Shell脚本变量引用问题 [英] Shell Script Variable Quoting Problem

问题描述

我有一个包含行的sh脚本

$PHP_COMMAND -r 'echo get_include_path();'

我无法编辑此脚本，但我需要最终的命令行(相当于)

php -d include_path='/path/with spaces/dir' -r 'echo get_include_path();'

如何实现此目标？

---

下面是演示该问题的脚本。

#!/bin/sh

# shell script quoting problem demonstration

# I need to be able to set a shell variable with a command with 
# some options, like so
PHP_COMMAND="php -d 'include_path=/path/with spaces/dir'"
# then use PHP_COMMAND to run something in another script, like this:
$PHP_COMMAND -r 'echo get_include_path();'
# the above fails when executed. However, if you copy/paste the output
# from this line and run it in the CLI, it works!
echo "$PHP_COMMAND -r 'echo get_include_path();'"
php -d include_path='/path/with spaces/dir' -r 'echo get_include_path();'
# what's going on?

# this is also interesting
echo "
--------------------"

# this works great, but only works if include_path doesn't need quoting
PHP_COMMAND="php -d include_path=/path/to/dir"
echo "$PHP_COMMAND -r 'echo get_include_path();'"
$PHP_COMMAND -r 'echo get_include_path();'

echo "
--------------------"

# this one doesn't when run in the sh script, but again if you copy/paste 
# the output it does work as expected.
PHP_COMMAND="php -d 'include_path=/path/to/dir'"
echo "$PHP_COMMAND -r 'echo get_include_path();'"
$PHP_COMMAND -r 'echo get_include_path();'

脚本也可在线使用：http://gist.github.com/276500

推荐答案

应该有一种更简单的方法，但一种解决方法是用双引号将整个命令行括起来，然后对其求值：

PHP_COMMAND="php -d 'include_path=/path/with spaces/dir'"
eval "$PHP_COMMAND -r 'echo get_include_path();'"

这篇关于Shell脚本变量引用问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！