在bash中创建对象数组

Question

是否可以在bash中创建一个对象数组？

这就是我的尝试:

declare -a identifications=(
  {
    email    = '...',
    password = '...'
  }
)

declare -a years=(
  '2011'
  '2012'
  '2013'
  '2014'
  '2015'
  '2016'
)

for identification in "${identifications[@]}"
do
  for year in "${years[@]}"
  do
    my_program --type=CNPJ --format=XLS --identification=${identification.email} --password=${identication.password} --competence=${year} --output="$identification - $year"
  done
done

显然,这不起作用,我找不到如何实现,因为我没有找到bash对象.

Answer 1

您可以使用关联数组(在Bash 4.0中引入)和namerefs(参见手册declare和Shell参数的第一段- 在Bash 4.3中引入)来做一些技巧:

#!/bin/bash

declare -A identification0=(
    [email]='test@abc.com'
    [password]='admin123'
)
declare -A identification1=(
    [email]='test@xyz.org'
    [password]='passwd1!'
)

declare -n identification
for identification in ${!identification@}; do
    echo "Email: ${identification[email]}"
    echo "Password: ${identification[password]}"
done

这打印

Email: test@abc.com
Password: admin123
Email: test@xyz.org
Password: passwd1!

declare -A 声明一个关联数组.

诀窍是将所有"对象"(关联数组)变量名称分配给相同的前缀,例如identification.该${!prefix@}符号扩展到开头的所有变量名prefix:

$ var1=
$ var2=
$ var3=
$ echo ${!var@}
var1 var2 var3

然后,要访问关联数组的键值对,我们使用nameref属性声明for循环的控制变量:

declare -n identification

这样循环

for identification in ${!identification@}; do

使identification行为好像它是扩展的实际变量${!identification@}.

尽管如此,在以下情况下做一些事情会更容易:

emails=('test@abc.com' 'test@xyz.org')
passwords=('admin123' 'passwd1!')
for (( i = 0; i < ${#emails[@]}; ++i )); do
    echo "Email: ${emails[i]}"
    echo "Password: ${passwords[i]}"
done

即,只需遍历包含您的信息的两个数组.

Answer 2

我倾向于使用 json 来创建对象。对我来说，这使它变得非常简单和灵活。

这是一个过于简单的例子。

我创建一个 json 文件：devices.json

{
          "backup" : [{
                "addr":"192.168.1.1",
                "username":"backuper",
                "dstfile":"firewallconfig",
                "ext":".cfg",
                "method":"ssh",
                "rotate":"no",
                "enabled":"yes"
            }, {
                "addr":"192.168.1.2",
                "username":"backuper",
                "dstfile":"routerconfig",
                "ext":".cfg",
                "method":"ssh",
                "rotate":"no",
                "enabled":"yes"
            }]
 }

Bash 脚本：task.sh

 # read the devices.json file and store it in the variable jsonlist
    jsonlist=$(jq -r '.backup' "devices.json")
        # inside the loop, you cant use the fuction _jq() to get values from each object.
        for row in $(echo "${jsonlist}" | jq -r '.[] | @base64'); do
            _jq()
            {
             echo ${row} | base64 --decode | jq -r ${1}
            }
         
        echo "backing up: $(_jq '.addr')"
        echo "using method: $(_jq '.method')"
        done

发布原始帖子的人可以通过谷歌搜索“using json with bash”或其他东西找到。