我想这样做:
foo=( )
foo[0]="bar"
foo[35]="baz"
for((i=0;i<${#foo[@]};i++))
do
echo "$i: ${foo[$i]}"
done
# Output:
# 0: bar
# 1:
然后我尝试通过它循环使用for in:
foo=( )
foo[0]="bar"
foo[35]="baz"
for i in ${foo[@]}
do
echo "?: $i"
done
# Output:
# ?: bar
# ?: naz
但这里我不知道下标值。
我知道你可以
foo=( )
foo[0]="bar"
foo[35]="baz"
declare -p foo
# Output:
# declare -a foo='([0]="bar" [35]="baz")'
但是，你不能用另一种方法吗?
当前回答
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
其他回答
INDEX=0
for i in $list; do
echo ${INDEX}_$i
let INDEX=${INDEX}+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
你总是可以使用迭代参数:
ITER=0
for I in ${FOO[@]}
do
echo ${I} ${ITER}
ITER=$(expr $ITER + 1)
done
在bash 4中，你可以使用关联数组:
declare -A foo
foo[0]="bar"
foo[35]="baz"
# for Zsh, change this to: for key in "${(k)foo[@]}"
for key in "${!foo[@]}"
do
echo "key: $key, value: ${foo[$key]}"
done
# output
# $ key: 0, value bar.
# $ key: 35, value baz.
在bash 3中，这是有效的(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 $value."
done
您会发现数组键的“${!”foo[@]}"(引用)，所以:
for i in "${!foo[@]}"; do
printf "%s\t%s\n" "$i" "${foo[$i]}"
done
这意味着索引将在$i中，而元素本身必须通过${foo[$i]}访问。
