如何将命令行参数转换为bash脚本数组?
我想说:
./something.sh arg1 arg2 arg3
并将其转换为
myArray=( arg1 arg2 arg3 )
这样我就可以使用myArray在脚本中进一步使用。
这之前的SO帖子接近，但没有进入如何创建一个数组:我如何在Bash中解析命令行参数?
我需要将参数转换为常规bash脚本数组;我意识到我可以使用其他语言(例如Python)，但需要在bash中这样做。我想我在寻找一个“追加”功能或类似的东西?
更新:我还想问如何检查零参数和分配默认数组值，感谢下面的答案，能够得到这个工作:
if [ "$#" -eq 0 ]; then
myArray=( defaultarg1 defaultarg2 )
else
myArray=( "$@" )
fi
当前回答
双引号的重要性值得强调。假设一个参数包含空格。
代码:
#!/bin/bash
printf 'arguments:%s\n' "$@"
declare -a arrayGOOD=( "$@" )
declare -a arrayBAAD=( $@ )
printf '\n%s:\n' arrayGOOD
declare -p arrayGOOD
arrayGOODlength=${#arrayGOOD[@]}
for (( i=1; i<${arrayGOODlength}+1; i++ ));
do
echo "${arrayGOOD[$i-1]}"
done
printf '\n%s:\n' arrayBAAD
declare -p arrayBAAD
arrayBAADlength=${#arrayBAAD[@]}
for (( i=1; i<${arrayBAADlength}+1; i++ ));
do
echo "${arrayBAAD[$i-1]}"
done
输出:
> ./bash-array-practice.sh 'The dog ate the "flea" -- and ' the mouse.
arguments:The dog ate the "flea" -- and
arguments:the
arguments:mouse.
arrayGOOD:
declare -a arrayGOOD='([0]="The dog ate the \"flea\" -- and " [1]="the" [2]="mouse.")'
The dog ate the "flea" -- and
the
mouse.
arrayBAAD:
declare -a arrayBAAD='([0]="The" [1]="dog" [2]="ate" [3]="the" [4]="\"flea\"" [5]="--" [6]="and" [7]="the" [8]="mouse.")'
The
dog
ate
the
"flea"
--
and
the
mouse.
>
其他回答
下面是另一个用法:
#!/bin/bash
array=( "$@" )
arraylength=${#array[@]}
for (( i=0; i<${arraylength}; i++ ));
do
echo "${array[$i]}"
done
并排查看数组和$@实际上是相同的。
代码:
#!/bin/bash
echo "Dollar-1 : $1"
echo "Dollar-2 : $2"
echo "Dollar-3 : $3"
echo "Dollar-AT: $@"
echo ""
myArray=( "$@" )
echo "A Val 0: ${myArray[0]}"
echo "A Val 1: ${myArray[1]}"
echo "A Val 2: ${myArray[2]}"
echo "A All Values: ${myArray[@]}"
输入:
./bash-array-practice.sh 1 2 3 4
输出:
Dollar-1 : 1
Dollar-2 : 2
Dollar-3 : 3
Dollar-AT: 1 2 3 4
A Val 0: 1
A Val 1: 2
A Val 2: 3
A All Values: 1 2 3 4
也许这可以有所帮助:
myArray=("$@")
你也可以通过省略'in'来迭代参数:
for arg; do
echo "$arg"
done
将是等价的
for arg in "${@}"; do
echo "$arg"
done
双引号的重要性值得强调。假设一个参数包含空格。
代码:
#!/bin/bash
printf 'arguments:%s\n' "$@"
declare -a arrayGOOD=( "$@" )
declare -a arrayBAAD=( $@ )
printf '\n%s:\n' arrayGOOD
declare -p arrayGOOD
arrayGOODlength=${#arrayGOOD[@]}
for (( i=1; i<${arrayGOODlength}+1; i++ ));
do
echo "${arrayGOOD[$i-1]}"
done
printf '\n%s:\n' arrayBAAD
declare -p arrayBAAD
arrayBAADlength=${#arrayBAAD[@]}
for (( i=1; i<${arrayBAADlength}+1; i++ ));
do
echo "${arrayBAAD[$i-1]}"
done
输出:
> ./bash-array-practice.sh 'The dog ate the "flea" -- and ' the mouse.
arguments:The dog ate the "flea" -- and
arguments:the
arguments:mouse.
arrayGOOD:
declare -a arrayGOOD='([0]="The dog ate the \"flea\" -- and " [1]="the" [2]="mouse.")'
The dog ate the "flea" -- and
the
mouse.
arrayBAAD:
declare -a arrayBAAD='([0]="The" [1]="dog" [2]="ate" [3]="the" [4]="\"flea\"" [5]="--" [6]="and" [7]="the" [8]="mouse.")'
The
dog
ate
the
"flea"
--
and
the
mouse.
>
实际上，参数列表可以用$1 $2…等。 这完全等价于:
${!i}
因此，参数列表可以通过set和${!I}是访问它们的正确方式:
$ set -- aa bb cc dd 55 ff gg hh ii jjj kkk lll
$ for ((i=0;i<=$#;i++)); do echo "$#" "$i" "${!i}"; done
12 1 aa
12 2 bb
12 3 cc
12 4 dd
12 5 55
12 6 ff
12 7 gg
12 8 hh
12 9 ii
12 10 jjj
12 11 kkk
12 12 lll
对于您的特定情况，可以使用这个(不需要数组)，在没有给出参数时设置参数列表:
if [ "$#" -eq 0 ]; then
set -- defaultarg1 defaultarg2
fi
这可以转化为更简单的表达式:
[ "$#" == "0" ] && set -- defaultarg1 defaultarg2
