我想写一个循环15个字符串的脚本(可能是数组?)这可能吗?

类似于:

for databaseName in listOfNames
then
  # Do something
end

当前回答

每个Bash脚本/会话的可能第一行:

say() { for line in "${@}" ; do printf "%s\n" "${line}" ; done ; }

使用例如:

$ aa=( 7 -4 -e ) ; say "${aa[@]}"
7
-4
-e

可以考虑:echo在此处将-e解释为选项

其他回答

循环遍历数组的方式取决于换行符的存在。使用分隔数组元素的换行符,数组可以称为“$array”,否则应称为“${array[@]}”。以下脚本将明确说明:

#!/bin/bash

mkdir temp
mkdir temp/aaa
mkdir temp/bbb
mkdir temp/ccc
array=$(ls temp)
array1=(aaa bbb ccc)
array2=$(echo -e "aaa\nbbb\nccc")

echo '$array'
echo "$array"
echo
for dirname in "$array"; do
    echo "$dirname"
done
echo
for dirname in "${array[@]}"; do
    echo "$dirname"
done
echo
echo '$array1'
echo "$array1"
echo
for dirname in "$array1"; do
    echo "$dirname"
done
echo
for dirname in "${array1[@]}"; do
    echo "$dirname"
done
echo
echo '$array2'
echo "$array2"
echo
for dirname in "$array2"; do
    echo "$dirname"
done
echo
for dirname in "${array2[@]}"; do
    echo "$dirname"
done
rmdir temp/aaa
rmdir temp/bbb
rmdir temp/ccc
rmdir temp

每个Bash脚本/会话的可能第一行:

say() { for line in "${@}" ; do printf "%s\n" "${line}" ; done ; }

使用例如:

$ aa=( 7 -4 -e ) ; say "${aa[@]}"
7
-4
-e

可以考虑:echo在此处将-e解释为选项

脚本或函数的隐式数组:

除了anubhava的正确答案:如果循环的基本语法是:

for var in "${arr[@]}" ;do ...$var... ;done

bash中有一个特殊情况:

当运行脚本或函数时,在命令行传递的参数将被分配给$@数组变量,您可以通过$1、$2、$3等进行访问。

可以通过以下方式填充(用于测试)

set -- arg1 arg2 arg3 ...

这个数组上的循环可以简单地写:

for item ;do
    echo "This is item: $item."
  done

请注意,中的保留工作不存在,也没有数组名称!

示例:

set -- arg1 arg2 arg3 ...
for item ;do
    echo "This is item: $item."
  done
This is item: arg1.
This is item: arg2.
This is item: arg3.
This is item: ....

注意,这与

for item in "$@";do
    echo "This is item: $item."
  done

然后进入脚本:

#!/bin/bash

for item ;do
    printf "Doing something with '%s'.\n" "$item"
  done

将其保存在脚本myscript.sh、chmod+x myscript.shh中,然后

./myscript.sh arg1 arg2 arg3 ...
Doing something with 'arg1'.
Doing something with 'arg2'.
Doing something with 'arg3'.
Doing something with '...'.

在函数中相同:

myfunc() { for item;do cat <<<"Working about '$item'."; done ; }

Then

myfunc item1 tiem2 time3
Working about 'item1'.
Working about 'tiem2'.
Working about 'time3'.

试试这个。它正在运行和测试。

for k in "${array[@]}"
do
    echo $k
done

# For accessing with the echo command: echo ${array[0]}, ${array[1]}

这些答案中没有一个包含计数器。。。

#!/bin/bash
## declare an array variable
declare -a array=("one" "two" "three")

# get length of an array
arraylength=${#array[@]}

# use for loop to read all values and indexes
for (( i=0; i<${arraylength}; i++ ));
do
  echo "index: $i, value: ${array[$i]}"
done

输出:

index: 0, value: one
index: 1, value: two
index: 2, value: three