启动在bash shell脚本动态变量（可变变量） [英] Initiating dynamic variables (variable variables) in bash shell script

问题描述

我是通过bash shell中使用PHP CLI。请查看<一个href=\"http://stackoverflow.com/questions/4377191/manipulating-an-array-printed-by-php-cli-in-shell-script\">Manipulating在shell脚本数组（通过PHP-CLI印刷）了解详情。

I am using PHP CLI through bash shell. Please check Manipulating an array (printed by php-cli) in shell script for details.

在下面的shell code我能够呼应键 - ，我得到的值对从PHP脚本。

In the following shell code I am able to echo the key- value pairs that I get from the PHP script.

IFS=":"

# parse php script output by read command
php $PWD'/test.php' | while read -r key val; do
    echo $key":"$val
done

以下是输出这一点 -

Following is the output for this -

BASE_PATH:/path/to/project/root
db_host:localhost
db_name:database
db_user:root
db_pass:root

现在我只想启动动态变量while循环中，这样我可以像 $ BASE_PATH 使用它们有值/路径/到/项目/根'， $ DB_HOST 有'localhost'的

Now I just want to initiate dynamic variables inside the while loop so that I can use them like $BASE_PATH having value '/path/to/project/root', $db_host having 'localhost'

我来自一个PHP的背景。我想类似 $$键= $ VAL 的PHP

I come from a PHP background. I would like something like $$key = $val of PHP

推荐答案

您可以尝试使用评估结构中的 BASH ：

You may try using the eval construct in BASH:

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

然后，就用它在你的循环。

Then, just use it in your loop.

编辑：这读取循环的创建一个子shell 的，不会让你使用它们的循环之外。您可能会重构读取循环，使子shell不创建：

this read loop creates a sub-shell which 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:

# 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

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