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

您可以使用${arrayName[@]}的语法

#!/bin/bash
# declare an array called files, that contains 3 values
files=( "/etc/passwd" "/etc/group" "/etc/hosts" )
for i in "${files[@]}"
do
    echo "$i"
done

其他回答

listOfNames="db_one db_two db_three"
for databaseName in $listOfNames
do
  echo $databaseName
done

或者只是

for databaseName in db_one db_two db_three
do
  echo $databaseName
done

我在GitHub更新中使用了这种方法,我发现它很简单。

## declare an array variable
arr_variable=("kofi" "kwame" "Ama")

## now loop through the above array
for i in "${arr_variable[@]}"
do
   echo "$i"


done
   

您可以使用带有三个表达式(C样式)的计数器遍历bash数组值,以读取循环语法的所有值和索引:

declare -a kofi=("kofi" "kwame" "Ama")
 
# get the length of the array
length=${#kofi[@]}

for (( j=0; j<${length}; j++ ));
do
  print (f "Current index %d with value %s\n" $j "${kofi[$j]}")
done

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'.

当然,这是可能的。

for databaseName in a b c d e f; do
  # do something like: echo $databaseName
done 

有关详细信息,请参阅、while和until的Bash循环。

单线循环,

 declare -a listOfNames=('db_a' 'db_b' 'db_c')
 for databaseName in ${listOfNames[@]}; do echo $databaseName; done;

你会得到这样的输出,

db_a
db_b
db_c