在 bash shell 脚本中初始化动态变量（变量变量）
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4385097/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Initiating dynamic variables (variable variables) in bash shell script
提问by Sandeepan Nath
I am using PHP CLI through bash shell. Please check Manipulating an array (printed by php-cli) in shell scriptfor details.
我正在通过 bash shell 使用 PHP CLI。有关详细信息，请检查在 shell 脚本中操作数组（由 php-cli 打印）。
In the following shell code I am able to echo the key- valuepairs that I get from the PHP script.
在下面的 shell 代码中，我可以回显从 PHP 脚本中获得的key-value对。
IFS=":"
# parse php script output by read command
php $PWD'/test.php' | while read -r key val; do
echo $key":"$val
done
Following is the output for this -
以下是此输出 -
BASE_PATH:/path/to/project/root
db_host:localhost
db_name:database
db_user:root
db_pass:root
Now I just want to initiate dynamic variables inside the while loop so that I can use them like $BASE_PATHhaving value '/path/to/project/root', $db_hosthaving 'localhost'
现在我只想在 while 循环中启动动态变量，这样我就可以像$BASE_PATH拥有 value一样使用它们 '/path/to/project/root'，$db_host拥有'localhost'
I come from a PHP background. I would like something like $$key = $valof PHP
我来自 PHP 背景。我想要类似$$key = $valPHP 的东西
回答by Paused until further notice.
Using evalintroduces security risks that must be considered. It's safer to use declare:
使用会eval引入必须考虑的安全风险。使用更安全declare：
# parse php script output by read command
while IFS=: read -r key val; do
echo $key":"$val
declare $key=$val
done < <(php $PWD'/test.php')
If you are using Bash 4, you can use associative arrays:
如果您使用的是 Bash 4，则可以使用关联数组：
declare -A some_array
# parse php script output by read command
while IFS=: read -r key val; do
echo $key":"$val
some_array[$key]=$val
done < <(php $PWD'/test.php')
Using process substition <()and redirecting it into the doneof the whileloop prevents the creation of a subshell. Setting IFS for only the readcommand eliminates the need to save and restore its value.
使用过程substition<()并重定向到done了中while环防止子shell的创建。仅为read命令设置 IFS消除了保存和恢复其值的需要。
回答by Martin Kosek
You may try using the evalconstruct in BASH:
您可以尝试eval在BASH 中使用构造：
key="BASE_PATH"
value="/path/to/project/root"
# Assign $value to variable named "BASE_PATH"
eval ${key}="${value}"
# Now you have the variable named BASE_PATH you want
# This will get you output "/path/to/project/root"
echo $BASE_PATH
Then, just use it in your loop.
然后，只需在您的循环中使用它。
EDIT: this read loop creates a sub-shellwhich will not allow you to use them outside of the loop. You may restructure the read loop so that the sub-shell is not created:
编辑：此读取循环创建了一个子外壳，不允许您在循环外使用它们。您可以重构读取循环，以便不创建子外壳：
# get the PHP output to a variable
php_output=`php test.php`
# parse the variable in a loop without creating a sub-shell
IFS=":"
while read -r key val; do
eval ${key}="${val}"
done <<< "$php_output"
echo $BASE_PATH
