bash sed 查找并用空格替换字符串
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sed find and replace a string with spaces
提问by Alex
Having the following in a file:
在文件中包含以下内容：
public $password = 'XYZ';
I'm trying to replace the password's value with a different one, through an automated deployment process from backup files.
我正在尝试通过备份文件的自动部署过程将密码的值替换为不同的值。
I have the regext that will match the string above in a file, but not much compatible with sed
我有一个与文件中的上述字符串相匹配的正则表达式，但与 sed
(public\s$password\s=\s'(.*)'?)
I also tried
我也试过
sed -i -e "s/public\s$password\s=\s'(.*)'/private\s$password\s=\s'jingle'" configuration.php
Any ideas?
有任何想法吗？
回答by ams
Try this:
尝试这个：
sed -i -e "s/public\s$password\s=\s'\(.*\)'/private $password = 'jingle'/" configuration.php
The problem was that you need to 'escape' the round brackets, and that \sdoesn't work in the output pattern. You also had missed the final /.
问题是您需要“转义”圆括号，这\s在输出模式中不起作用。你也错过了决赛/。
