bash 使用 sed 修改 key="Value" 配置
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Modify key="Value" config with sed
提问by SebastianK
I am trying to write a script that configures a config file used by a nother script. I am trying to use sed like this
我正在尝试编写一个脚本来配置另一个脚本使用的配置文件。我正在尝试像这样使用 sed
sed -c -i "s/\($TARGET_KEY *= *\).*/$REPLACEMENT_VALUE/" $CONFIG_FILE
But it's not working as it is intended to it strips the quotation marks and i cant figure out how to write it so it dont.
但它不起作用，因为它旨在去除引号，我无法弄清楚如何编写它所以它不会。
the second problem is that when i run this on Mac OS the out put is an error:
第二个问题是，当我在 Mac OS 上运行它时，输出是一个错误：
sed: illegal option -- c
usage: sed script [-Ealn] [-i extension] [file ...]
sed [-Ealn] [-i extension] [-e script] ... [-f script_file] ... [file ...]
I am new to usage of sed so please forgive my lack of skills in this area.
我是 sed 的新手，所以请原谅我在这方面缺乏技能。
采纳答案by Kent
see the test below, I didn't add "-i", just print the output. you can add -i if you need:
看下面的测试，我没有加“-i”，只是打印输出。如果需要，您可以添加 -i：
kent$ cat c.conf
key1="value1"
foo = "fooValue"
bar="barValue"
kent$ echo $k1
foo
kent$ echo $v1
foo_new
kent$ sed -r "s/($k1 *= *\").*/$v1\"/" c.conf
key1="value1"
foo = "foo_new"
bar="barValue"
回答by OpenSauce
Have you tried escaping the quotes? This works for me (on Cygwin):
你有没有试过转义引号？这对我有用（在 Cygwin 上）：
~$ echo -e "key1=\"value1\"\nkey2=\"value2\""
key1="value1"
key2="value2"
~$ TARGET_KEY=key2
~$ REPLACEMENT_VALUE=new_val
~$ echo -e "key1=\"value1\"\nkey2=\"value2\"" | sed "s/\($TARGET_KEY *= *\"\).*/$REPLACEMENT_VALUE\"/"
key1="value1"
key2="new_val"
~$
