在 bash 中的数组运算符中
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In array operator in bash
提问by ???u
Is there a way to test whether an array contains a specified element?
有没有办法测试一个数组是否包含指定的元素？
e.g., something like:
例如，类似于：
array=(one two three)
if [ "one" in ${array} ]; then
...
fi
回答by Frank
A for loop will do the trick.
for 循环可以解决问题。
array=(one two three)
for i in "${array[@]}"; do
if [[ "$i" = "one" ]]; then
...
break
fi
done
回答by pepoluan
Try this:
尝试这个：
array=(one two three)
if [[ "${array[*]}" =~ "one" ]]; then
echo "'one' is found"
fi
回答by user unknown
I got an function 'contains' in my .bashrc-file:
我的 .bashrc 文件中有一个函数“包含”：
contains ()
{
param=;
shift;
for elem in "$@";
do
[[ "$param" = "$elem" ]] && return 0;
done;
return 1
}
It works well with an array:
它适用于数组：
contains on $array && echo hit || echo miss
miss
contains one $array && echo hit || echo miss
hit
contains onex $array && echo hit || echo miss
miss
But doesn't need an array:
但不需要数组：
contains one four two one zero && echo hit || echo miss
hit
回答by William Pursell
I like using grep for this:
我喜欢为此使用 grep：
if echo ${array[@]} | grep -qw one; then
# "one" is in the array
...
fi
(Note that both -qand -ware non-standard options to grep: -wtells it to work on whole words only, and -q("quiet") suppresses all output.)
（请注意，-q和-w都是 grep 的非标准选项：-w告诉它只处理整个单词，并且-q("quiet") 抑制所有输出。）
回答by Szépe Viktor
In_array() {
local NEEDLE=""
local ELEMENT
shift
for ELEMENT; do
if [ "$ELEMENT" == "$NEEDLE" ]; then
return 0
fi
done
return 1
}
declare -a ARRAY=( "elem1" "elem2" "elem3" )
if In_array "elem1" "${ARRAY[@]}"; then
...
A nice and elegant version of the above.
上面的一个漂亮而优雅的版本。
回答by drysdam
array="one two three"
if [ $(echo "$array" | grep one | wc -l) -gt 0 ] ;
then echo yes;
fi
If that's ugly, you could hide it away in a function.
如果这很丑陋，您可以将其隐藏在函数中。
回答by ghostdog74
if you just want to check whether an element is in array, another approach
如果你只是想检查一个元素是否在数组中，另一种方法
case "${array[@]/one/}" in
"${array[@]}" ) echo "not in there";;
*) echo "found ";;
esac
