Shell脚本变量引用问题 [英] Shell Script Variable Quoting Problem
本文介绍了Shell脚本变量引用问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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();'"
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