1F'}" # same as code above
tmp=${tmp#$'\x1F'} # remove initial 0x1f
IFS=$'\x1F' read -ra array <<< "$tmp" # assign array
declare -p array # verification
I have created a hex to ASCII converter for strings in bash. The application I'm on changes characters (anything but [0-9],[A-Z],[a-z]) , in a string to its corresponding %hexadecimal. Eg: / changes to %2F in a string
I want to retain the ASCII characters as it is. Below is my code:
NAME=%2fhome%40%21%23
C_NAME=""
for (( i=0; i<${#NAME}; i++ )); do
CHK=$(echo "{NAME:$i:1}" | grep -v "\%" &> /dev/null;echo $?)
if [[ ${CHK} -eq 0 ]]; then
C_NAME=`echo "$C_NAME${NAME:$i:1}"`
else
HEX=`echo "${NAME:$i:3}" | sed "s/%//"`
C_NAME=`echo -n "$C_NAME";printf "\x$HEX"`
continue 2
fi
done
echo "$C_NAME"
OUTPUT:
/2fhome#40!21#23
EXPECTED:
/home#!#
So basically the conversion is happening, but not in place. Its retaining the hex values as well, which tells me the continue 2 statement is probably not working as I expect in my code. Any workarounds please.
You only have one loop so I assume you expected that continue 2 skips the current and next iteration of the current loop, however, the documentation help continue clearly states
continue [n]
[...]
If N is specified, resumes the Nth enclosing loop.
There is no built-in to skip the current and also the next iteration of the current loop, but in your case you can use (( i += 2 )) instead of continue 2.
Using the structure of your script with some simplifications and corrections:
#!/bin/bash
name=%2fhome%40%21%23
c_name=""
for (( i=0; i<${#name}; i++ )); do
c=${name:i:1}
if [[ $c != % ]]; then
c_name=$c_name$c
else
hex=${name:i+1:2}
printf -v c_name "%s\x$hex" "$c_name"
(( i += 2 )) # stolen from Dudi Boy's answer
fi
done
echo "$c_name"
Always use lower case or mixed case variables to avoid the chance of name collisions with shell or environment variables
Always use $() instead of backticks
Most of the echo commands you use aren't necessary
You can avoid using sed and grep
Variables should never be included in the format string of printf but it can't be avoided easily here (you could use echo -e "\x$hex" instead though)
You can do math inside parameter expansions
% doesn't need to be escaped in your grep command
You could eliminate the $hex variable if you used its value directly:
printf -v c_name "%s\x${name:i+1:2}" "$c_name"
I really enjoyed your exercise and decided to solve it with awk (my current study).
Hope you like it as well.
cat script.awk
BEGIN {RS = "%[[:xdigit:]]+"} { # redefine record separtor to RegEx (gawk specific)
decNum = strtonum("0x"substr(RT, 2)); # remove prefix # from record separator, convert hex num to dec
outputStr = outputStr""$0""sprintf("%c", decNum); # reconstruct output string
}
END {print outputStr}
The output
echo %2fhome%40%21%23 |awk -f script.awk
/home#!#
#!/bin/bash
# this is a sample value
hash=d7dd933b5bb968b6ba9ee40548b1b27a
# retrieve all letters from this hash
count=0
for (( i=0; i<${#hash}; i++)); do
if [[ ${hash:i:1} == [a-f] ]] ; then
code[$count]=${hash:i:1}
count=$((count + 1))
echo ${code[i]}
#echo ${hash:i:1}
fi
done
Instead of printing all characters in the hash (as I expect), this is printing only the first two characters, followed by newlines. (Eventually, I intend to take only the first two characters extracted from hash, but this is not an immediate goal).
What's wrong here?
In order to filter out all the letters from the hash string, you can modify your for loop as follows:
code=()
count=0
for (( i = 0; i < ${#hash}; i++ )); do
if [[ ${hash:i:1} == [a-zA-Z] ]]; then
code[$count]="${hash:i:1}"
((count++))
fi
done
Here, ${hash:i:1} selects each character in hash individually. If the character is in the range of letters a-zA-Z, meaning it is a letter, then it is stored in the array code, and count is accumulated.
Consider the following full code as an example (assume it is contained in a file called script):
#!/bin/bash
hash="$1" # $1 means first argument given to this script file
printf "hash = %s\n" "$hash"
code=() # initialize code array
count=0
for (( i = 0; i < ${#hash}; i++ )); do
if [[ ${hash:i:1} == [a-zA-Z] ]]; then
code[$count]="${hash:i:1}"
((count++))
fi
done
printf "All letters are: "
printf "%s" "${code[#]}"
printf "\n"
printf "First two letters are: %c%c \n" "${code[0]}" "${code[1]}"
Tests:
$ ./script dd7d933b5bb968b6ba9ee40548b1b27a
hash = dd7d933b5bb968b6ba9ee40548b1b27a
All letters are: dddbbbbbaeebba
First two letters are: dd
$ ./script slfj2948347slddkshfsl2348sldfjsf
hash = slfj2948347slddkshfsl2348sldfjsf
All letters are: slfjslddkshfslsldfjsf
First two letters are: sl
Parameter expansion provides numerous string-manipulation primitives, including search-and-replace, which can be readily used for this purpose:
s=1a2b3c4d5e6f7g
code=${s//[![:alpha:]]/}
echo "$code"
...should emit:
abcdefg
...if you wanted to emit only the first two characters in code, that would be:
echo "${code:0:2}"
Why this works
[:alpha:] is a POSIX character class which includes characters consider alphabet members in the current locale.
[![:alpha:]] is a glob expression which matches any single character not in that class
${var//value/replacement} expands the variable named var, changing all instances of value to replacement
Thus, ${s//[![:alpha:]]/} expands the variable s, changing all characters which are not alphabetic in nature to the empty string.
OK so Ive been at this for a couple days,im new to this whole bash UNIX system thing i just got into it but I am trying to write a script where the user inputs an integer and the script will take that integer and print out a triangle using the integer that was inputted as a base and decreasing until it reaches zero. An example would be:
reverse_triangle.bash 4
****
***
**
*
so this is what I have so far but when I run it nothing happens I have no idea what is wrong
#!/bin/bash
input=$1
count=1
for (( i=$input; i>=$count;i-- ))
do
for (( j=1; j>=i; j++ ))
do
echo -n "*"
done
echo
done
exit 0
when I try to run it nothing happens it just goes to the next line. help would be greatly appreciated :)
As I said in a comment, your test is wrong: you need
for (( j=1; j<=i; j++ ))
instead of
for (( j=1; j>=i; j++ ))
Otherwise, this loop is only executed when i=1, and it becomes an infinite loop.
Now if you want another way to solve that, in a much better way:
#!/bin/bash
[[ $1 = +([[:digit:]]) ]] || { printf >&2 'Argument must be a number\n'; exit 1; }
number=$((10#$1))
for ((;number>=1;--number)); do
printf -v spn '%*s' "$number"
printf '%s\n' "${spn// /*}"
done
Why is it better? first off, we check that the argument is really a number. Without this, your code is subject to arbitrary code injection. Also, we make sure that the number is understood in radix 10 with 10#$1. Otherwise, an argument like 09 would raise an error.
We don't really need an extra variable for the loop, the provided argument is good enough. Now the trick: to print n times a pattern, a cool method is to store n spaces in a variable with printf: %*s will expand to n spaces, where n is the corresponding argument found by printf.
For example:
printf '%s%*s%s\n' hello 42 world
would print:
hello world
(with 42 spaces).
Editor's note: %*s will NOT generally expand to n spaces, as evidenced by above output, which contains 37 spaces.
Instead, the argument that * is mapped to,42, is the field width for the sfield, which maps to the following argument,world, causing string world to be left-space-padded to a length of 42; since world has a character count of 5, 37 spaces are used for padding.
To make the example work as intended, use printf '%s%*s%s\n' hello 42 '' world - note the empty string argument following 42, which ensures that the entire field is made up of padding, i.e., spaces (you'd get the same effect if no arguments followed 42).
With printf's -v option, we can store any string formatted by printf into a variable; here we're storing $number spaces in spn. Finally, we replace all spaces by the character *, using the expansion ${spn// /*}.
Yet another possibility:
#!/bin/bash
[[ $1 = +([[:digit:]]) ]] || { printf >&2 'Argument must be a number\n'; exit 1; }
printf -v s '%*s' $((10#1))
s=${s// /*}
while [[ $s ]]; do
printf '%s\n' "$s"
s=${s%?}
done
This time we construct the variable s that contains a bunch of * (number given by user), using the previous technique. Then we have a while loop that loops while s is non empty. At each iteration we print the content of s and we remove a character with the expansion ${s%?} that removes the last character of s.
Building on gniourf_gniourf's helpful answer:
The following is simpler and performs significantly better:
#!/bin/bash
count=$1 # (... number-validation code omitted for brevity)
# Create the 1st line, composed of $count '*' chars, and store in var. $line.
printf -v line '%.s*' $(seq $count)
# Count from $count down to 1.
while (( count-- )); do
# Print a *substring* of the 1st line based on the current value of $count.
printf "%.${count}s\n" "$line"
done
printf -v line '*%.s' $(seq $count) is a trick that prints * $count times, thanks to %.s* resulting in * for each argument supplied, irrespective of the arguments' values (thanks to %.s, which effectively ignores its argument). $(seq $count) expands to $count arguments, resulting in a string composed of $count * chars. overall, which - thanks to -v line, is stored in variable $line.
printf "%.${count}s\n" "$line" prints a substring from the beginning of $line that is $count chars. long.
I want to split the string and construct the array. I tried the below code:
myString="first column:second column:third column"
set -A myArray `echo $myString | awk 'BEGIN{FS=":"}{for (i=1; i<=NF; i++) print $i}'`
# Following is just to make sure that array is constructed properly
i=0
while [ $i -lt ${#myArray[#]} ]
do
echo "Element $i:${myArray[$i]}"
(( i=i+1 ))
done
exit 0
It produces the following result:
Element 0:first
Element 1:column
Element 2:second
Element 3:column
Element 4:third
Element 5:column
This is not what I want it to be. When I construct the array, I want that array to contain only three elements.
Element 0:first column
Element 1:second column
Element 2:third column
Can you please advise?
Here is how I would approach this problem: use the IFS variable to tell the shell (bash) that you want to split the string into colon-separated tokens.
$ cat split.sh
#!/bin/sh
# Script to split fields into tokens
# Here is the string where tokens separated by colons
s="first column:second column:third column"
IFS=":" # Set the field separator
set $s # Breaks the string into $1, $2, ...
i=0
for item # A for loop by default loop through $1, $2, ...
do
echo "Element $i: $item"
((i++))
done
Run it:
$ ./split.sh
Element 0: first column
Element 1: second column
Element 2: third column
if you definitely want to use arrays in Bash, you can try this way
$ myString="first column:second column:third column"
$ myString="${myString//:/ }" #remove all the colons
$ echo "${myString}"
first column second column third column
$ read -a myArr <<<$myString
$ echo ${myArr[#]}
first column second column third column
$ echo ${myArr[1]}
column
$ echo ${myArr[2]}
second
otherwise, the "better" method is to use awk entirely
Note that saving and restoring IFS as I often seen in these solutions has the side effect that if IFS wasn't set, it ends up changed to being an empty string which causes weird problems with subsequent splitting.
Here's the solution I came up with based on Anton Olsen's extended to handle >2 values separated by a colon. It handles values in the list that have spaces correctly, not splitting on the space.
colon_list=${1} # colon-separate list to split
while true ; do
part=${colon_list%%:*} # Delete longest substring match from back
colon_list=${colon_list#*:} # Delete shortest substring match from front
parts[i++]=$part
# We are done when there is no more colon
if test "$colon_list" = "$part" ; then
break
fi
done
# Show we've split the list
for part in "${parts[#]}"; do
echo $part
done
Ksh or Bash
#! /bin/sh
myString="first column:second column:third column"
IFS=: A=( $myString )
echo ${A[0]}
echo ${A[1]}
Looks like you've already found the solution, but note that you can do away with awk entirely:
myString="first column:second column:third column"
OIFS="$IFS"
IFS=':'
myArray=($myString)
IFS=$OIFS
i=0
while [ $i -lt ${#myArray[#]} ]
do
echo "Element $i:${myArray[$i]}"
(( i=i+1 ))
done