I want to do something like this:
foo=( )
foo[0]="bar"
foo[35]="baz"
for((i=0;i<${#foo[@]};i++))
do
echo "$i: ${foo[$i]}"
done
# Output:
# 0: bar
# 1:
Then i tried to loop through it using for in:
foo=( )
foo[0]="bar"
foo[35]="baz"
for i in ${foo[@]}
do
echo "?: $i"
done
# Output:
# ?: bar
# ?: naz
but here I don't know the index value.
I know you could something like
foo=( )
foo[0]="bar"
foo[35]="baz"
declare -p foo
# Output:
# declare -a foo='([0]="bar" [35]="baz")'
but, can't you do it in another way?
This question is related to
bash
In bash 4, you can use associative arrays:
declare -A foo
foo[0]="bar"
foo[35]="baz"
for key in "${!foo[@]}"
do
echo "key: $key, value: ${foo[$key]}"
done
# output
# $ key: 0, value bar.
# $ key: 35, value baz.
In bash 3, this works (also works in zsh):
map=( )
map+=("0:bar")
map+=("35:baz")
for keyvalue in "${map[@]}" ; do
key=${keyvalue%%:*}
value=${keyvalue#*:}
echo "key: $key, value $value."
done
INDEX=0
for i in $list; do
echo ${INDEX}_$i
let INDEX=${INDEX}+1
done
I've added one value with spaces:
foo=()
foo[12]="bar"
foo[42]="foo bar baz"
foo[35]="baz"
I, for quickly dump bash arrays or associative arrays I use
This one line command:
paste <(printf "%s\n" "${!foo[@]}") <(printf "%s\n" "${foo[@]}")
Will render:
12 bar
35 baz
42 foo bar baz
printf "%s\n" "${!foo[@]}"
will print all keys separated by a newline,printf "%s\n" "${foo[@]}"
will print all values separated by a newline,paste <(cmd1) <(cmd2)
will merge output of cmd1
and cmd2
line by line.This could be tunned by -d
switch:
paste -d : <(printf "%s\n" "${!foo[@]}") <(printf "%s\n" "${foo[@]}")
12:bar
35:baz
42:foo bar baz
or even:
paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "'%s'\n" "${foo[@]}")
foo[12]='bar'
foo[35]='baz'
foo[42]='foo bar baz'
declare -A bar=([foo]=snoopy [bar]=nice [baz]=cool [foo bar]='Hello world!')
paste -d = <(printf "bar[%s]\n" "${!bar[@]}") <(printf '"%s"\n' "${bar[@]}")
bar[foo bar]="Hello world!"
bar[foo]="snoopy"
bar[bar]="nice"
bar[baz]="cool"
Unfortunely, there is at least one condition making this not work anymore: when variable do contain newline:
foo[17]=$'There is one\nnewline'
Command paste
will merge line-by-line, so output will become wrong:
paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "'%s'\n" "${foo[@]}")
foo[12]='bar'
foo[17]='There is one
foo[35]=newline'
foo[42]='baz'
='foo bar baz'
For this work, you could use %q
instead of %s
in second printf
command (and whipe quoting):
paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "%q\n" "${foo[@]}")
Will render perfect:
foo[12]=bar
foo[17]=$'There is one\nnewline'
foo[35]=baz
foo[42]=foo\ bar\ baz
From man bash
:
%q causes printf to output the corresponding argument in a format that can be reused as shell input.
you can always use iteration param:
ITER=0
for I in ${FOO[@]}
do
echo ${I} ${ITER}
ITER=$(expr $ITER + 1)
done
users=("kamal" "jamal" "rahim" "karim" "sadia")
index=()
t=-1
for i in ${users[@]}; do
t=$(( t + 1 ))
if [ $t -eq 0 ]; then
for j in ${!users[@]}; do
index[$j]=$j
done
fi
echo "${index[$t]} is $i"
done
Source: Stackoverflow.com