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This question already has answers here:
Passing An Array From One Bash Script to Another
(3 answers)
How to pass an array argument to the Bash script
(8 answers)
Closed 2 years ago.
I use create.sh to create random passwords, where each pw is appened to an array which i echo out and of which i give the expanded version (ie. [#]) to the select script.
In the select script i cant access the array. I echoed its length, which said to be 1. even though 2 elements where in it when i printed seperated by a space.
Now i cant access the index of 1 because only index 0 exists which is all elements concatenated by a whitespace.
Thanks in advance.
Robert
create.sh
SEQ_LEN=1
MAX_CHAR=25
PASSWORDS=()
COUNTER=0
while getopts "x:l:" OPTION; do
case $OPTION in
x )
SEQ_LEN="$OPTARG"
;;
l )
MAX_CHAR="$OPTARG"
;;
\? )
echo "correct usage: create [-x] [SEQ_LEN] [-l] [MAX_CHAR]"
exit 1
;;
esac
done
shift "$(($OPTIND -1))"
for VAR in $(seq 1 $SEQ_LEN)
do
PASS=$(openssl rand -base64 48 | cut -c1-$MAX_CHAR)
PASSWORDS[$COUNTER]="$PASS"
COUNTER=$(($COUNTER +1))
done
echo "${PASSWORDS[#]}"
select.sh
#!/bin/bash
ARRAY=()
INDEX=0
while getopts "a:i:" OPTION; do
case $OPTION in
a )
ARRAY=$OPTARG
;;
i )
INDEX=$OPTARG
;;
\? )
echo "Correct Usage: select [-a] [PASSWORD_ARRAY] [-i] [SELECTED_INDEX]"
;;
esac
done
echo "$ARRAY $INDEX" # looks identical
echo "${#ARRAY[#]}" # is 1 - should be 2
Thanks for the ans wers and helpful advice.
Since bash is so great, it turns out you can just do
ARRAY=( $OPTARG )
and it splits the $OPTARG string by " ", turning it into an "array"
"Obviously" you can just for loop through the string and append to a new variable, that being declared as -a array. source
as seen below:
sentence="This is a sentence."
for word in $sentence
do
echo $word
done
I need to remove an element from an array in bash shell.
Generally I'd simply do:
array=("${(#)array:#<element to remove>}")
Unfortunately the element I want to remove is a variable so I can't use the previous command.
Down here an example:
array+=(pluto)
array+=(pippo)
delete=(pluto)
array( ${array[#]/$delete} ) -> but clearly doesn't work because of {}
Any idea?
The following works as you would like in bash and zsh:
$ array=(pluto pippo)
$ delete=pluto
$ echo ${array[#]/$delete}
pippo
$ array=( "${array[#]/$delete}" ) #Quotes when working with strings
If need to delete more than one element:
...
$ delete=(pluto pippo)
for del in ${delete[#]}
do
array=("${array[#]/$del}") #Quotes when working with strings
done
Caveat
This technique actually removes prefixes matching $delete from the elements, not necessarily whole elements.
Update
To really remove an exact item, you need to walk through the array, comparing the target to each element, and using unset to delete an exact match.
array=(pluto pippo bob)
delete=(pippo)
for target in "${delete[#]}"; do
for i in "${!array[#]}"; do
if [[ ${array[i]} = $target ]]; then
unset 'array[i]'
fi
done
done
Note that if you do this, and one or more elements is removed, the indices will no longer be a continuous sequence of integers.
$ declare -p array
declare -a array=([0]="pluto" [2]="bob")
The simple fact is, arrays were not designed for use as mutable data structures. They are primarily used for storing lists of items in a single variable without needing to waste a character as a delimiter (e.g., to store a list of strings which can contain whitespace).
If gaps are a problem, then you need to rebuild the array to fill the gaps:
for i in "${!array[#]}"; do
new_array+=( "${array[i]}" )
done
array=("${new_array[#]}")
unset new_array
You could build up a new array without the undesired element, then assign it back to the old array. This works in bash:
array=(pluto pippo)
new_array=()
for value in "${array[#]}"
do
[[ $value != pluto ]] && new_array+=($value)
done
array=("${new_array[#]}")
unset new_array
This yields:
echo "${array[#]}"
pippo
This is the most direct way to unset a value if you know it's position.
$ array=(one two three)
$ echo ${#array[#]}
3
$ unset 'array[1]'
$ echo ${array[#]}
one three
$ echo ${#array[#]}
2
This answer is specific to the case of deleting multiple values from large arrays, where performance is important.
The most voted solutions are (1) pattern substitution on an array, or (2) iterating over the array elements. The first is fast, but can only deal with elements that have distinct prefix, the second has O(n*k), n=array size, k=elements to remove. Associative array are relative new feature, and might not have been common when the question was originally posted.
For the exact match case, with large n and k, possible to improve performance from O(nk) to O(n+klog(k)). In practice, O(n) assuming k much lower than n. Most of the speed up is based on using associative array to identify items to be removed.
Performance (n-array size, k-values to delete). Performance measure seconds of user time
N K New(seconds) Current(seconds) Speedup
1000 10 0.005 0.033 6X
10000 10 0.070 0.348 5X
10000 20 0.070 0.656 9X
10000 1 0.043 0.050 -7%
As expected, the current solution is linear to N*K, and the fast solution is practically linear to K, with much lower constant. The fast solution is slightly slower vs the current solution when k=1, due to additional setup.
The 'Fast' solution: array=list of input, delete=list of values to remove.
declare -A delk
for del in "${delete[#]}" ; do delk[$del]=1 ; done
# Tag items to remove, based on
for k in "${!array[#]}" ; do
[ "${delk[${array[$k]}]-}" ] && unset 'array[k]'
done
# Compaction
array=("${array[#]}")
Benchmarked against current solution, from the most-voted answer.
for target in "${delete[#]}"; do
for i in "${!array[#]}"; do
if [[ ${array[i]} = $target ]]; then
unset 'array[i]'
fi
done
done
array=("${array[#]}")
Here's a one-line solution with mapfile:
$ mapfile -d $'\0' -t arr < <(printf '%s\0' "${arr[#]}" | grep -Pzv "<regexp>")
Example:
$ arr=("Adam" "Bob" "Claire"$'\n'"Smith" "David" "Eve" "Fred")
$ echo "Size: ${#arr[*]} Contents: ${arr[*]}"
Size: 6 Contents: Adam Bob Claire
Smith David Eve Fred
$ mapfile -d $'\0' -t arr < <(printf '%s\0' "${arr[#]}" | grep -Pzv "^Claire\nSmith$")
$ echo "Size: ${#arr[*]} Contents: ${arr[*]}"
Size: 5 Contents: Adam Bob David Eve Fred
This method allows for great flexibility by modifying/exchanging the grep command and doesn't leave any empty strings in the array.
Partial answer only
To delete the first item in the array
unset 'array[0]'
To delete the last item in the array
unset 'array[-1]'
To expand on the above answers, the following can be used to remove multiple elements from an array, without partial matching:
ARRAY=(one two onetwo three four threefour "one six")
TO_REMOVE=(one four)
TEMP_ARRAY=()
for pkg in "${ARRAY[#]}"; do
for remove in "${TO_REMOVE[#]}"; do
KEEP=true
if [[ ${pkg} == ${remove} ]]; then
KEEP=false
break
fi
done
if ${KEEP}; then
TEMP_ARRAY+=(${pkg})
fi
done
ARRAY=("${TEMP_ARRAY[#]}")
unset TEMP_ARRAY
This will result in an array containing:
(two onetwo three threefour "one six")
Here's a (probably very bash-specific) little function involving bash variable indirection and unset; it's a general solution that does not involve text substitution or discarding empty elements and has no problems with quoting/whitespace etc.
delete_ary_elmt() {
local word=$1 # the element to search for & delete
local aryref="$2[#]" # a necessary step since '${!$2[#]}' is a syntax error
local arycopy=("${!aryref}") # create a copy of the input array
local status=1
for (( i = ${#arycopy[#]} - 1; i >= 0; i-- )); do # iterate over indices backwards
elmt=${arycopy[$i]}
[[ $elmt == $word ]] && unset "$2[$i]" && status=0 # unset matching elmts in orig. ary
done
return $status # return 0 if something was deleted; 1 if not
}
array=(a 0 0 b 0 0 0 c 0 d e 0 0 0)
delete_ary_elmt 0 array
for e in "${array[#]}"; do
echo "$e"
done
# prints "a" "b" "c" "d" in lines
Use it like delete_ary_elmt ELEMENT ARRAYNAME without any $ sigil. Switch the == $word for == $word* for prefix matches; use ${elmt,,} == ${word,,} for case-insensitive matches; etc., whatever bash [[ supports.
It works by determining the indices of the input array and iterating over them backwards (so deleting elements doesn't screw up iteration order). To get the indices you need to access the input array by name, which can be done via bash variable indirection x=1; varname=x; echo ${!varname} # prints "1".
You can't access arrays by name like aryname=a; echo "${$aryname[#]}, this gives you an error. You can't do aryname=a; echo "${!aryname[#]}", this gives you the indices of the variable aryname (although it is not an array). What DOES work is aryref="a[#]"; echo "${!aryref}", which will print the elements of the array a, preserving shell-word quoting and whitespace exactly like echo "${a[#]}". But this only works for printing the elements of an array, not for printing its length or indices (aryref="!a[#]" or aryref="#a[#]" or "${!!aryref}" or "${#!aryref}", they all fail).
So I copy the original array by its name via bash indirection and get the indices from the copy. To iterate over the indices in reverse I use a C-style for loop. I could also do it by accessing the indices via ${!arycopy[#]} and reversing them with tac, which is a cat that turns around the input line order.
A function solution without variable indirection would probably have to involve eval, which may or may not be safe to use in that situation (I can't tell).
Using unset
To remove an element at particular index, we can use unset and then do copy to another array. Only just unset is not required in this case. Because unset does not remove the element it just sets null string to the particular index in array.
declare -a arr=('aa' 'bb' 'cc' 'dd' 'ee')
unset 'arr[1]'
declare -a arr2=()
i=0
for element in "${arr[#]}"
do
arr2[$i]=$element
((++i))
done
echo "${arr[#]}"
echo "1st val is ${arr[1]}, 2nd val is ${arr[2]}"
echo "${arr2[#]}"
echo "1st val is ${arr2[1]}, 2nd val is ${arr2[2]}"
Output is
aa cc dd ee
1st val is , 2nd val is cc
aa cc dd ee
1st val is cc, 2nd val is dd
Using :<idx>
We can remove some set of elements using :<idx> also. For example if we want to remove 1st element we can use :1 as mentioned below.
declare -a arr=('aa' 'bb' 'cc' 'dd' 'ee')
arr2=("${arr[#]:1}")
echo "${arr2[#]}"
echo "1st val is ${arr2[1]}, 2nd val is ${arr2[2]}"
Output is
bb cc dd ee
1st val is cc, 2nd val is dd
http://wiki.bash-hackers.org/syntax/pe#substring_removal
${PARAMETER#PATTERN} # remove from beginning
${PARAMETER##PATTERN} # remove from the beginning, greedy match
${PARAMETER%PATTERN} # remove from the end
${PARAMETER%%PATTERN} # remove from the end, greedy match
In order to do a full remove element, you have to do an unset command with an if statement. If you don't care about removing prefixes from other variables or about supporting whitespace in the array, then you can just drop the quotes and forget about for loops.
See example below for a few different ways to clean up an array.
options=("foo" "bar" "foo" "foobar" "foo bar" "bars" "bar")
# remove bar from the start of each element
options=("${options[#]/#"bar"}")
# options=("foo" "" "foo" "foobar" "foo bar" "s" "")
# remove the complete string "foo" in a for loop
count=${#options[#]}
for ((i = 0; i < count; i++)); do
if [ "${options[i]}" = "foo" ] ; then
unset 'options[i]'
fi
done
# options=( "" "foobar" "foo bar" "s" "")
# remove empty options
# note the count variable can't be recalculated easily on a sparse array
for ((i = 0; i < count; i++)); do
# echo "Element $i: '${options[i]}'"
if [ -z "${options[i]}" ] ; then
unset 'options[i]'
fi
done
# options=("foobar" "foo bar" "s")
# list them with select
echo "Choose an option:"
PS3='Option? '
select i in "${options[#]}" Quit
do
case $i in
Quit) break ;;
*) echo "You selected \"$i\"" ;;
esac
done
Output
Choose an option:
1) foobar
2) foo bar
3) s
4) Quit
Option?
Hope that helps.
There is also this syntax, e.g. if you want to delete the 2nd element :
array=("${array[#]:0:1}" "${array[#]:2}")
which is in fact the concatenation of 2 tabs. The first from the index 0 to the index 1 (exclusive) and the 2nd from the index 2 to the end.
POSIX shell script does not have arrays.
So most probably you are using a specific dialect such as bash, korn shells or zsh.
Therefore, your question as of now cannot be answered.
Maybe this works for you:
unset array[$delete]
What I do is:
array="$(echo $array | tr ' ' '\n' | sed "/itemtodelete/d")"
BAM, that item is removed.
This is a quick-and-dirty solution that will work in simple cases but will break if (a) there are regex special characters in $delete, or (b) there are any spaces at all in any items. Starting with:
array+=(pluto)
array+=(pippo)
delete=(pluto)
Delete all entries exactly matching $delete:
array=(`echo $array | fmt -1 | grep -v "^${delete}$" | fmt -999999`)
resulting in
echo $array -> pippo, and making sure it's an array:
echo $array[1] -> pippo
fmt is a little obscure: fmt -1 wraps at the first column (to put each item on its own line. That's where the problem arises with items in spaces.) fmt -999999 unwraps it back to one line, putting back the spaces between items. There are other ways to do that, such as xargs.
Addendum: If you want to delete just the first match, use sed, as described here:
array=(`echo $array | fmt -1 | sed "0,/^${delete}$/{//d;}" | fmt -999999`)
Actually, I just noticed that the shell syntax somewhat has a behavior built-in that allows for easy reconstruction of the array when, as posed in the question, an item should be removed.
# let's set up an array of items to consume:
x=()
for (( i=0; i<10; i++ )); do
x+=("$i")
done
# here, we consume that array:
while (( ${#x[#]} )); do
i=$(( $RANDOM % ${#x[#]} ))
echo "${x[i]} / ${x[#]}"
x=("${x[#]:0:i}" "${x[#]:i+1}")
done
Notice how we constructed the array using bash's x+=() syntax?
You could actually add more than one item with that, the content of a whole other array at once.
In ZSH this is dead easy (note this uses more bash compatible syntax than necessary where possible for ease of understanding):
# I always include an edge case to make sure each element
# is not being word split.
start=(one two three 'four 4' five)
work=(${(#)start})
idx=2
val=${work[idx]}
# How to remove a single element easily.
# Also works for associative arrays (at least in zsh)
work[$idx]=()
echo "Array size went down by one: "
[[ $#work -eq $(($#start - 1)) ]] && echo "OK"
echo "Array item "$val" is now gone: "
[[ -z ${work[(r)$val]} ]] && echo OK
echo "Array contents are as expected: "
wanted=("${start[#]:0:1}" "${start[#]:2}")
[[ "${(j.:.)wanted[#]}" == "${(j.:.)work[#]}" ]] && echo "OK"
echo "-- array contents: start --"
print -l -r -- "-- $#start elements" ${(#)start}
echo "-- array contents: work --"
print -l -r -- "-- $#work elements" "${work[#]}"
Results:
Array size went down by one:
OK
Array item two is now gone:
OK
Array contents are as expected:
OK
-- array contents: start --
-- 5 elements
one
two
three
four 4
five
-- array contents: work --
-- 4 elements
one
three
four 4
five
To avoid conflicts with array index using unset - see https://stackoverflow.com/a/49626928/3223785 and https://stackoverflow.com/a/47798640/3223785 for more information - reassign the array to itself: ARRAY_VAR=(${ARRAY_VAR[#]}).
#!/bin/bash
ARRAY_VAR=(0 1 2 3 4 5 6 7 8 9)
unset ARRAY_VAR[5]
unset ARRAY_VAR[4]
ARRAY_VAR=(${ARRAY_VAR[#]})
echo ${ARRAY_VAR[#]}
A_LENGTH=${#ARRAY_VAR[*]}
for (( i=0; i<=$(( $A_LENGTH -1 )); i++ )) ; do
echo ""
echo "INDEX - $i"
echo "VALUE - ${ARRAY_VAR[$i]}"
done
exit 0
[Ref.: https://tecadmin.net/working-with-array-bash-script/ ]
How about something like:
array=(one two three)
array_t=" ${array[#]} "
delete=one
array=(${array_t// $delete / })
unset array_t
#/bin/bash
echo "# define array with six elements"
arr=(zero one two three 'four 4' five)
echo "# unset by index: 0"
unset -v 'arr[0]'
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
arr_delete_by_content() { # value to delete
for i in ${!arr[*]}; do
[ "${arr[$i]}" = "$1" ] && unset -v 'arr[$i]'
done
}
echo "# unset in global variable where value: three"
arr_delete_by_content three
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
echo "# rearrange indices"
arr=( "${arr[#]}" )
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
delete_value() { # value arrayelements..., returns array decl.
local e val=$1; new=(); shift
for e in "${#}"; do [ "$val" != "$e" ] && new+=("$e"); done
declare -p new|sed 's,^[^=]*=,,'
}
echo "# new array without value: two"
declare -a arr="$(delete_value two "${arr[#]}")"
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
delete_values() { # arraydecl values..., returns array decl. (keeps indices)
declare -a arr="$1"; local i v; shift
for v in "${#}"; do
for i in ${!arr[*]}; do
[ "$v" = "${arr[$i]}" ] && unset -v 'arr[$i]'
done
done
declare -p arr|sed 's,^[^=]*=,,'
}
echo "# new array without values: one five (keep indices)"
declare -a arr="$(delete_values "$(declare -p arr|sed 's,^[^=]*=,,')" one five)"
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
# new array without multiple values and rearranged indices is left to the reader
This question already has answers here:
Remove an element from a Bash array
(19 answers)
Closed 6 years ago.
I have an array list in a bash script, and a variable var. I know that $var appears in ${list[#]}, but have no easy way of determining its index. I'd like to remove it from list.
This answer achieves something very close to what I need, except that list retains an empty element where $var once was. Note, e.g.:
$ list=(one two three)
$ var="two"
$ list=( "${list[#]/$var}" )
$ echo ${list[#]}
one three
$ echo ${#list[#]}
3
The same thing happens if I use delete=( "$var" ) and replace $var for $delete in the third line. Also, doing list=( "${list[#]/$var/}" ) makes no difference either.
(I'll note that, experimenting with the comment to that answer, I managed to match only whole words using list=( "${list[#]/%$var}" ), omitting the #.)
I also saw this answer proposing a nice trick to keep track of index and use unset, but that is unfeasible in my case. Finally, the same issue also appeared here, except that OP was satisfied with the result and probably didn't run into the problem empty elements create for me later on in my script, when I iterate through list. I tried to negate that problem by using expansion as follows, without any apparent effect:
for item in "${list[#]}"; do
if [ -n ${item:+'x'} ];then
...
fi
done
It's the same when I do [ ${#item} > 0 ], and I'm running out of ideas. Suggestions?
EDIT:
I have no understanding of why this happens, but #l0b0's comment made me notice something. Using the above preamble, I get:
$ for item in "${list[#]}"; do echo "Here!"; done
Here!
Here!
Here!
but:
$ for item in ${list[#]}; do echo "Here!"; done
Here!
Here!
I'm not sure I can omit the quotes in my script, though, as items are considerably more complicated there (file names and paths, both containing spaces and odd characters).
You can delete an element from existing array though the whole process isn't very straightforward and may appear like a hack.
#!/bin/bash
list=( "one" "two" "three" "four" "five" )
var1="two"
var2="four"
printf "%s\n" "Before:"
for (( i=0; i<${#list[#]}; i++ )); do
printf "%s = %s\n" "$i" "${list[i]}";
done
for (( i=0; i<${#list[#]}; i++ )); do
if [[ ${list[i]} == $var1 || ${list[i]} == $var2 ]]; then
list=( "${list[#]:0:$i}" "${list[#]:$((i + 1))}" )
i=$((i - 1))
fi
done
printf "\n%s\n" "After:"
for (( i=0; i<${#list[#]}; i++ )); do
printf "%s = %s\n" "$i" "${list[i]}";
done
This script outputs:
Before:
0 = one
1 = two
2 = three
3 = four
4 = five
After:
0 = one
1 = three
2 = five
Key part of the script is:
list=( "${list[#]:0:$i}" "${list[#]:$((i + 1))}" )
Here we re-construct your existing array by specifying the index and length to remove the element from array completely and re-order the indices.
If you want to delete the array element & shift the indices, you can use answer by l0b0 or JS웃.
However, if you don't want to shift the indices, you can use below script-let:
(Particularly useful for associative arrays)
$ list=(one two three)
$ delete_me=two
$ for i in ${!list[#]};do
if [ "${list[$i]}" == "$delete_me" ]; then
unset list[$i]
fi
done
$ for i in ${!list[#]};do echo "$i = ${list[$i]}"; done
0 = one
2 = three
If you want to shift the indices to make them continuous, re-construct the array as this:
$ list=("${list[#]}")
$ for i in ${!list[#]};do echo "$i = ${list[$i]}"; done
0 = one
1 = three
If you want to remove by value and shift the indexes I think you have to create a new array:
list=(one two three)
new_list=() # Not strictly necessary, but added for clarity
var="two"
for item in ${list[#]}
do
if [ "$item" != "$var" ]
then
new_list+=("$item")
fi
done
list=("${new_list[#]}")
unset new_list
Test:
$ echo "${list[#]}"
one three
$ echo "${#list[#]}"
2
The code
SourceFolder[0]=""
SourceFolder[1]="inbound2"
SourceFolder[2]="inbound3"
for i in "${!SourceFolder[#]}"
do
if [ -z "${SourceFolder[$i]}"]; then
${SourceFolder[$i]} = "TEST"
fi
done
${SourceFolder[$i]} = "TEST" - doesn't work
it says
=: command not found
How to change value in current loop index in an array?
Because of the first space = is not interpreted as an assignment. There is a full explanation on So.
Btw ${SourceFolder[$i]} evaluate the array element, which is not what you want to do. For instance for the first one it is the empty string.
Replaces with SourceFolder[$i]=
You must change indexnumber in the your array:
ARRAYNAME[indexnumber]=value
ok, you have array is:
array=(one two three)
you can add count to your script for initialize and change element for array indexnumber, example:
#!/bin/bash
count=0
array=(one two three)
for i in ${array[#]}
do
echo "$i"
array[$count]="$i-indexnumber-is-$count"
count=$((count + 1))
echo $count
done
echo ${array[*]}
Result:
bash test-arr.sh
one
1
two
2
three
3
one-indexnumber-is-0 two-indexnumber-is-1 three-indexnumber-is-2
How do I create an array in unix shell scripting?
The following code creates and prints an array of strings in shell:
#!/bin/bash
array=("A" "B" "ElementC" "ElementE")
for element in "${array[#]}"
do
echo "$element"
done
echo
echo "Number of elements: ${#array[#]}"
echo
echo "${array[#]}"
Result:
A
B
ElementC
ElementE
Number of elements: 4
A B ElementC ElementE
in bash, you create array like this
arr=(one two three)
to call the elements
$ echo "${arr[0]}"
one
$ echo "${arr[2]}"
three
to ask for user input, you can use read
read -p "Enter your choice: " choice
Bourne shell doesn't support arrays. However, there are two ways to handle the issue.
Use positional shell parameters $1, $2, etc.:
$ set one two three
$ echo $*
one two three
$ echo $#
3
$ echo $2
two
Use variable evaluations:
$ n=1 ; eval a$n="one"
$ n=2 ; eval a$n="two"
$ n=3 ; eval a$n="three"
$ n=2
$ eval echo \$a$n
two
#!/bin/bash
# define a array, space to separate every item
foo=(foo1 foo2)
# access
echo "${foo[1]}"
# add or changes
foo[0]=bar
foo[2]=cat
foo[1000]=also_OK
You can read the ABS "Advanced Bash-Scripting Guide"
The Bourne shell and C shell don't have arrays, IIRC.
In addition to what others have said, in Bash you can get the number of elements in an array as follows:
elements=${#arrayname[#]}
and do slice-style operations:
arrayname=(apple banana cherry)
echo ${arrayname[#]:1} # yields "banana cherry"
echo ${arrayname[#]: -1} # yields "cherry"
echo ${arrayname[${#arrayname[#]}-1]} # yields "cherry"
echo ${arrayname[#]:0:2} # yields "apple banana"
echo ${arrayname[#]:1:1} # yields "banana"
Try this :
echo "Find the Largest Number and Smallest Number of a given number"
echo "---------------------------------------------------------------------------------"
echo "Enter the number"
read n
i=0
while [ $n -gt 0 ] #For Seperating digits and Stored into array "x"
do
x[$i]=`expr $n % 10`
n=`expr $n / 10`
i=`expr $i + 1`
done
echo "Array values ${x[#]}" # For displaying array elements
len=${#x[*]} # it returns the array length
for (( i=0; i<len; i++ )) # For Sorting array elements using Bubble sort
do
for (( j=i+1; j<len; j++ ))
do
if [ `echo "${x[$i]} > ${x[$j]}"|bc` ]
then
t=${x[$i]}
t=${x[$i]}
x[$i]=${x[$j]}
x[$j]=$t
fi
done
done
echo "Array values ${x[*]}" # Displaying of Sorted Array
for (( i=len-1; i>=0; i-- )) # Form largest number
do
a=`echo $a \* 10 + ${x[$i]}|bc`
done
echo "Largest Number is : $a"
l=$a #Largest number
s=0
while [ $a -gt 0 ] # Reversing of number, We get Smallest number
do
r=`expr $a % 10`
s=`echo "$s * 10 + $r"|bc`
a=`expr $a / 10`
done
echo "Smallest Number is : $s" #Smallest Number
echo "Difference between Largest number and Smallest number"
echo "=========================================="
Diff=`expr $l - $s`
echo "Result is : $Diff"
echo "If you try it, We can get it"
Your question asks about "unix shell scripting", but is tagged bash. Those are two different answers.
The POSIX specification for shells does not have anything to say about arrays, as the original Bourne shell did not support them. Even today, on FreeBSD, Ubuntu Linux, and many other systems, /bin/sh does not have array support. So if you want your script to work in different Bourne-compatible shells, you shouldn't use them. Alternatively, if you are assuming a specific shell, then be sure to put its full name in the shebang line, e.g. #!/usr/bin/env bash.
If you are using bash or zsh, or a modern version of ksh, you can create an array like this:
myArray=(first "second element" 3rd)
and access elements like this
$ echo "${myArray[1]}" # for bash/ksh; for zsh, echo $myArray[2]
second element
You can get all the elements via "${myArray[#]}". You can use the slice notation ${array[#]:start:length} to restrict the portion of the array referenced, e.g. "${myArray[#]:1}" to leave off the first element.
The length of the array is ${#myArray[#]}. You can get a new array containing all the indexes from an existing array with "${!myArray[#]}".
Older versions of ksh before ksh93 also had arrays, but not the parenthesis-based notation, nor did they support slicing. You could create an array like this, though:
set -A myArray -- first "second element" 3rd
You can try of the following type :
#!/bin/bash
declare -a arr
i=0
j=0
for dir in $(find /home/rmajeti/programs -type d)
do
arr[i]=$dir
i=$((i+1))
done
while [ $j -lt $i ]
do
echo ${arr[$j]}
j=$((j+1))
done
An array can be loaded in twoways.
set -A TEST_ARRAY alpha beta gamma
or
X=0 # Initialize counter to zero.
-- Load the array with the strings alpha, beta, and gamma
for ELEMENT in alpha gamma beta
do
TEST_ARRAY[$X]=$ELEMENT
((X = X + 1))
done
Also, I think below information may help:
The shell supports one-dimensional arrays. The maximum number of array
elements is 1,024. When an array is defined, it is automatically
dimensioned to 1,024 elements. A one-dimensional array contains a
sequence of array elements, which are like the boxcars connected
together on a train track.
In case you want to access the array:
echo ${MY_ARRAY[2] # Show the third array element
gamma
echo ${MY_ARRAY[*] # Show all array elements
- alpha beta gamma
echo ${MY_ARRAY[#] # Show all array elements
- alpha beta gamma
echo ${#MY_ARRAY[*]} # Show the total number of array elements
- 3
echo ${#MY_ARRAY[#]} # Show the total number of array elements
- 3
echo ${MY_ARRAY} # Show array element 0 (the first element)
- alpha
If you want a key value store with support for spaces use the -A parameter:
declare -A programCollection
programCollection["xwininfo"]="to aquire information about the target window."
for program in ${!programCollection[#]}
do
echo "The program ${program} is used ${programCollection[${program}]}"
done
http://linux.die.net/man/1/bash "Associative arrays are created using declare -A name. "
There are multiple ways to create an array in shell.
ARR[0]="ABC"
ARR[1]="BCD"
echo ${ARR[*]}
${ARR[*]} prints all elements in the array.
Second way is:
ARR=("A" "B" "C" "D" 5 7 "J")
echo ${#ARR[#]}
echo ${ARR[0]}
${#ARR[#]} is used to count length of the array.
To read the values from keybord and insert element into array
# enter 0 when exit the insert element
echo "Enter the numbers"
read n
while [ $n -ne 0 ]
do
x[$i]=`expr $n`
read n
let i++
done
#display the all array elements
echo "Array values ${x[#]}"
echo "Array values ${x[*]}"
# To find the array length
length=${#x[*]}
echo $length
A Simple way :
arr=("sharlock" "bomkesh" "feluda" ) ##declare array
len=${#arr[*]} #determine length of array
# iterate with for loop
for (( i=0; i<len; i++ ))
do
echo ${arr[$i]}
done
In ksh you do it:
set -A array element1 element2 elementn
# view the first element
echo ${array[0]}
# Amount elements (You have to substitute 1)
echo ${#array[*]}
# show last element
echo ${array[ $(( ${#array[*]} - 1 )) ]}