bash Bash通过args选择数组元素的位置
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Bash Select position of array element by args
提问by C?t?lin George Fe?til?
I want make a bash script wich return position of element array by give an arg. See code bellow , i use
我想制作一个 bash 脚本，通过给定一个参数来返回元素数组的位置。请参阅下面的代码，我使用
#!/bin/bash
args=("$@")
echo ${args[0]}
test_array=('AA' 'BB' 'CC' 'DD' 'EE')
echo $test_array
elem_array=${#test_array[@]}
for args in $test_array
do
echo
done
Finnally i should have output like:
最后我应该有如下输出：
$script.sh DD
4
采纳答案by jfg956
Trying to avoid complex tools:
尽量避免使用复杂的工具：
test_array=('AA' 'BB' 'CC' 'D D' 'EE')
OLD_IFS="$IFS"
IFS="
"
element=$(grep -n '^D D$' <<< "${test_array[*]}" | cut -d ":" -f 1)
IFS="$OLD_IFS"
echo $element
However, it consumes 2 processes. If we allow ourselves sed, we could do it with a single process:
但是，它消耗了2个进程。如果我们允许自己sed，我们可以用一个过程来完成：
test_array=('AA' 'BB' 'CC' 'D D' 'EE')
OLD_IFS="$IFS"
IFS="
"
element=$(sed -n -e '/^D D$/=' <<< "${test_array[*]}")
IFS="$OLD_IFS"
echo $element
Update:
更新：
As pointed out by thkala in the comments, this solution is broken in 3 cases. Be careful not to use it if:
正如 thkala 在评论中指出的那样，该解决方案在 3 种情况下被破坏。如果出现以下情况，请注意不要使用它：
- You want zero indexed offset.
- You have newlines in your array elements.
- And you have a sparse array, or have other keys than integers.
- 你想要零索引偏移。
- 您的数组元素中有换行符。
- 并且您有一个稀疏数组，或者有除整数以外的其他键。
回答by thkala
#!/bin/bash
A=(AA BB CC DD EE)
for i in "${!A[@]}"; do
if [[ "${A[i]}" = "" ]]; then
echo "$i"
fi
done
Note the "${!A[@]}"notation that gives the list of valid indexes in the array. In general you cannot just go from 0 to "${#A[@]}" - 1, because the indexes are not necessarily contiguous. There can be gaps in the index range if there were gaps in the array element assignments or if some elements have been unset.
请注意"${!A[@]}"给出数组中有效索引列表的符号。一般来说，你不能只是从 0 到"${#A[@]}" - 1，因为索引不一定是连续的。如果数组元素分配中存在间隙或某些元素未设置，则索引范围中可能存在间隙。
The script above will output all indexes of the array for which its content is equal to the first command line argument of the script.
上面的脚本将输出数组的所有索引，其内容等于脚本的第一个命令行参数。
EDIT:
编辑：
In your question, you seem to want the result as a one-based array index. In that case you can just increment the result by one:
在您的问题中，您似乎希望结果作为基于 1 的数组索引。在这种情况下，您可以将结果加一：
#!/bin/bash
A=(AA BB CC DD EE)
for i in "${!A[@]}"; do
if [[ "${A[i]}" = "" ]]; then
let i++;
echo "$i"
fi
done
Keep in mind, though, that this index will have to be decrementedbefore being used with a zero-based array.
但请记住，该索引在与基于零的数组一起使用之前必须递减。
回答by dogbane
Loop over the array and keep track of the position. When you find the element matching the input argument, print out the position of the element. You need to add one to the position, because arrays have zero-based indexing.
循环遍历数组并跟踪位置。当找到与输入参数匹配的元素时，打印出该元素的位置。您需要在位置上加 1，因为数组具有从零开始的索引。
#! /bin/sh
arg=
echo $arg
test_array=('AA' 'BB' 'CC' 'DD' 'EE')
element_count=${#test_array[@]}
index=0
while [ $index -lt $element_count ]
do
if [ "${test_array[index]}" = "$arg" ]
then
echo $((index+1))
break
fi
((index++))
done
回答by AlexanderKL
Without loop:
无循环：
#!/bin/bash
index() {
local IFS=$'\n';
echo "${*:2}" | awk '5
== "'"${1//\"/\\"}"'" { print NR-1; exit; }'
}
array=("D A D" "A D" bBb "D WW" D "\" D \"" e1e " D " E1E D AA "" BB)
element=${array[5]}
index "$element" "${array[@]}"
Output:
输出：
##代码##