bash 循环数组，打印索引和值
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Looping over arrays, printing both index and value
提问by Tyilo
I want to do something like this:
我想做这样的事情：
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:
然后我尝试使用 for in 循环遍历它：
foo=( )
foo[0]="bar"
foo[35]="baz"
for i in ${foo[@]}
do
echo "?: $i"
done
# Output:
# ?: bar
# ?: naz
but here I don't know the index value.
但在这里我不知道索引值。
I know you could something like
我知道你可以像
foo=( )
foo[0]="bar"
foo[35]="baz"
declare -p foo
# Output:
# declare -a foo='([0]="bar" [35]="baz")'
but, can't you do it in another way?
但是，你不能换一种方式吗？
回答by glenn Hymanman
You would find the array keys with "${!foo[@]}"(reference), so:
你会用"${!foo[@]}"( reference)找到数组键，所以：
for i in "${!foo[@]}"; do
printf "%s\t%s\n" "$i" "${foo[$i]}"
done
Which means that indices will be in $iwhile the elements themselves have to be accessed via ${foo[$i]}
这意味着索引将在，$i而元素本身必须通过访问${foo[$i]}
回答by Eyal Ch
you can always use iteration param:
你总是可以使用迭代参数：
ITER=0
for I in ${FOO[@]}
do
echo ${I} ${ITER}
ITER=$(expr $ITER + 1)
done
回答by Aruy Aruy
INDEX=0
for i in $list; do
echo ${INDEX}_$i
let INDEX=${INDEX}+1
done
回答by mattmc3
In bash 4, you can use associative arrays:
在 bash 4 中，您可以使用关联数组：
declare -A foo
foo[0]="bar"
foo[35]="baz"
for key in "${!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):
在 bash 3 中，这有效（在 zsh 中也有效）：
map=( )
map+=("0:bar")
map+=("35:baz")
for keyvalue in "${map[@]}" ; do
key=${keyvalue%%:*}
value=${keyvalue#*:}
echo "key: $key, value $value."
done
回答by F. Hauri
Simple one linetrick for dumping array
倾倒数组的简单一行技巧
I've added one value with spaces:
我用空格添加了一个值：
foo=()
foo[12]="bar"
foo[42]="foo bar baz"
foo[35]="baz"
I, for quickly dump basharrays or associative arraysI use
我，为了快速转储我使用的bash数组或关联数组
This one line command:
这一行命令：
paste <(printf "%s\n" "${!foo[@]}") <(printf "%s\n" "${foo[@]}")
paste <(printf "%s\n" "${!foo[@]}") <(printf "%s\n" "${foo[@]}")
Will render:
将呈现：
12 bar
35 baz
42 foo bar baz
Explained
解释
printf "%s\n" "${!foo[@]}"will print all keysseparated by a newline,printf "%s\n" "${foo[@]}"will print all valuesseparated by a newline,paste <(cmd1) <(cmd2)will merge output ofcmd1andcmd2line by line.
printf "%s\n" "${!foo[@]}"将打印由换行符分隔的所有键，printf "%s\n" "${foo[@]}"将打印由换行符分隔的所有值，paste <(cmd1) <(cmd2)将逐行合并cmd1和 的输出cmd2。
Tunning
调谐
This could be tunned by -dswitch:
这可以通过-d开关调整：
paste -d : <(printf "%s\n" "${!foo[@]}") <(printf "%s\n" "${foo[@]}")
12:bar
35:baz
42:foo bar baz
or even:
甚至：
paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "'%s'\n" "${foo[@]}")
foo[12]='bar'
foo[35]='baz'
foo[42]='foo bar baz'
Associative array will work same:
关联数组的工作方式相同：
declare -A bar=([foo]=snoopy [bar]=nice [baz]=cool [foo bar]='Hello world!')
paste -d = <(printf "bar[%s]\n" "${!bar[@]}") <(printf '"%s"\n' "${bar[@]}")
bar[foo bar]="Hello world!"
bar[foo]="snoopy"
bar[bar]="nice"
bar[baz]="cool"
Issue with newlinesor special chars
问题与换行符或特殊字符
Unfortunely, there is at least one condition making this not work anymore: when variable do contain newline:
不幸的是，至少有一个条件使它不再起作用：当变量确实包含换行符时：
foo[17]=$'There is one\nnewline'
Command pastewill merge line-by-line, so output will become wrong:
命令paste会逐行合并，所以输出会出错：
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 %qinstead of %sin second printfcommand (and whipe quoting):
对于这项工作，您可以使用%q而不是%s在第二个printf命令中（和whipe 引用）：
paste -d = <(printf "foo[%s]\n" "${!foo[@]}") <(printf "%q\n" "${foo[@]}")
Will render perfect:
将完美呈现：
foo[12]=bar
foo[17]=$'There is one\nnewline'
foo[35]=baz
foo[42]=foo\ bar\ baz
From man bash:
来自man bash：
%q causes printf to output the corresponding argument in a format that can be reused as shell input.
%q causes printf to output the corresponding argument in a format that can be reused as shell input.
回答by code4mk
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
